Ch 6 Trigonometry

6

Trigonometry TERMINOLOGY Angle of depression: The angle between the horizontal and the line of sight when looking down to an object below Angle of elevation: The angle between the horizontal and the line of sight when looking up to an object above Angles of any magnitude: Angles can be measured around a circle at the centre to find the trigonometric ratios of angles of any size from 0c to 360c and beyond Bearing: The direction relative to north. Bearings may be written as true bearings (clockwise from North) or as compass bearings (using N, S, E and W)

Complementary angles: Two or more angles that add up to 90c Cosecant: The reciprocal ratio of sine (sin). It is the hypotenuse over the opposite side in a right triangle Cotangent: The reciprocal ratio of tangent (tan). It is the adjacent over the opposite side in a right triangle Secant: The reciprocal ratio of cosine (cos). It is the hypotenuse over the adjacent side in a right triangle Trigonometric identities: A statement that is true for all trigonometric values in the domain. Relationships between trigonometric ratios

Chapter 6 Trigonometry

291

INTRODUCTION TRIGONOMETRY IS USED IN many fields, such as building, surveying and navigating. Wave theory also uses trigonometry. This chapter revises basic right-angled triangle problems and applies them to real-life situations. Some properties of trigonometric ratios, angles greater than 90c and trigonometric equations are introduced. You will also study trigonometry in non-right-angled triangles.

DID YOU KNOW? Ptolemy (Claudius Ptolemaeus), in the second century, wrote He¯ mathe¯matike¯ syntaxis (or Almagest as it is now known) on astronomy. This is considered to be the first treatise on trigonometry, but was based on circles and spheres rather than on triangles. The notation ‘chord of an angle’ was used rather than sin, cos or tan. Ptolemy constructed a table of sines from 0c to 90c in steps of a quarter of a degree. He also calculated a value of r to 5 decimal places, and established the relationship for sin (X ! Y ) and cos (X ! Y ) .

Trigonometric Ratios In similar triangles, pairs of corresponding angles are equal and sides are in proportion. For example:

You studied similar triangles in Geometry in Chapter 4.

In any triangle containing an angle of 30c, the ratio of AB:AC = 1:2. Similarly, the ratios of other corresponding sides will be equal. These ratios of sides form the basis of the trigonometric ratios. In order to refer to these ratios, we name the sides in relation to the angle being studied:

• the hypotenuse is the longest side, and is always opposite the right angle • the opposite side is opposite the angle marked in the triangle • the adjacent side is next to the angle marked

ch6.indd 291

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Maths In Focus Mathematics Extension 1 Preliminary Course

The opposite and adjacent sides vary according to where the angle is marked. For example:

The trigonometric ratios are

You can learn these by their initials SOH, CAH, TOA.

What about Some Old Hags Can’t Always Hide Their Old Age?

Sine

sin i =

Cosine

cos i =

Tangent tan i =

opposite hypotenuse adjacent hypotenuse opposite adjacent

As well as these ratios, there are three inverse ratios,

Cosecant cosec i =

1 sin i

sec i =

1 cos i

Cotangent cot i =

1 tan i

Secant

hypotenuse p opposite hypotenuse f= p adjacent adjacent f= p opposite f=

DID YOU KNOW? Trigonometry, or triangle measurement, progressed from the study of geometry in ancient Greece. Trigonometry was seen as applied mathematics. It gave a tool for the measurement of planets and their motion. It was also used extensively in navigation, surveying and mapping, and it is still used in these fields today. Trigonometry was crucial in the setting up of an accurate calendar, since this involved measuring the distances between the Earth, sun and moon.


293

EXAMPLES 1. Find sin a, tan a and sec a.

Solution AB = hypotenuse = 5 BC = opposite side = 3 AC = adjacent side = 4 opposite sin a = hypotenuse 3 = 5 opposite tan a = adjacent 3 = 4 1 sec a = cos a hypotenuse = adjacent 5 = 4 2. If sin i =

2 , find the exact ratios of cos i, tan i and cot i. 7

Solution To find the other ratios you need to find the adjacent side.

By Pythagoras’ theorem: c2 = a2 + b2 72 = a2 + 22 49 = a 2 + 4 45 = a 2 à=

45 CONTINUED

294


cos i = = tan i =

adjacent hypotenuse 45 7 opposite

adjacent 2 = 45 1 cot i = tan i 45 = 2

Complementary angles

In D ABC, if+B = i, then +A = 90c - i

(angle sum of a Δ) a (90c - i) = c b cos (90c - i) = c a tan (90c - i) = b c sec (90c - i) = b c cosec (90c - i) = a b cot (90c - i) = a

b i= c a cos i = c b tan i = a c sec i = a c cosec i = b a cot i = b

sin

sin

From these ratios come the results.

sin i = cos (90° - i) cos i = sin (90° - i) sec i = cosec (90° - i) cosec i = sec (90° - i) tan i = cot (90° - i) cot i = tan (90° - i)


295

EXAMPLES 1. Simplify tan 50c - cot 40c.

Solution tan 50c = cot ] 90c - 50c g = cot 40c ` tan 50c - cot 40c = tan 50c - tan 50c =0

Check this answer on your calculator.

2. Find the value of m if sec 55c = cosec ] 2m - 15 g c.

Solution sec 55c = cosec ] 90c - 55c g = cosec 35c ` 2m - 15 = 35 2m = 50 m = 25

Check this by substituting m into the equation.

6.1 Exercises 1.

Write down the ratios of cos i, sin i and tan i.

2.

Find sin b, cot b and sec b.

3.

Find the exact ratios of sin b, tan b and cos b.

4.

Find exact values for cos x, tan x and cosec x.

296


5. 6.

7.

Hint: Change 0.7 to a fraction.

8.

9.

4 , find cos i and sin i. 3 2 If cos i = , find exact values for 3 tan i, sec i and sin i. If tan i =

1 If sin i = , find the exact ratios 6 of cos i and tan i. If cos i = 0.7, find exact values for tan i and sin i. D ABC is a right-angled isosceles triangle with +ABC = 90c and AB = BC = 1. (a) Find the exact length of AC. (b) Find +BAC. (c) From the triangle, write down the exact ratios of sin 45c, cos 45c and tan 45c.

10.

(c) Write down the exact ratios of sin 60c, cos 60c and tan 60c. 11. Show sin 67c = cos 23c. 12. Show sec 82c = cosec 8c. 13. Show tan 48c = cot 42c. 14. Simplify (a) cos 61c + sin 29c (b) sec i - cosec ] 90c - i g (c) tan 70c + cot 20c - 2 tan 70c (d)

sin 55c cos 35c

(e)

cot 25c + tan 65c cot 25c

15. Find the value of x if sin 80c = cos ] 90 - x g c. 16. Find the value of y if tan 22c = cot ^ 90 - y h c. 17. Find the value of p if cos 49c = sin ^ p + 10 h c. 18. Find the value of b if sin 35c = cos ] b + 30 g c. 19. Find the value of t if cot ] 2t + 5 g c = tan ] 3t - 15 g c.

(a) Using Pythagoras’ theorem, find the exact length of AC. (b) Write down the exact ratios of sin 30c, cos 30c and tan 30c.

20. Find the value of k if tan ] 15 - k g c = cot ] 2k + 60 g c.

Trigonometric ratios and the calculator Angles are usually given in degrees and minutes. In this section you will practise rounding off angles and finding trigonometric ratios on the calculator. Angles are usually given in degrees and minutes in this course. The calculator uses degrees, minutes and seconds, so you need to round off. 60 minutes = 1 deg ree (60l = 1c) 60 sec onds = 1 min ute (60m = 1l) In normal rounding off, you round up to the next number if the number to the right is 5 or more. Angles are rounded off to the nearest degree by rounding up if there are 30 minutes or more. Similarly, angles are rounded off to the nearest minute by rounding up if there are 30 seconds or more.


297

EXAMPLES Round off to the nearest minute. 1. 23c 12l 22m

Solution 23c 12l 22m = 23c 12l 2. 59c 34l 41m

Solution 59c 34l 41m = 59c 35l 3. 16c 54l 30m Because 30 seconds is half a minute, we round up to the next minute.

Solution 16c 54l 30m = 16c 55l

% , ,,

KEY Some calculators have

This key changes decimal angles into degrees, minutes and seconds

deg or dms keys.

and vice versa.

EXAMPLES 1. Change 58c 19l into a decimal.

Solution Press 58 % , ,, 19 % , ,, = % , ,, So 58c 19l = 58.31666667 2. Change 45.236c into degrees and minutes.

Solution Press 45.236 = SHIFT % , ,, So 45.236c = 45c14l

If your calculator does not give these answers, check the instructions for its use.

298


In order to use trigonometry in right-angled triangle problems, you need to find the ratios of angles on your calculator.

EXAMPLES 1. Find cos 58c 19l, correct to 3 decimal places.

Solution

If your calculator doesn't give this answer, check that it is in degree mode.

Press COS 58 % , ,, 19 % , ,, = So cos 58c19l = 0.525 2. Find sin 38c14l, correct to 3 decimal places.

Solution Press SIN 38 % , ,, 14 % , ,, = So sin 38c 14l = 0.619 3. If tani = 0.348, find i in degrees and minutes.

Solution This is the reverse of finding trigonometric ratios. To find the angle, given the ratio, use the inverse key ^ tan - 1 h . Press SHIFT TAN - 1 0.348 = SHIFT % , ,, tan i = 0.348 i = tan - 1 (0.348) = 19c11l 4. Find i in degrees and minutes if cos i = 0.675.

Solution Press SHIFT COS - 1 0.675 = SHIFT % , ,, cos i = 0.675 i = cos - 1 (0.675) = 47c 33l

6.2 Exercises 1.

Round off to the nearest degree. (a) 47° 13l 12m (b) 81° 45l 43m (c) 19° 25l 34m (d) 76° 37l 19m (e) 52° 29l 54m

2.

Round off to the nearest minute. (a) 47° 13l 12m (b) 81° 45l 43m (c) 19° 25l 34m (d) 76° 37l 19m (e) 52° 29l 54m


3.

Change to a decimal. (a) 77c45l (b) 65c30l (c) 24c51l (d) 68c21l (e) 82c31l

5.

Find correct to 3 decimal places. (a) sin 39c25l (b) cos 45c 51l (c) tan18c43l (d) sin 68c06l (e) tan 54c20l

4.

Change into degrees and minutes. (a) 59.53c (b) 72.231c (c) 85.887c (d) 46.9c (e) 73.213c

6.

Find i in degrees and minutes if (a) sin i = 0.298 (b) tan i = 0.683 (c) cos i = 0.827 (d) tan i = 1.056 (e) cos i = 0.188

Right-angled Triangle Problems Trigonometry is used to find an unknown side or angle of a triangle.

Finding a side We can use trigonometry to find a side of a right-angled triangle.

EXAMPLES 1. Find the value of x, correct to 1 decimal place.

Solution cos i =

adjacent

hypotenuse x cos 23° 49l = 11.8 x 11.8 # cos 23° 49l = 11.8 # 11.8 11.8 cos 23° 49l = x `

10.8 cm = x ^ to 1 decimal point h CONTINUED

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300


2. Find the value of y, correct to 3 significant figures.

Solution sin i =

opposite

hypotenuse 9.7 sin 41c 15l = y 9.7 y # sin 41c 15l = y # y y sin 41c 15l = 9.7 y sin 41c 15l 9.7 = sin 41c 15l sin 41c 15l 9.7 y= sin 41c 15l = 14.7 m ^ to 3 significant figures h

6.3 Exercises 1.

Find the values of all pronumerals, correct to 1 decimal place. (a)

(c)

(b) (d)


(e) (l)

4.7 cm

x

(f)

37c22l 72c18l

(m) x

6.3 cm

(g) (n) 63c14l

23 mm

x

(o)

39c47l

(h) 3.7 m

(i) (p) k

(j)

46c5l

14.3 cm

(q)

5.4 cm

(k)

31c12l

y

h

x

4.8 m

74c29l

301

302


68c41l

(r)

d

0.45 m

6.2 cm 73c

4.

x

(s)

5.75 cm

19c17l

17.3 m

(t)

Hamish is standing at an angle of 67c from a goalpost and 12.8 m away as shown. How far does he need to kick a football for it to reach the goal?

x

12.8 m

67c b

5.

6c3l

Square ABCD with side 6 cm has line CD produced to E as shown so that +EAD = 64c 12l. Evaluate the length, correct to 1 decimal place, of (a) CE (b) AE E

2.

A roof is pitched at 60c. A room built inside the roof space is to have a 2.7 m high ceiling. How far in from the side of the roof will the wall for the room go? 64c12l

D

A

2.7 m 60c

x

3.

A diagonal in a rectangle with breadth 6.2 cm makes an angle of 73c with the vertex as shown. Find the length of the rectangle correct to 1 decimal place.

C

6.

6 cm

B

A right-angled triangle with hypotenuse 14.5 cm long has one interior angle of 43c 36l. Find the lengths of the other two sides of the triangle.


7.

8.

9.

A right-angled triangle ABC with the right angle at A has +B = 56c44l and AB = 26 mm. Find the length of the hypotenuse. A triangular fence is made for a garden inside a park. Three holes A, B and C for fence posts are made at the corners so that A and B are 10.2 m apart, AB and CB are perpendicular, and angle CAB is 59c 54l. How far apart are A and C? Triangle ABC has +BAC = 46c and +ABC = 54c. An altitude is drawn from C to meet AB at point D. If the altitude is 5.3 cm long, find, correct to 1 decimal place, the length of sides (a) AC (b) BC (c) AB

(a) Find the length of the side of the rhombus. (b) Find the length of the other diagonal. 11. Kite ABCD has diagonal BD = 15.8 cm as shown. If +ABD =57c29l and +DBC = 72c51l, find the length of the other diagonal AC. A

B

57c29l

72c51l

D

15.8 cm

C

10. A rhombus has one diagonal 12 cm long and the diagonal makes an angle of 28c 23l with the side of the rhombus.

Finding an angle Trigonometry can also be used to find one of the angles in a right-angled triangle.

EXAMPLES 1. Find the value of i, in degrees and minutes.

CONTINUED

303

304


Solution cos i =

adjacent

hypotenuse 5.8 = 7.3 5.8 ` i = cos - 1 c m 7.3 = 37c 23l

2. Find the value of a, in degrees and minutes.

Solution tan a =

opposite

adjacent 4 = .9 2 .1 4 .9 ` a = tan - 1 c m 2 .1 = 66c 48l

6.4 Exercises 1.

Find the value of each pronumeral, in degrees and minutes. (a)

(b)


(c)

(i)

(j) (d)

(e)

3.8 cm

(k)

2.4 cm

a

(l)

i

(f) 8.3 cm

5.7 cm

(m)

i 6.9 mm

(g)

11.3 mm

(n)

(h)

i

3m

7m

305

306


(o)

b 20 m

5.1 cm

i

11.6 cm

3.

(p) 15 m

A field is 13.7 m wide and Andre is on one side. There is a gate on the opposite side and 5.6 m along from where Andre is. At what angle will he walk to get to the gate?

a

Andre

13 m

(q)

4.4 cm

i

12.3 m

i

7.6 cm 13.7 m

(r)

a

5.6 m

14.3 cm

8.4 cm

(s)

4.

i

Gate

A 60 m long bridge has an opening in the middle and both sides open up to let boats pass underneath. The two parts of the bridge floor rise up to a height of 18 m. Through what angle do they move?

3m

18 m

5m

(t)

c

i 18.9 cm

10.3 cm

2.

A kite is flying at an angle of i above the ground as shown. If the kite is 12.3 m above the ground and has 20 m of string, find angle i.

5.

60 m

An equilateral triangle ABC with side 7 cm has an altitude AD that is 4.5 cm long. Evaluate the angle the altitude makes with vertex A ]+DAB g.


6.

7.

Rectangle ABCD has dimensions 18 cm # 7 cm. A line AE is drawn so that E bisects DC. (a) How long is line AE? (Answer to 1 decimal place). (b) Evaluate +DEA. A 52 m tall tower has wire stays on either side to minimise wind movement. One stay is 61.3 m long and the other is 74.5 m long as shown. Find the angles that the tower makes with each stay. a b 61.3 m

A

B

5 cm

D 1 cm

C

E

(a) Find +BEC. (b) Find the length of the rectangle. 10. A diagonal of a rhombus with side 9 cm makes an angle of 16c with the side as shown. Find the lengths of the diagonals.

74.5 m

16c

52 m

9 cm

8.

(a) The angle from the ground up to the top of a pole is 41c when standing 15 m on one side of it. Find the height h of the pole, to the nearest metre. (b) If Seb stands 6 m away on the other side, find angle i.

h

11. (a) Kate is standing at the side of a road at point A, 15.9 m away from an intersection. She is at an angle of 39c from point B on the other side of the road. What is the width w of the road? (b) Kate walks 7.4 m to point C. At what angle is she from point B? B

i 6m

9.

15 m

41c

Rectangle ABCD has a line BE drawn so that +AEB = 90c and DE = 1 cm. The width of the rectangle is 5 cm.

w A

39c 7.4 m

C

i 15.9 m

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308


Applications DID YOU KNOW? The Leaning Tower of Pisa was built as a belfry for the cathedral nearby. Work started on the tower in 1174, but when it was only half completed the soil underneath one side of it subsided. This made the tower lean to one side. Work stopped, and it wasn’t until 100 years later that architects found a way of completing the tower. The third and fifth storeys were built close to the vertical to compensate for the lean. Later a vertical top storey was added.

The tower is about 55 m tall and 16 m in diameter. It is tilted about 5 m from the vertical, and tilts by an extra 0.6 cm each year.

Class Investigation Discuss some of the problems with the Leaning Tower of Pisa. • Find the angle at which it is tilted from the vertical. • Work out how far it will be tilted in 10 years. • Use research to find out if the tower will fall over, and if so, when.

Angle of elevation The angle of elevation is used to measure the height of tall objects that cannot be measured directly, for example a tree, cliff, tower or building.


Class Exercise Stand outside the school building and look up to the top of the building. Think about which angle your eyes pass through to look up to the top of the building.

The angle of elevation, i, is the angle measured when looking from the ground up to the top of the object. We assume that the ground is horizontal.

EXAMPLE The angle of elevation of a tree from a point 50 m out from its base is 38c 14l. Find the height of the tree, to the nearest metre.

Solution We assume that the tree is vertical!

A clinometer is used to measure the angle of elevation or depression.

tan 38c 14l =

h 50

50 # tan 38c 14l = 50 #

h 50

50 tan 38c 14l = h 39 Z h So the tree is 39 m tall, to the nearest metre.

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310


Angle of depression The angle of depression is the angle formed when looking down from a high place to an object below.

Class Exercise If your classroom is high enough, stand at the window and look down to something below the window. If the classroom is not high enough, find a hill or other high place. Through which angle do your eyes pass as you look down?

The angle of depression, i, is the angle measured when looking down from the horizontal to an object below.

EXAMPLES 1. The angle of depression from the top of a 20 m building to a boy below is 61c 39l. How far is the boy from the building, to 1 decimal place?

Solution


+DAC = +ACB = 61c 39l 20 tan 61c 39l = x 20 x # tan 61c 39l = x # x x tan 61c 39l = 20 x tan 61c 39l 20 = tan 61c 39l tan 61c 39l 20 x= tan 61c 39l Z 10.8

(alternate angles, AD < BC)

So the boy is 10.8 m from the building. 2. A bird sitting on top of an 8 m tall tree looks down at a possum 3.5 m out from the base of the tree. Find the angle of elevation to the nearest minute.

Solution B

A

i

8m

C

3.5 m

The angle of depression is i Since AB < DC +BDC = i tan i =

8 3.5

8 m 3 .5 = 66c 22l

` i = tan - 1 c

] horizontal lines g ^ alternate angles h

D

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312


Bearings Bearings can be described in different ways: For example, N70c W:

Start at north and measure 70o around towards the west.

True bearings measure angles clockwise from north

EXAMPLES We could write 315o T for true bearings.

1. Sketch the diagram when M is on a bearing of 315c from P.

Solution

Measure clockwise, starting at north.

2. X is on a bearing of 030c from Y. Sketch this diagram.

Solution

All bearings have 3 digits so 30° becomes 030° for a bearing.

3. A house is on a bearing of 305c from a school. What is the bearing of the school from the house?


Solution The diagram below shows the bearing of the house from the school. North House

School 305c

To find the bearing of the school from the house, draw in North from the house and use geometry to find the bearing as follows: N2

N1 H

S 305c

The bearing of the school from the house is +N 2 HS. +N 1 SH = 360c - 305c = 55c

^ angle of revolution h

+N 2 HS = 180c - 55c = 125c

(cointerior angles, N 2 H < N 1 S)

So the bearing of the school from the house is 125c.

4. A plane leaves Sydney and flies 100 km due east, then 125 km due north. Find the bearing of the plane from Sydney, to the nearest degree.

CONTINUED

313

314


Solution

125 100 = 1.25

tan x =

x = tan - 1 (1.25) (to the nearest degree) = 51c i = 90c - xc = 90c - 51c = 39c So the bearing of the plane from Sydney is 039°. 5. A ship sails on a bearing of 140° from Sydney for 250 km. How far east of Sydney is the ship now, to the nearest km?

Solution

A navigator on a ship uses a sextant to measure angles.

Could you use a different triangle for this question?

i = 140c - 90c = 50c x cos 50c = 250 x 250 # cos 50c = 250 # 250 250 cos 50c = x 161 Z x So the ship is 161 km east of Sydney, to the nearest kilometre.


6.5 Exercises 1.

2.

Draw a diagram to show the bearing in each question. (a) A boat is on a bearing of 100c from a beach house. (b) Jamie is on a bearing of 320c from a campsite. (c) A seagull is on a bearing of 200c from a jetty. (d) Alistair is on a bearing of 050c from the bus stop. (e) A plane is on a bearing of 285c from Broken Hill. (f) A farmhouse is on a bearing of 012c from a dam. (g) Mohammed is on a bearing of 160c from his house. (h) A mine shaft is on a bearing of 080c from a town. (i) Yvonne is on a bearing of 349c from her school. (j) A boat ramp is on a bearing of 280c from an island.

North

(b)

Y

West

East 35c X

South

(c)

North

X 10c

Y

West

East

South

(d) X

Find the bearing of X from Y in each question in 3 figure (true) bearings. North (a)

North

23c West

Y

East

South

(e) 112c

North

Y

West X

Y

X South

East

315

316


3.

Jack is on a bearing of 260c from Jill. What is Jill’s bearing from Jack?

4.

A tower is on a bearing of 030c from a house. What is the bearing of the house from the tower?

5.

Tamworth is on a bearing of 340c from Newcastle. What is the bearing of Newcastle from Tamworth?

6.

7.

8.

The angle of elevation from a point 11.5 m away from the base of a tree up to the top of the tree is 42c 12l. Find the height of the tree to one decimal point. Geoff stands 25.8 m away from the base of a tower and measures the angle of elevation as 39c 20l. Find the height of the tower to the nearest metre. A wire is suspended from the top of a 100 m tall bridge tower down to the bridge at an angle of elevation of 52c. How long is the wire, to 1 decimal place?

10. A plane leaves Melbourne and flies on a bearing of 065c for 2500 km. (a) How far north of Melbourne is the plane? (b) How far east of Melbourne is it? (c) What is the bearing of Melbourne from the plane? 11. The angle of elevation of a tower is 39c 44l when measured at a point 100 m from its base. Find the height of the tower, to 1 decimal place. 12. Kim leaves his house and walks for 2 km on a bearing of 155c . How far south is Kim from his house now, to 1 decimal place? 13. The angle of depression from the top of an 8 m tree down to a rabbit is 43c 52l. If an eagle is perched in the top of the tree, how far does it need to fly to reach the rabbit, to the nearest metre? 14. A girl rides a motorbike through her property, starting at her house. If she rides south for 1.3 km, then rides west for 2.4 km, what is her bearing from the house, to the nearest degree? 15. A plane flies north from Sydney for 560 km, then turns and flies east for 390 km. What is its bearing from Sydney, to the nearest degree?

9.

A cat crouches at the top of a 4.2 m high cliff and looks down at a mouse 1.3 m out from the foot (base) of the cliff. What is the angle of depression, to the nearest minute?

16. Find the height of a pole, correct to 1 decimal place, if a 10 m rope tied to it at the top and stretched out straight to reach the ground makes an angle of elevation of 67c13l.


17. The angle of depression from the top of a cliff down to a boat 100 m out from the foot of the cliff is 59c42l. How high is the cliff, to the nearest metre? 18. A group of students are bushwalking. They walk north from their camp for 7.5 km, then walk west until their bearing from camp is 320c. How far are they from camp, to 1 decimal place? 19. A 20 m tall tower casts a shadow 15.8 m long at a certain time of day. What is the angle of elevation from the edge of the shadow up to the top of the tower at this time?

20 m

15.8 m

20. A flat verandah roof 1.8 m deep is 2.6 m up from the ground. At a certain time of day, the sun makes an angle of elevation of 72c 25l. How much shade is provided on the ground by the verandah roof at that time, to 1 decimal place?

21. Find the angle of elevation of a 15.9 m cliff from a point 100 m out from its base. 22. A plane leaves Sydney and flies for 2000 km on a bearing of 195c. How far due south of Sydney is it? 23. The angle of depression from the top of a 15 m tree down to a pond is 25c41l. If a bird is perched in the top of the tree, how far does it need to fly to reach the pond, to the nearest metre? 24. A girl starting at her house, walks south for 2.7 km then walks east for 1.6 km. What is her bearing from the house, to the nearest degree? 25. The angle of depression from the top of a tower down to a car 250 m out from the foot of the tower is 38c19l. How high is the tower, to the nearest metre? 26. A hot air balloon flies south for 3.6 km then turns and flies east until it is on a bearing of 127c from where it started. How far east does it fly? 27. A 24 m wire is attached to the top of a pole and runs down to the ground where the angle of elevation is 22c 32l. Find the height of the pole.

317

318


28. A train depot has train tracks running north for 7.8 km where they meet another set of tracks going east for 5.8 km into a station. What is the bearing of the depot from the station, to the nearest degree? 29. Jessica leaves home and walks for 4.7 km on a bearing of 075c. She then turns and walks for 2.9 km on a bearing of 115c and she is then due east of her home. (a) How far north does Jessica walk? (b) How far is she from home?

30. Builder Jo stands 4.5 m out from the foot of a building and looks up at to the top of the building where the angle of elevation is 71c. Builder Ben stands at the top of the building looking down at his wheelbarrow that is 10.8 m out from the foot of the building on the opposite side from where Jo is standing. (a) Find the height of the building. (b) Find the angle of depression from Ben down to his wheelbarrow.

Exact Ratios A right-angled triangle with one angle of 45° is isosceles. The exact length of its hypotenuse can be found.

Pythagoras’ theorem is used to find the length of the hypotenuse.

c2 = a2 + b2 AC 2 = 1 2 + 1 2 =2 AC =

2

This means that the trigonometric ratios of 45c can be written as exact ratios.

1 2 1 cos 45c = 2 tan 45c = 1 sin 45c =


319

This angle is commonly used; for example, 45° is often used for the pitch of a roof. The triangle with angles of 60° and 30° can also be written with exact sides.

Halve the equilateral triangle to get TABD.

AD 2 = 2 2 - 1 2 =3 AD =

3

3 2 1 cos 60° = 2 tan 60° = 3

sin 30c =

sin 60° =

1 2

3 2 1 tan 30c = 3

It may be easier to remember the triangle rather than all these ratios.

cos 30c =

DID YOU KNOW? The ratios of all multiples of these angles follow a pattern: A

0c

30c

45c

60c

90c

120c

135c

150c

sin A

0 2

1 2

2 2

3 2

4 2

3 2

2 2

1 2

cos A

4 2

3 2

2 2

1 2

0 2

- 1 2

- 2 2

- 3 2

The rules of the pattern are: • for sin A, when you reach 4, reverse the numbers • for cos A, when you reach 0, change signs and reverse

320


EXAMPLES 1. Find the exact value of sec 45°.

Solution 1 cos 45° 1 = 1 2 = 2

sec 45° =

2. A boat ramp is to be made with an angle of 30c and base length 5 m. What is the exact length of the surface of the ramp?

Solution 5 cos 30c = x x cos 30c = 5 5 cos 30c 5 = 3 2 2 =5# 3 10 = 3 10 3 = 3

x=

So the exact length of the ramp is

10 3 m. 3


321

6.6 Exercises Find the exact value in all questions, with rational denominator where relevant. 1.

(b)

Evaluate (a) sin 60c + cos 60c (b) cos 2 45c + sin 2 45c (c) cosec 45c (d) 2 sec 60c (e) cot 30c + cot 60c

cos 2 45c = (cos 45c) 2

(c)

(f) tan 60c - tan 30c (g) sin 2 60c + sin 2 45c (h) sin 45c cos 30c + cos 45c sin 30c (i) 3 tan 30c tan 45c + tan 60c (j) 1 - tan 45c tan 60c

3.

(k) cos 30c cos 60c - sin 30c sin 60c 4. (l) cos 2 30c + sin 2 30c (m) 2 sec 45c - cosec 30c 2 sin 60c sin 45c (o) 1 + tan 2 30c (n)

(p)

1 - cos 45c 1 + cos 45c

(q)

cot 30c sec 60c

(r) sin 2 45c - 1

2.

If the tent in the previous question was pitched at an angle of 60c, how high would the pole need to be?

6.

The angle of elevation from a point 10 m out from the base of a tower to the top of the tower is 30c. Find the exact height of the tower, with rational denominator.

2 - tan 60c sec 2 45c

Find the exact value of all pronumerals (a)

A 2-person tent is pitched at an angle of 45c. Each side of the tent is 2 m long. A pole of what height is needed for the centre of the tent?

5.

(s) 5 cosec 2 60c (t)

A 2.4 m ladder reaches 1.2 m up a wall. At what angle is it resting against the wall?

322


7.

the floor. How far out from the wall is it?

The pitch of a roof is 45c and spans a length of 12 m.

Find the exact length of AC.

9.

(a) What is the length l of the roof? (b) If a wall is placed inside the roof one third of the way along from the corner, what height will the wall be? 8.

A 1.8 m ladder is placed so that it makes a 60c angle where it meets

10. The angle of depression from the top of a 100 m cliff down to a boat at the foot of the cliff is 30c. How far out from the cliff is the boat?

Angles of Any Magnitude The angles in a right-angled triangle are always acute. However, angles greater than 90c are used in many situations, such as in bearings. Negative angles are also used in areas such as engineering and science. We can use a circle to find trigonometric ratios of angles of any magnitude (size) up to and beyond 360c.

Investigation 1. (a) Copy and complete the table for these acute angles (between 0c and 90c). x

0c

10c

20c

30c

40c

50c

60c

70c

80c

90c

sin x cos x tan x (b) Copy and complete the table for these obtuse angles (between 90c and 180c). x sin x cos x tan x

100c

110c

120c

130c

140c

150c

160c

170c

180c


(c) Copy and complete the table for these reflex angles (between 180c and 270c). x

190c

200c

210c

220c

230c

240c

250c

260c

270c

350c

360c

sin x cos x tan x (d) Copy and complete the table for these reflex angles (between 270c and 360c). x

280c

290c

300c

310c

320c

330c

340c

sin x cos x tan x 2. What do you notice about their signs? Can you see any patterns? Could you write down any rules for the sign of sin, cos and tan for different angle sizes? 3. Draw the graphs of y = sin x, y = cos x and y = tan x for 0c # x # 360c. For y = tan x, you may need to find the ratios of angle close to and either side of 90c and 270c.

Drawing the graphs of the trigonometric ratios can help us to see the change in signs as angles increase. We divide the domain 0c to 360c into 4 quadrants:

1st quadrant: 0c to 90c 2nd quadrant: 90c to 180c 3rd quadrant: 180c to 270c 4th quadrant: 270c to 360c

EXAMPLES 1. Describe the sign of sin x in each section (quadrant) of the graph y = sin x.

Solution We can sketch the graph using the table below or using the values from the tables in the investigation above for more accuracy. x

0c

90c

180c

y

0

1

0

270c -1

360c 0 CONTINUED

323

324


y

1 y = sin x

90c

180c

270c

360c

x

-1

The graph is above the x-axis for the first 2 quadrants, then below for the 3rd and 4th quadrants. This means that sin x is positive in the 1st and 2nd quadrants and negative in the 3rd and 4th quadrants. 2. Describe the sign of cos x in each section (quadrant) of the graph of y = cos x.


0c

90c

y

1

0

180c -1

270c

360c

0

1

y

y = cos x

1

90c

180c

270c

360c

x

-1

The graph is above the x-axis in the 1st quadrant, then below for the 2nd and 3rd quadrants and above again for the 4th quadrant.


325

This means that cos x is positive in the 1st and 4th quadrants and negative in the 2nd and 3rd quadrants. 3. Describe the sign of tan x in each section (quadrant) of the graph y = tan x.


0c

90c

180c

270c

360c

y

0

No result

0

No result

0

Neither tan 90c nor tan 270c exists (we say that they are undefined). Find the tan of angles close to these angles, for example tan 89c 59l and tan 90c 01l, tan 279c 59l and tan 270c 01l. There are asymptotes at 90c and 270c. On the left of 90c and 270c, tan x is positive and on the right, the ratio is negative. y

90c

180c

270c

360c

x

y = tan x

The graph is above the x-axis in the 1st quadrant, below for the 2nd, above for the 3rd and below for the 4th quadrant. This means that tan x is positive in the 1st and 3rd quadrants and negative in the 2nd and 4th quadrants.

To show why these ratios have different signs in different quadrants, we look at angles around a unit circle (a circle with radius 1 unit). We use congruent triangles when finding angles of any magnitude. Page 326 shows an example of congruent triangles all with angles of 20c inside a circle with radius 1 unit.

You will see why these ratios are undefined later on in this chapter.

326


y

1 unit

1 unit 20c 20c

20c 20c 1 unit

x

1 unit

If we divide the circle into 4 quadrants, we notice that the x- and y-values have different signs in different quadrants. This is crucial to notice when looking at angles of any magnitude and explains the different signs you get when finding sin, cos and tan for angles greater than 90c.

Quadrant 1 Looking at the first quadrant (see diagram below), notice that x and y are both positive and that angle i is turning anticlockwise from the x-axis. y First quadrant The angle at the x-axis is 0 and the angle at the y-axis is 90c, with all other angles in this quadrant between these two angles.

(x, y) 1 unit

y

i x

x

Point (x, y) forms a triangle with sides 1, x and y, so we can find the trigonometric ratios for angle i.


327

y 1 =y

sin i =

x 1 =x

cos i =

y tan i = x

Investigation Since cos i = x and sin i = y, we can write the point (x, y) as (cos i, sin i). The polar coordinates (cos i, sin i) give a circle. The polar coordinates 6 A sin ] ai + c g, B sin ] bi g @ form a shape called a Lissajous figure. These are sometimes called a Bowditch curve and they are often used as logos, for example the ABC logo. Use the Internet to research these and other similar shapes. Use a graphics calculator or a computer program such as Autograph to draw other graphs with polar coordinates using variations of sin i and cos i.

Quadrant 2 In the second quadrant, angles are between 90c and 180c. If we take the 1st quadrant coordinates (x, y), where x 2 0 and y 2 0 and put them in the 2nd quadrant, we notice that all x values are negative in the second quadrant and y values are positive. So the point in the 2nd quadrant will be (-x, y) y 90c Second quadrant (-x, y) y 180c

1 unit

x

i

180c- i

0c

x

These are called polar coordinates.

328


Since cos i = x, cos i will negative in the 2nd quadrant. Since sin i = y , sin i will be positive in the 2nd quadrant. y tan i = x so it will be negative (a positive number divided by a negative number). To have an angle of i in the triangle, the angle around the circle is 180c - i.

Quadrant 3 In the third quadrant, angles are between 180c and 270c. y

90c

180c

180c + i

y

0c

i

x

x

1 unit

(-x, -y) Third quadrant

270c

Notice that x and y are both negative in the third quadrant, so cos i and sin i will be both negative. y tan i = x so will be positive (a negative divided by a negative number). To have an angle of i in the triangle, the angle around the circle is 180c + i.

Quadrant 4 In the fourth quadrant, angles are between 270c and 360c. y 90c

180c

i 360c - i

0c

x

1 unit

360c y (x, -y)

270c

Fourth quadrant

x


329

While y remains negative in the fourth quadrant, x is positive again, so sin i is negative and cos i is positive. y tan i = x so will be negative (a negative divided by a positive number) For an angle i in the triangle, the angle around the circle is 360c - i.

ASTC rule Putting all of these results together gives a rule for all four quadrants that we usually call the ASTC rule. y 90c

2nd quadrant

1st quadrant

180c - i

You could remember this rule as All Stations To Central or A Silly Trigonometry Concept, or you could make up your own!

i

S

A 0c

180c

360c

T

180c + i

C

3rd quadrant

x

360c - i 4th quadrant

270c

A: ALL ratios are positive in the 1st quadrant S: Sin is positive in the 2nd quadrant (cos and tan are negative) T: Tan is positive in the 3rd quadrant (sin and cos are negative) C: Cos is positive in the 4th quadrant (sin and tan are negative) This rule also works for the reciprocal trigonometric ratios. For example, where cos is positive, sec is also positive, where sin is positive, so is cosec and where tan is positive, so is cot. We can summarise the ASTC rules for all 4 quadrants:

First quadrant: Angle i: sin i is positive cos i is positive tan i is positive

330


Second quadrant: Angle 180c - i: sin ] 180c - i g = sin i cos ] 180c - i g = - cos i tan ] 180c - i g = - tan i Third quadrant: Angle 180c + i: sin ] 180c + i g = - sin i cos ] 180c + i g = - cos i tan ] 180c + i g = tan i Fourth quadrant: Angle 360c - i: sin ] 360c - i g = - sin i cos ] 360c - i g = cos i tan ] 360c - i g = - tan i

EXAMPLES 1. Find all quadrants where (a) sin i 2 0 (b) cos i 1 0 (c) tan i 1 0 and cos i 2 0

Solution (a) sin i 2 0 means sin i is positive. Using the ASTC rule, sin i is positive in the 1st and 2nd quadrants. (b) cos i is positive in the 1st and 4th quadrants, so cos i is negative in the 2nd and 3rd quadrants. (c) tan i is positive in the 1st and 3rd quadrants so tan i is negative in the 2nd and 4th quadrants. Also cos i is positive in the 1st and 4th quadrants. So tan i 1 0 and cos i 2 0 in the 4th quadrant.


331

2. Find the exact ratio of tan 330c.

Solution First we find the quadrant that 330c is in. It is in the 4th quadrant. y

330c

30c

x

The angle inside the triangle in the 4th quadrant is 30c and tan is negative in the 4th quadrant.

Notice that 360c - 30c = 330c.

tan 330c = - tan 30c 1 =3

30c

:3

2

60c

1

3. Find the exact value of sin 225c.

Solution The angle in the triangle in the 3rd quadrant is 45c and sin is negative in the 3rd quadrant. CONTINUED

Notice that 180c + 45c = 225c.

332


y

225c

x

45c

sin 225c = - sin 45c 1 =2

45c

:2

45c

1

1

4. Find the exact value of cos 510c.

Solution To find cos 510c, we move around the circle more than once. y

30c

150c 510c

510c - 360c = 150c So 510c = 360c + 150c

x


The angle is in the 2nd quadrant where cos is negative. The triangle has 30c in it.

30c

cos 510c = - cos 30c =-

333

3 2

:3

2

60c

Notice that 180c - 30c = 150c.

1

5. Simplify cos (180c + x).

Solution 180c + x is an angle in the 3rd quadrant where cos is negative. So cos ] 180c + x g = - cos x 6. If sin x = -

3 and cos x 2 0, find the value of tan x and sec x. 5

Solution sin x 1 0 in the 3rd and 4th quadrants and cos x 2 0 in the 1st and 4th quadrants. So sin x 1 0 and cos x 2 0 in the 4th quadrant. This means that tan x 1 0 and sec x 2 0. sin x =

sec x is the reciprocal of cos x so is positive in the 4th quadrant.

opposite hypotenuse

So the opposite side is 3 and the hypotenuse is 5. y

x

x 5

3

This is a 3-4-5 triangle.

By Pythagoras’ theorem, the adjacent side is 4. CONTINUED

334


3 4 1 sec x = cos x 5 = 4

So tan x = -

The ASTC rule also works for negative angles. These are measured in the opposite way (clockwise) from positive angles as shown. y -270c

2nd quadrant -(180c+ i )

S

1st quadrant

A

-360c 0

-180c

-(180c- i )

-(360c- i )

T

C

3rd quadrant

x

-i 4th quadrant

-90c

The only difference with this rule is that the angles are labelled differently.

EXAMPLE Find the exact value of tan (-120c).

Solution Notice that - (180c - 60c) = -120c.

Moving around the circle the opposite way, the angle is in the 3rd quadrant, with 60c in the triangle. y

60c

120c

x


Tan is positive in the 3rd quadrant. tan ] -120c g = tan 60c =

3 30c

:3

2

60c

1

6.7 Exercises 1.

Find all quadrants where (a) cos i 2 0 (b) tan i 2 0 (c) sin i 2 0 (d) tan i 1 0 (e) sin i 1 0 (f) cos i 1 0 (g) sin i 1 0 and tan i 2 0 (h) cos i 1 0 and tan i 2 0 (i) sin i 2 0 and tan i 1 0 (j) sin i 1 0 and tan i 1 0

2.

(a) Which quadrant is the angle 240c in? (b) Find the exact value of cos 240c.

3.

(a) Which quadrant is the angle 315c in? (b) Find the exact value of sin 315c.

4.

(a) Which quadrant is the angle 120c in? (b) Find the exact value of tan 120c .

5.

(a) Which quadrant is the angle -225c in? (b) Find the exact value of sin (-225c).

6.

(a) Which quadrant is the angle -330c in? (b) Find the exact value of cos (-330c).

7.

Find the exact value of each ratio. (a) tan 225c (b) cos 315c (c) tan 300c (d) sin 150c (e) cos 120c (f) sin 210c (g) cos 330c (h) tan 150c (i) sin 300c (j) cos 135c

8.

Find the exact value of each ratio. (a) cos (-225c) (b) cos (-210c) (c) tan (-300c) (d) cos (-150c) (e) sin (-60c) (f) tan (-240c) (g) cos (-300c) (h) tan (-30c) (i) cos (-45c) (j) sin (-135c)

335

336


9.

Find the exact value of (a) cos 570c (b) tan 420c (c) sin 480c (d) cos 660c (e) sin 690c (f) tan 600c (g) sin 495c (h) cos 405c (i) tan 675c (j) sin 390c 3 and cos i 1 0, find 4 sin i and cos i as fractions.

10. If tan i =

Use Pythagoras’ theorem to find the third side.

4 11. Given sin i = and tan i 1 0, 7 find the exact value of cos i and tan i. 5 12. If sin x 1 0 and tan x = - , find 8 the exact value of cos x and cosec x. 2 and tan x 1 0, 5 find the exact value of cosec x, cot x and tan x.

13. Given cos x =

14. If cos x 1 0 and sin x 1 0, find cos x and sin x in surd form with 5 rational denominator if tan x = . 7

4 and 9 270c 1 i 1 360c, find the exact

15. If sin i = -

value of tan i and sec i. 3 and 8 180° 1 i 1 270°, find the exact value of tan x, sec x and cosec x.

16. If cos i = -

17. Given sin x = 0.3 and tan x 1 0, (a) express sin x as a fraction (b) find the exact value of cos x and tan x. 18. If tan a = - 1.2 and 270° 1 i 1 360°, find the exact values of cot a, sec a and cosec a. 19. Given that cos i = - 0.7 and 90c 1 i 1 180c , find the exact value of sin i and cot i. 20. Simplify (a) sin ] 180c - i g (b) cos ] 360c - x g (c) (d) (e) (f) (g) (h)

tan ^ 180c + b h sin ] 180c + a g tan ] 360c - i g sin ] - i g cos ] - a g tan ] - x g

Trigonometric Equations This is called the principle solution.

Whenever you find an unknown angle in a triangle, you solve a trigonometric equation e.g. cos x = 0.34. You can find this on your calculator. Now that we know how to find the trigonometric ratios of angles of any magnitude, there can be more than one solution to a trigonometric equation if we look at a larger domain.


337

EXAMPLES 1. Solve cos x =

3 in the domain 0° # x # 360°. 2

Solution 3 is a positive ratio and cos is positive in the 1st and 4th quadrants. 2 So there are two possible answers. In the 1st quadrant, angles are in the form of i and in the 4th quadrant angles are in the form of 360c - i. cos 30c =

3 2

30c

But there is also a solution in the 4th quadrant where the angle is 360c - i. 3 2 x = 30c , 360c - 30c = 30c , 330c

:3

2

For cos x =

60c

1

2. Solve 2 sin 2 x - 1 = 0 for 0c # x # 360c.

Solution 2 sin 2 x - 1 = 0 2 sin 2 x = 1 1 sin 2 x = 2 sin x = !

This is called the principle solution.

1

2 1 =! 2 Since the ratio could be positive or negative, there are solutions in all 4 quadrants. 1st quadrant: angle i 2nd quadrant: angle 180c - i 3rd quadrant: angle 180c + i 4th quadrant: angle 360c - i CONTINUED

338


1 2 x = 45c , 180c - 45c , 180c + 45c , 360c - 45c = 45c , 135c , 225c , 315c

sin 45c =

45c

:2

45c

3. Solve tan x =

1

1

3 for - 180c # x # 180c.

Solution 3 is a positive ratio and tan is positive in the 1st and 3rd quadrants. So there are two possible answers. In the domain - 180c # x # 180c, we use positive angles for 0c # x # 180c and negative angles for - 180c # x # 0c. y 90c

2nd quadrant

1st quadrant

180c - i

i

S

A

180c

0c

-180c -(180c - i)

x

0c

T

C

3rd quadrant

-i 4th quadrant

-90c

In the 1st quadrant, angles are in the form of i and in the 3rd quadrant angles are in the form of - ^ 180c - i h . tan 60c = 3 But there is also a solution in the 3rd quadrant where the angle is - ^ 180c - i h . For tan x = 3 x = 60c , - ] 180c - 60c g = 30c , -120c

30c

:3

2

60c

1


339

4. Solve 2 sin 2x - 1 = 0 for 0c # x # 360c.

Solution Notice that the angle is 2x but the domain is for x. If 0c # x # 360c then we multiply each part by 2 to get the domain for 2x. 0c # 2x # 720c This means that we can find the solutions by going around the circle twice!

30c

2 sin 2x - 1 = 0 2 sin 2x = 1

:3

2

1 2 1 sin 30c = 2 sin 2x =

60c

1

Sin is positive in the 1st and 2nd quadrants. First time around the circle, 1st quadrant is i and the 2nd quadrant is 180c - i. Second time around the circle, we add 360c to the angles. So 1st quadrant answer is 360c + i and the 2nd quadrant answer is 360c + ] 180c - i g or 540c - i. So 2x = 30c , 180c - 30c, 360c + 30c , 540c - 30c = 30c , 150c , 390c , 510c ` x = 15c , 75c , 195c , 255c

The trigonometric graphs can also help solve some trigonometric equations.

EXAMPLE Solve cos x = 0 for 0c # x # 360c. cos 90c = 0 However, looking at the graph of y = cos x shows that there is another solution in the domain 0c # x # 360c. y

1

90c -1

For cos x = 0 x = 90c, 270c

180c

270c 360c

x

Notice that these solutions lie inside the original domain of 0c # x # 360c.

340


Investigation Here are the 3 trigonometric graphs that you explored earlier in the chapter. y = sin x

y = cos x

y = tan x

Use the values in the sin, cos and tan graphs to find values for the inverse trigonometric functions in the tables below and then sketch the inverse trigonometric functions. For example sin 270° = -1 1 So cosec 270c = -1 = -1 Some values will be undefined, so you will need to find values near them in order to see where the graph goes. y = cosec x x sin x cosec x

0c

90c

180c

270c

360c


y = sec x x

0c

90c

180c

270c

360c

0c

90c

180c

270c

360c

cos x sec x y = cot x x tan x cot x

Here are the graphs of the inverse trigonometric functions. y = cosec x

y = sec x

y = cot x

y y = cotx 1

0 -1

90c

180c 270c

360c

x 360c

341

342


6.8 Exercises 1.

Solve for 0c # i # 360c. (a) sin i = 0.35 1 (b) cos i = 2 (c) tan i = - 1 3 (d) sin i = 2 1 (e) tan i = 3 (f) 2 cos i = 3 (g) tan 2i =

3

(h) 2 sin 3i = - 1 (i) 2 cos 2i - 1 = 0 (j) tan 2 3i = 1 2.

Solve for -180c # i # 180c. (a) cos i = 0.187 1 (b) sin i = 2 (c) tan i = 1 3 (d) sin i = 2 1 (e) tan i = 3 (f) 3 tan 2 i = 1 (g) tan 2i = 1 (h) 2 sin 2 3i = 1 (i) tan i + 1 = 0 (j) tan 2 2i = 3

3.

Sketch y = cos x for 0c # x # 360c .

4.

Evaluate sin 270c .

5.

Sketch y = tan x for 0c # x # 360c .

6.

Solve tan x = 0 for 0c # x # 360c .

7.

Evaluate cos 180c .

8.

Find the value of sin 90c .

9.

Solve cos x = 1 for 0c # x # 360c .

10. Sketch y = sin x for -180c # x # 180c . 11. Evaluate cos 270c. 12. Solve sin x + 1 = 0 for 0c # x # 360c . 13. Solve cos 2 x = 1 for 0c # x # 360c . 14. Solve sin x = 0 for 0c # x # 360c . 15. Solve sin x = 1 for - 360c # x # 360c . 16. Sketch y = sec x for 0c # x # 360c . 17. Sketch y = cot x for 0c # x # 360c .

Trigonometric Identities Trigonometric identities are statements about the relationships of trigonometric ratios. You have already met some of these—the reciprocal ratios, complementary angles and the rules for the angle of any magnitude.


Reciprocal ratios

1 sin i 1 sec i = cos i 1 cot i = tan i

cosec i =

Complementary angles

sin i = cos ] 90c - i g cosec i = sec ] 90c - i g tan i = cot ] 90c - i g

Angles of any magnitude

sin ] 180c - i g = sin i cos ] 180c - i g = - cos i tan ] 180c - i g = - tan i sin (180c + i) = - sin i cos (180c + i) = - cos i tan (180c + i) = tan i sin (360c - i) = - sin i cos (360c - i) = cos i tan (360c - i) = - tan i sin (- i) = - sin i cos (- i) = cos i tan (- i) = - tan i

In this section you will learn some other identities, based on the unit circle. In the work on angles of any magnitude, we defined sin i as the y-coordinate of P and cos i as the x-coordinate of P.

343

344


y tan i = x sin i = cos i

tan i =

sin i cos i

cot i =

cos i sin i

1 tan i cos i = sin i

cot i =

Pythagorean identities The circle has equation x 2 + y 2 = 1. Substituting x = cos i and y = sin i into x 2 + y 2 = 1 gives Remeber that cos 2 i means (cos i) 2.

cos 2 i + sin 2 i = 1

This is an equation so can be rearranged to give sin 2 i = 1 - cos 2 i cos 2 i = 1 - sin 2 i There are two other identities that can be derived from this identity.

1 + tan 2 i = sec 2 i


345

Proof cos 2 i + sin 2 i = 1 cos 2 i sin 2 i 1 + = cos 2 i cos 2 i cos 2 i 1 + tan 2 i = sec 2 i This identity can be rearranged to give tan 2 i = sec 2 i - 1 1 = sec 2 i - tan 2 i cot 2 i + 1 = cosec 2 i

Proof cos 2 i + sin 2 i = 1 cos 2 i sin 2 i 1 + = 2 2 sin i sin i sin 2 i 2 cot i + 1 = cosec 2 i This identity can be rearranged to give

These are called Pythagorean identities since the equation of the circle comes from Pythagoras’ rule (see Chapter 5).

cot 2 i = cosec 2 i - 1 1 = cosec 2 i - cot 2 i

EXAMPLES 1. Simplify sin i cot i.

Solution sin i cot i = sin i # = cos i

cos i sin i

2. Simplify sin ^ 90c - b h sec b where b is an acute angle.

Solution sin ^ 90c - b h sec b = cos b # =1

1 cos b

CONTINUED

346


3. Simplify

sin 4 i + sin 2 i cos 2 i .

Solution sin 4 i + sin 2 i cos 2 i = sin 2 i ^ sin 2 i + cos 2 i h = sin 2 i ] 1 g = sin 2 i = sin i 4. Prove cot x + tan x = cosec x sec x.

Solution LHS = cot x + tan x cos x sin x = + sin x cos x cos 2 x + sin 2 x sin x cos x 1 = sin x cos x 1 1 = # cos x sin x = cosec x sec x = RHS =

` cot x + tan x = cosec x sec x 5. Prove that

1 - cos x 1 = . 1 + cos x sin 2 x

Solution 1 - cos x sin 2 x 1 - cos x = 1 - cos 2 x 1 - cos x = ] 1 + cos x g ] 1 - cos x g 1 = 1 + cos x = RHS

LHS =

`

1 - cos x 1 = 2 1 cos x + sin x


6.9 Exercises 1.

2.

Simplify (a) sin ] 90c - i g (b) tan ] 360c - i g (c) cos ] - i g (d) cot ] 90c - i g (e) sec ] 180c + a g

= cosec 2 x - cot 2 x (e) ] sin x - cos x g3 = sin x - cos x - 2 sin 2 x cos x +2 sin x cos 2 x (f) cot i + 2 sec i 1 - sin 2 i + 2 sin i sin i cos i (g) cos 2 ] 90c - i g cot i

Simplify (a) tan i cos i (b) tan i cosec i (c) sec x cot x (d) 1 - sin 2 x 1 - cos a cot 2 x + 1 1 + tan 2 x sec 2 i - 1 5 cot 2 i + 5 1 (j) cosec 2 x (k) sin 2 a cosec 2 a (l) cot i - cot i cos 2 i (e) (f) (g) (h) (i)

3.

(d) sec 2 x - tan 2 x

=

= sin i cos i (h) (cosec x + cot x) (cosec x - cot x) = 1

2

Prove that (a) cos 2 x - 1 = - sin 2 x 1 + sin i (b) sec i + tan i = cos i 3 2 (c) 3 + 3 tan a = 1 - sin 2 a

1 - sin 2 i cos 2 i cos 2 i 2 = tan i + cos 2 i 1 + cot b (j) - cos b cosec b sec b = tan b + cot b (i)

4.

If x = 2 cos i and y = 2 sin i, show that x 2 + y 2 = 4.

5.

Show that x 2 + y 2 = 81 if x = 9 cos i and y = 9 sin i.

Non-right-angled Triangle Results A non-right-angled triangle is named so that its angles and opposite sides have the same pronumeral. There are two rules in trigonometry that refer to nonright-angled triangles. These are the sine rule and the cosine rule.

347

348


Sine rule

sin A sin B sin C a = b = c

Use this rule for finding an angle.

Use this rule for finding a side.

a c b = = sin A sin B sin C

or

Proof

In TABC, draw perpendicular AD and call it h. From TABD, h sin B = c ` h = c sin B

(1)

From TACD, h b h = b sin C

sin C = `

From (1) and (2), c sin B = b sin C sin B sin C = c b Similarly, drawing a perpendicular from C it can be proven that sin A sin B a = b .

(2)


349

EXAMPLES The sine rule uses 2 sides and 2 angles, with 1 unknown.

1. Find the value of x, correct to 1 decimal place.

Solution Name the sides a and b, and angles A and B. a b = sin A sin B 10.7 x = sin 43c 21l sin 79c 12l 10.7 x sin 43c 21l # = sin 43c 21l # sin 43c 21l sin 79c 12l 10.7 sin 43c 21l x= sin 79c 12l Z 7.5 cm 2. Find the value of y, to the nearest whole number. You can rename the triangle ABC or just make sure you put sides with their opposite angles together.

Solution +Y = 180c - (53c + 24c ) = 103c

You need to find +Y first, as it is opposite y.

a b = sin A sin B y 8 = sin 103c sin 53c y 8 = sin 103c # sin 103c # sin 103c sin 53c 8 sin 103c y= sin 53c Z 10

CONTINUED

350


3. Find the value of i, in degrees and minutes.

Solution sin A sin B a = b sin i sin 86c 11l = 6.7 8.3 sin i sin 86c 11l = 6.7 # 6.7 # 6 .7 8.3 6.7 sin 86c 11l sin i = 8 .3 - 1 6.7 sin 86c 11l i = sin c m 8 .3 Z 53c39l

Since sin x is positive in the first 2 quadrants, both acute angles (between 0c and 90c) and obtuse angles (between 90c and 180c) give positive sin ratios. e.g. sin 50c = 0.766 and sin 130c = 0.766 This affects the sine rule, since there is no way of distinguishing between an acute angle and an obtuse angle. When doing a question involving an obtuse angle, we need to use the 2nd quadrant angle of 180c - i rather than relying on the calculator to give the correct answer.

EXAMPLE Angle i is obtuse. Find the value of i, in degrees and minutes.


Solution sin A sin B a = b sin i sin 15c 49l = 5.4 11.9 sin i sin 15c 49l = 11.9 # 11.9 # 5.4 11.9 11.9 sin 15c 49l sin i = 5.4 - 1 11.9 sin 15c 49l m i = sin c 5.4 = 36c 55l ^ acute angle h But i is obtuse ` i = 180c - 36c 55l = 143c 05l

6.10 1.

Exercises

Evaluate all pronumerals, correct to 1 decimal place.

(c)

(a)

(d)

(b)

(e)

351

352


2.

BC = 4.6 cm and +ACB = 33c 47l.

Find the value of all pronumerals, in degrees and minutes. (a)

4.

Triangle EFG has +FEG = 48c , +EGF = 32c and FG = 18.9 mm. Find the length of (a) the shortest side (b) the longest side..

5.

Triangle XYZ has +XYZ = 51c , +YXZ = 86c and XZ = 2.1 m. Find the length of (a) the shortest side (b) the longest side.

6.

Triangle XYZ has XY = 5.4 cm, +ZXY = 48c and +XZY = 63c. Find the length of XZ.

7.

Triangle ABC has BC = 12.7 m, +ABC = 47c and +ACB = 53c as shown. Find the lengths of (a) AB (b) AC.

The shortest side is opposite the smallest angle and the longest side is opposite the largest angle.

(b)

(c)

(d)

A

B

(e) (i is obtuse)

C

9.

Triangle ABC is isosceles with AB = AC. BC is produced to D as shown. If AB = 8.3 cm, +BAC = 52c and +ADC = 32c find the length of

i

Triangle ABC has an obtuse angle at A. Evaluate this angle to the nearest minute if AB = 3.2 cm,

53c

Triangle PQR has sides PQ = 15 mm, QR = 14.7 mm and +PRQ = 62c 29l. Find to the nearest minute (a) +QPR (b) +PQR.

3.7

3.

12.7 m

8.

4.9 21c31l

47c


(a) AD (b) BD. A 52c

8.3 cm

B

32c

C

10. Triangle ABC is equilateral with side 63 mm. A line is drawn from A to BC where it meets BC at D and +DAB = 26c 15l. Find the length of (a) AD (b) DC. D

Cosine rule c 2 = a 2 + b 2 - 2ab cos C

Similarly a 2 = b 2 + c 2 - 2bc cos A b 2 = a 2 + c 2 - 2ac cos B

Proof A

c

b

C

x

p

D

a-x

B

In triangle ABC, draw perpendicular CD with length p and let CD = x. Since BC = a, BD = a - x From triangle ACD b2 = x2 + p2 x cos C = b ` b cos C = x

(1)

(2)

From triangle DAB c2 = p2 + ] a - x g 2 = p 2 + a 2 - 2ax + x 2 = p 2 + x 2 + a 2 - 2ax

(3)

353

354


Substitute (1) into (3): c 2 = b 2 + a 2 - 2ax

(4)

Substituting (2) into (4): c 2 = b 2 + a 2 - 2a ] b cos C g = b 2 + a 2 - 2ab cos C

DID YOU KNOW? Pythagoras’ theorem is a special case of the cosine rule when the triangle is right angled. c 2 = a 2 + b 2 - 2ab cos C When C = 90c c 2 = a 2 + b 2 - 2ab cos 90c = a 2 + b 2 - 2ab ] 0 g = a2 + b2

EXAMPLE Find the value of x, correct to the nearest whole number. The cosine rule uses 3 sides and 1 angle, with 1 unknown.

Solution c 2 = a 2 + b 2 - 2ab cos C x 2 = 5.6 2 + 6.4 2 - 2 (5.6) (6.4) cos 112c 32l Z 99.79 x = 99.79 Z 10 Press 5.6 x 2 + 6.4 x 2 - 2 # 5.6 # 6.4 # cos 112 % , ,, 32 % , ,, =

=


355

When finding an unknown angle, it is easier to change the subject of this formula to cos C. c 2 = a 2 + b 2 - 2ab cos C c 2 + 2ab cos C = a 2 + b 2 - 2ab cos C + 2ab cos C c 2 + 2ab cos C = a 2 + b 2 c 2 - c 2 + 2ab cos C = a 2 + b 2 - c 2 2ab cos C = a 2 + b 2 - c 2 2ab cos C a2 + b2 - c2 = 2ab 2ab cos C =

a2 + b2 - c2 2ab

Similarly cos A =

b2 + c2 - a2 2bc

cos B =

a +c -b 2ac 2

2

Subtract the square of the side opposite the unknown angle.

2

EXAMPLES 1. Find i, in degrees and minutes.

Solution a2 + b2 - c2 2ab 52 + 62 - 32 cos i = 2 ]5 g]6 g 52 = 60 52 m i = cos - 1 c 60 Z 29c 56l

cos C =

2. Evaluate +BAC in degrees and minutes. A 6.1 cm

4.5 cm B

8.4 cm

C

CONTINUED

356


Solution a2 + b2 - c2 2ab 4.5 2 + 6.1 2 - 8.4 2 cos +BAC = 2 ] 4. 5 g ] 6 .1 g = - 0.2386 cos C =

Notice that the negative sign tells us that the angle will be obtuse.

+BAC = cos- 1 ] - 0.2386 g = 103c 48l

6.11 1.

Exercises

Find the value of all pronumerals, correct to 1 decimal place.

(e)

(a) 2.

Evaluate all pronumerals correct to the nearest minute (a)

(b)

(b) (c)

(c) (d)


YZ = 5.9 cm. Find the value of all angles, to the nearest minute.

(d)

7.

Isosceles trapezium MNOP has MP = NO = 12 mm, MN = 8.9 mm, OP = 15.6 mm and +NMP = 119c 15l. (a) Find the length of diagonal NP. (b) Find +NOP.

8.

Given the figure below, find the length of (a) AC (b) AD.

(e)

3.

Kite ABCD has AB = 12.9 mm, CD = 23.8 mm and +ABC = 125c as shown. Find the length of diagonal AC.

B

42 c8 l 8.4 cm

B 12.9 mm

125 c

A

101 c38 l

C

3.7 cm

C

A

23.8 mm

9.9 cm

D

4.

5.

6.

Parallelogram ABCD has sides 11 cm and 5 cm, and one interior angle 79c 25l. Find the length of the diagonals. Quadrilateral ABCD has sides AB = 12 cm, BC = 10.4 cm, CD = 8.4 cm and AD = 9.7 cm with +ABC = 63c 57l. (a) Find the length of diagonal AC (b) Find +DAC (c) Find +ADC. Triangle XYZ is isosceles with XY = XZ = 7.3 cm and

D

9.

In a regular pentagon ABCDE with sides 8 cm, find the length of diagonal AD.

10. A regular hexagon ABCDEF has sides 5.5 cm. (a) Find the length of AD. (b) Find +ADF.

357

358


Applications The sine and cosine rules can be used in solving problems.

Use the sine rule to find: 1. a side, given one side and two angles 2. an angle, given two sides and one angle Use the cosine rule to find: 1. a side, given two sides and one angle 2. an angle, given three sides

EXAMPLES 1. The angle of elevation of a tower from point A is 72c. From point B, 50 m further away from the tower than A, the angle of elevation is 47c. (a) Find the exact length of AT. (b) Hence, or otherwise, find the height h of the tower to 1 decimal place.

Solution

Use TBTA to find AT.

(a) +BAT = 180c - 72c = 108c +BTA = 180c - ] 47c + 108c g = 25c a b = sin A sin B 50 AT = sin 47c sin 25c 50 sin 47c AT = ` sin 25c

^ straight angle h (angle sum of T)


h AT h = AT sin 72c 50 sin 47c = # sin 72c sin 25c Z 82.3 m

359

(b) sin 72c = `

Use right-angled TATO to find h. Do not use the sine rule.

2. A ship sails from Sydney for 200 km on a bearing of 040c , then sails on a bearing of 157c for 345 km. (a) How far from Sydney is the ship, to the nearest km? (b) What is the bearing of the ship from Sydney, to the nearest degree?

Solution

(a) +SAN = 180c - 40c = 140c

^ cointerior angles h

` +SAB = 360c - (140c + 157c) = 63c

^ angle of revolution h

c 2 = a 2 + b 2 - 2ab cos C x 2 = 200 2 + 345 2 - 2 (200) (345) cos 63c Z 96374.3 x = 96374.3 Z 310 So the ship is 310 km from Sydney. sin A sin B a = b sin i sin 63c = 345 310 345 sin 63c ` sin i = 310 Z 0.99 i Z 82c ( b)

The bearing from Sydney = 40c + 82c = 122c

To find the bearing, measure +TSB.

360


6.12

Exercises

1.

Find the lengths of the diagonals of a parallelogram with adjacent sides 5 cm and 8 cm and one of its angles 32c 42l.

2.

A car is broken down to the north of 2 towns. The car is 39 km from town A and 52 km from town B. If A is due west of B and the 2 towns are 68 km apart, what is the bearing of the car from (a) town A (b) town B, to the nearest degree?

3.

7.

A boat is sinking 1.3 km out to sea from a marina. Its bearing is 041c from the marina and 324c from a rescue boat. The rescue boat is due east of the marina. (a) How far, correct to 2 decimal places, is the rescue boat from the sinking boat? (b) How long will it take the rescue boat, to the nearest minute, to reach the other boat if it travels at 80 km/h?

8.

The angle of elevation of the top of a flagpole from a point a certain distance away from its base is 20c. After walking 80 m towards the flagpole, the angle of elevation is 75c. Find the height of the flagpole, to the nearest metre.

9.

A triangular field ABC has sides AB = 85 m and AC = 50 m. If B is on a bearing of 065c from A and C is on a bearing of 166c from A, find the length of BC, correct to the nearest metre.

The angle of elevation to the top of a tower is 54c 37l from a point 12.8 m out from its base. The tower is leaning at an angle of 85c 58l as shown. Find the height of the tower.

54 c37 l

4.

from one post and 11 m from the other, find the angle within which the ball must be kicked to score a goal, to the nearest degree.

12.8 m

85c58 l

A triangular park has sides 145.6 m, 210.3 m and 122.5 m. Find the size of the largest interior angle of the park.

5.

A 1.5 m high fence leans outwards from a house at an angle of 102c. A boy sits on top of the fence and the angle of depression from him down to the house is 32c 44l . How far from the fence is the house?

6.

Football posts are 3.5 m apart. If a footballer is standing 8 m

10. (a) Find the exact value of AC in the diagram. (b) Hence, or otherwise, find the angle i, correct to the nearest minute.


11. Find the value of h, correct to 1 decimal place.

16. Rhombus ABCD with side 8 cm has diagonal BD 11.3 cm long. Find +DAB. 17. Zeke leaves school and runs for 8.7 km on a bearing of 338c, then turns and runs on a bearing of 061c until he is due north of school. How far north of school is he?

12. A motorbike and a car leave a service station at the same time. The motorbike travels on a bearing of 080c and the car travels for 15.7 km on a bearing of 108c until the bearing of the motorbike from the car is 310c. How far, correct to 1 decimal place, has the motorbike travelled? 13. A submarine is being followed by two ships, A and B, 3.8 km apart, with A due east of B. If A is on a bearing of 165c from the submarine and B is on a bearing of 205c from the submarine, find the distance from the submarine to both ships. 14. A plane flies from Dubbo on a bearing of 139c for 852 km, then turns and flies on a bearing of 285cuntil it is due west of Dubbo. How far from Dubbo is the plane, to the nearest km? 15. A triangular roof is 16.8 m up to its peak, then 23.4 m on the other side with a 125c angle at the peak as shown. Find the length of the roof.

125 c 16.8 m

23.4 m

18. A car drives due east for 83.7 km then turns and travels for 105.6 km on a bearing of 029c. How far is the car from its starting point? 19. The figure below shows the diagram that a surveyor makes to measure a triangular piece of land. Find its perimeter.

13.9 m 58 c1l 11.4 m

14.3 m 132 c31l

20. A light plane leaves Sydney and flies for 1280 km on a bearing of 050c. It then turns and flies for 3215 km on a bearing of 149c. How far is the plane from Sydney, to the nearest km? 21. Trapezium ABCD has AD ; BC, with AB = 4.6 cm, BC = 11.3 cm, CD = 6.4 cm, +DAC = 23c 30l and +ABC = 78c . (a) Find the length of AC. (b) Find +ADC to the nearest minute. 22. A plane leaves Adelaide and flies for 875 km on a bearing of 056c. It then turns and flies on a bearing of i for 630 km until it is due east of Adelaide. Evaluate i to the nearest degree.

361

362


23. Quadrilateral ABCD has AB = AD = 7.2 cm, BC = 8.9 cm and CD = 10.4 cm, with +DAB = 107c (a) Find the length of diagonal BD. (b) Find +BCD. 24. Stig leaves home and travels on a bearing of 248c for 109.8 km. He then turns and travels for 271.8 km on a bearing of 143c. Stig then turns and travels home on a bearing of a. (a) How far does he travel on the final part of his journey? (b) Evaluate a.

25. A wall leans inwards and makes an angle of 88c with the floor. (a) A 4 m long ladder leans against the wall with its base 2.3 m out from the wall. Find the angle that the top of the ladder makes with the wall. (b) A longer ladder is placed the same distance out from the wall and its top makes an angle of 31c with the wall. (i) How long is this ladder? (ii) How much further does it reach up the wall than the first ladder?

Area To find the area of a triangle, you need to know its perpendicular height. Trigonometry allows us to find this height in terms of one of the angles in the triangle.

A=

Similarly, 1 ac sin B 2 1 A = bc sin A 2 A=

Proof From D BCD, h sin C = a ` h = a sin C 1 bh 2 1 = ba sin C 2

A=

1 ab sin C 2


363

EXAMPLE Find the area of D ABC correct to 2 decimal places.

To find the area, use 2 sides and their included angle.

Solution 1 ab sin C 2 1 = (4.3) (5.8) sin 112c 34l 2 Z 11.52 units 2

A=

6.13 1.

Exercises

Find the area of each triangle correct to 1 decimal place. (a)

(c)

(d)

(b)

(e)

364


2.

3.

Calculate the exact area of D ABC.

Find the area of DOAB correct to 1 decimal place (O is the centre of the circle).

7.

Find the area of a regular hexagon with sides 4 cm, to the nearest cm 2 .

8.

Calculate the area of a regular pentagon with sides 12 mm.

9.

The figure below is made from a rectangle and isosceles triangle with AE = AB as shown. A 84c

E

4.

5.

Find the area of a parallelogram with sides 3.5 cm and 4.8 cm, and one of its internal angles 67c 13l, correct to 1 decimal place.

B

14.3 cm

D

Find the area of kite ABCD, correct to 3 significant figures.

10.5 cm

C

(a) Find the length of AE. (b) Find the area of the figure. 10. Given the following figure, A 58c

B

6.

Find the area of the sail, correct to 1 decimal place.

44c

9.4 cm

C

36c 6.7 cm

D

(a) Find the length of AC (b) Find the area of triangle ACD (c) Find the area of triangle ABC.


365

Trigonometry in Three Dimensions EXAMPLES 1. From point X, 25 m due south of the base of a tower, the angle of elevation is 47c. Point Y is 15 m due east of the tower. Find: (a) the height, h, of the tower, correct to 1 decimal place (b) the angle of elevation, i, of the tower from point Y.

Solution (a) From D XTO h 25 25 tan 47c = h 26.8 = h tan 47c =

So the tower is 26.8 m high. (b) From DYTO 26.8 tan i = 15 `

26.8 m 15 = 60c 46l

i = tan - 1 c

So the angle of elevation from Y is 60c 46l. 2. A cone has a base diameter of 18 cm and a slant height of 15 cm. Find the vertical angle at the top of the cone.

Solution The radius of the base is 9 cm. 9 sin i = 15 9 ` i = sin - 1 c m 15 = 36c 52l Vertical angle = 2i = 73c 44l

Use the full value of 26.80921775 for a more accurate answer to (b).

366


6.14 1.

2.

3.

Exercises

A gymnastics bar is supported by wires as shown below. (a) If one wire is inclined at an angle of 55c to the horizontal and is 1.4 m out from the base of the bar, find the height of the bar, to the nearest metre. (b) The second wire is inclined at an angle of 68c to the horizontal. How long is the wire (to 1 decimal place)? (c) The third wire is 2.2 m long. What is its angle of elevation?

A pole has two supporting ropes, 2.5 m and 3.1 m long. (a) If the 3.1 m rope makes an angle of elevation of 38c , find the length of the pole, correct to 1 decimal place. (b) What angle of elevation does the other rope make?

A 25 cm #11 cm # 8 cm cardboard box contains an insert (the shaded area) made of foam. (a) Find the area of foam in the insert, to the nearest cm 2 . (b) Find the angle, i, the insert makes at the corner of the box.

4.

A cone has radius 7 cm and a slant height of 13 cm. Find the vertical angle at the top of the cone, in degrees and minutes.

5.

From a point 15 m due north of a tower, the angle of elevation of the tower is 32c (a) Find the height of the tower, to the nearest metre. (b) Find the angle of elevation of the tower at a point 20 m due east of the tower.

6.

A pole is seen from two points A and B. The angle of elevation from A is 58c . If +CAB = 52c and +ABC = 34c , and A and B are 100 m apart, find: (a) how far A is from the foot of the pole, to the nearest metre. (b) the height of the pole, to 1 decimal place.


7.

Two straight paths up to the top of a cliff are inclined at angles of 25c and 22c to the horizontal.

(a) If path 1 is 114 m long, find the height of the cliff, to the nearest metre. (b) Find the length of path 2, to 1 decimal place. (c) If the paths meet at 47c at the base of the cliff, find their distance apart at the top of the cliff, correct to 1 decimal place. 8.

9.

A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?

10. A hot air balloon flying at 950 m/h at a constant altitude of 3000 m is observed to have an angle of elevation of 78c . After 20 minutes, the angle of elevation is 73c . Calculate the angle through which the observer has turned during those 20 minutes.

David walks along a straight road. At one point he notices a tower on a bearing of 053c with an angle of elevation of 21c. After walking 230 m, the tower is on a bearing of 342c , with an angle of elevation of 26c . Find the height of the tower correct to the nearest metre.

Sums and Differences of Angles Sums and differences Angles can be expressed as sums or differences of other angles. This enables us to simplify or evaluate some angles that normally would be too hard to simplify.

cos ^ x - y h = cos x cos y + sin x sin y

367

368


Proof

Remember: cos i = x -coordinate and sin i = y -coordinate.

By the distance formula: d 2 = _ x 2 - x 1 i2 + _ y 2 - y 1 i2 AB 2 = ^ cos x - cos y h2 + ^ sin x - sin y h2 = cos 2 x - 2 cos x cos y + cos 2 y + sin 2 x - 2 sin x sin y + sin 2 y = (cos 2 x + sin 2 x) + (cos 2 y + sin 2 y) - 2 cos x cos y - 2 sin x sin y = 2 - 2 (cos x cos y + sin x sin y)

(1)

By the cosine rule: c 2 = a 2 + b 2 - 2ab cos C AB 2 = 1 2 + 1 2 - 2 (1) (1) cos (x - y) = 2 - 2 cos (x - y) From (1) and (2): 2 - 2 cos ^ x - y h = 2 - 2 ^ cos x cos y + sin x sin y h ` cos ^ x - y h = cos x cos y + sin x sin y

cos ^ x + y h = cos x cos y - sin x sin y

Proof Substitute - y for y. cos (x - y) = cos x cos y + sin x sin y cos (x - (- y )) = cos x cos (- y ) + sin x sin (- y ) cos (x + y) = cos x cos y + sin x (- sin y ) cos (x + y) = cos x cos y - sin x sin y

sin ^ x + y h = sin x cos y + cos x sin y

(2)


369

Proof Substitute 90c - x for x. cos (x - y) = cos x cos y + sin x sin y cos (90c - x - y) = cos (90c - x) cos y + sin (90c- x) sin y cos (90c - (x + y)) = sin x cos y + cos x sin y sin (x + y) = sin x cos y + cos x sin y

sin ^ x - y h = sin x cos y - cos x sin y

Proof Substitute - y for y. sin (x + y) = sin x cos y + cos x sin y sin (x + (- y)) = sin x cos (- y) + cos x sin (- y) sin (x - y) = sin x cos y + cos x (- sin y) sin (x - y) = sin x cos y - cos x sin y

tan ^ x + y h =

tan x + tan y 1 - tan x tan y

Proof tan (x + y) =

sin ^ x + y h

cos ^ x + y h sin x cos y + cos x sin y = cos x cos y - sin x sin y sin x cos y + cos x sin y cos x cos y = cos x cos y - sin x sin y cos x cos y tan x + tan y tan (x + y) = 1 - tan x tan y

tan ^ x - y h =

tan x - tan y 1 + tan x tan y

Divide top and bottom by cos x cos y.

370


Proof Substitute -y for y. tan (x + y) = tan (x + (- y)) = tan (x - y) =

tan x + tan y 1 - tan x tan y tan x + tan ^ - y h 1 - tan x tan ^ - y h tan x - tan y

1 - tan x ^ - tan y h tan x - tan y tan (x - y) = 1 + tan x tan y

EXAMPLES 1. Simplify sin 2i cos i - cos 2i sin i.

Solution sin 2i cos i - cos 2i sin i = sin (2i - i) = sin i 2. Find the exact value of cos 75c .

Solution cos 75c = cos (30c+ 45c) = cos 30c cos 45c - sin 30c sin 45c = =

3 1 1 1 # - # 2 2 2 2 3-1

=

2 2 3-1

# 2 2 6- 2 = 4

2 2

3. Simplify cos ] i + 60c g + sin ] i + 60c g .

Solution cos (i + 60c) + sin (i + 60c) = cos i cos 60c - sin i sin 60c + sin i cos 60c + cos i sin 60c 3 3 1 1 - sin i # + sin i # + cos i # 2 2 2 2 3 3 1 1 n + sin i d = cos i d + + n 2 2 2 2 1+ 3n 1- 3n = cos i d + sin i d 2 2 = cos i #


371

Ratios of double angles By using the sum of angles, we can find the trigonometric ratios for double angles.

sin 2x = 2 sin x cos x

Proof sin 2x = sin (x + x) = sin x cos x + cos x sin x = 2 sin x cos x

cos 2x = cos 2 x - sin 2 x = 1 - 2 sin 2 x = 2 cos 2 x - 1

Proof cos 2x = cos (x + x) = cos x cos x - sin x sin x = cos 2 x - sin 2 x = (1 - sin 2 x) - sin 2 x = 1 - 2 sin 2 x = 1 - 2 (1 - cos 2 x) = 2 cos 2 x - 1

tan 2x =

Proof tan 2x = tan (x + x) tan x + tan x = 1 - tan x tan x 2 tan x tan 2x = 1 - tan 2 x

Remember: sin 2 x + cos 2 x = 1.

2 tan x 1 - tan 2 x

372


EXAMPLES 1. Simplify cos 2 2i - sin 2 2i.

Solution cos 2 2i - sin 2 2i = cos 2 (2i) = cos 4i 2. If sin x =

4 , find the exact value of sin 2x. 7

Solution AC 2 = 7 2 - 4 2 = 33 ` AC = 33 sin 2x = 2 sin x cos x 4 # 7 8 33 = 49 =2#

33 7

PROBLEM 1 ] 3 sin i - sin 3i g to 4 draw up a table of sine ratios. Can you prove this relation?

Ulug Beg (1393–1449) used the relation sin 3 i =

6.15 1.

Exercises

Expand (a) sin ] a - b g (b) (c) (d) (e) (f) (g) (h) (i) (j)

cos ^ p + q h tan ^ a + b h sin (x + 20c) tan ] 48c + x g cos ] 2i - a g cos ( x + 75c) tan ^ 5x - 7y h sin ^ 4a - b h tan ^ a - 3b h

2.

Simplify (a) sin a cos b + cos a sin b tan 36c + tan 29c (b) 1 - tan 36c tan 29c (c) cos 28c cos 27c - sin 28c sin 27c (d) sin 2x cos 3y + cos 2x sin 3y tan 3i - tan i (e) 1 + tan 3i tan i (f) sin 74c cos 42c - cos 74c sin 42c (g) sin ] a + b g + sin ] a - b g (h) sin ^ x + y h - sin ^ x - y h (i) cos ^ x - y h - cos ^ x + y h (j) cos ] m + n g + cos ] m - n g


3.

4.

Find the exact value of (a) sin 75c (b) cos 15c (c) tan 75c (d) tan 105c (e) cos 105c (f) sin 15c (g) sin 105c (h) tan 285c (i) sin (x + 30c) + cos (x + 30c) (j) cos ^ 45c - y h + cos ^ 45c + y h Simplify tan ^ x + y h + tan ^ x - y h 1 - tan ^ x + y h tan ^ x - y h

.

3 2 and cos y = , find 4 3 the exact value of (a) sin ^ x + y h (b) cos ^ x - y h (c) tan ^ x + y h

5.

If sin x =

6.

By taking 2i = i + i, find an expression for (a) sin 2i (b) cos 2i (c) tan 2i

7.

By writing 3i as 2i + i, find an expression in terms of i for (a) sin 3i (b) cos 3i (c) tan 3i tan 7i - tan 3i . 1 + tan 7i tan 3i (b) Find an expression for sin 4i in terms of 7i and 3i.

3 5 and cos y = , find 5 13 the value of (a) cos x (b) sin y (c) sin ^ x - y h (d) tan y (e) tan ^ x + y h

11. If sin x =

12. (a) Write an expression for cos ^ x + y h + cos ^ x - y h . (b) Hence write an expression for cos 50c cos 65c. 13. Find an expression for (a) sin ^ x + y h + sin ^ x - y h (b) (c) (d) (e) (f)

cos ^ x + y h - cos ^ x - y h sin ^ x + y h - sin ^ x - y h cos ^ x + y h + sin ^ x - y h tan ^ x + y h + tan ^ x - y h tan ^ x + y h - tan ^ x - y h

14. Expand (a) sin 2b (b) tan 2i (c) cos 2i (d) sin (x + 2y) (e) cos (2a + b ) (f) tan (x + 2y) (g) sin (2i - d ) (h) cos (i - 2c) (i) tan (x - 2z) (j) sin (2x - 2y)

15. Simplify (a) 2 cos 3x sin 3x (b) cos 2 7y - sin 2 7y 2 tan 5i (c) 1 - tan 2 5i 9. Find an expression for cos 9x in (d) 1 - 2 sin 2 y terms of 2x and 7x. (e) sin 6i cos 6i 10. Find the exact value of (f) ] sin x + cos x g2 (a) cos 23c cos 22c - sin 23c sin 22c (g) 2 cos 2 3a - 1 tan 85c - tan 25c (b) (h) 1 - 2 sin 2 40c 1 + tan 85c tan 25c 2 tan b (c) sin 180c cos 60c (i) 1 - tan 2 b + cos 180c sin 60c (j) ] sin 3x - cos 3x g2 (d) cos 290c cos 80c 8.

(a) Simplify

+ sin 290c sin 80c tan 11c + tan 19c (e) 1 - tan 11c tan 19c

373

374


16. Find the exact value of (a) cos 22.5c sin 22.5c (b) cos 2 30c - sin 2 30c 2 tan 15c (c) 1 - tan 2 15c (d) 2 sin 75c cos 75c 2 tan 120c (e) 1 - tan 2 120c (f) 1 - 2 sin 2 165c (g) 2 cos 2 22.5c - 1 2 tan i (h) where i = 112.5c 1 - tan 2 i (i) sin 67.5c cos 67.5c (j) 2 cos 105c sin 105c 5 , find the exact value 8 of cos 2x and sin 2x.

17. If cos x =

3 12 and tan b = , find 5 5 the exact values of (a) sin ^ a + b h (b) cos 2a (c) sin 2b (d) tan ^ a - b h

18. If sin a =

19. Express sin 4i in terms of i. 20. (a) Simplify

sin 2x . 1 + cos 2x

(b) Hence, find the exact value of tan 15c. 1 21. Find the exact value of tan 22 c 2 by using the expression for tan 2x. 22. Prove (a) sin 2 i = (b) tan

1 sin 2i tan i 2

i 1 - cos i = 2 sin i

23. Show that sin 2 7i - sin 2 4i = sin 11i sin 3i. 24. Prove that cos 3i = 4 cos 3 i - 3 cos i. 25. Find an expression for sin 3x in terms of sin x.

Further Trigonometric Equations Some trigonometric equations are difficult to solve. However, there are some expressions that can be used to solve them.

Ratios in terms of tan

If tan

i 2 i 2t = t, then tan i = 2 1 - t2

Proof 2 tan A 1 - tan 2 A i 2 tan 2 ` tan i = where i = 2A i 1 - tan 2 2 2t = 1 - t2 tan 2A =


If tan

375

i 2t = t, then sin i = 2 1 + t2

Proof tan

i t =t= 2 1

The hypotenuse is 1 + t 2 by Pythagoras’ theorem.

sin 2A = 2 sin A cos A i i ` sin i = 2 sin cos where i = 2A 2 2 t 1 = 2e oe o 2 1+t 1 + t2 2t = 1 + t2

If tan

These ratios for sin and cos

i 2

i come from the 2 triangle above.

i 1 - t2 = t, then cos i = 2 1 + t2

Proof cos 2A = cos 2 A - sin 2 A i i where i = 2A ` cos i = cos 2 - sin 2 2 2 2 2 t 1 =e -e o o 1 + t2 1 + t2 2 t 1 = 2 1+t 1 + t2 1 - t2 = 1 + t2

i The ratios for cos and 2 i sin come from the 2 previous triangle.

376


EXAMPLES 1. Find the exact value of

Solution sin i =

So

2t 1 + t2

2 tan 15c . 1 + tan 2 15c

where t = tan

i 2

2 tan 15c = sin 30c 1 + tan 2 15c 1 = 2

2. Prove that cot

i i - 2 cot i = tan . 2 2

Solution i - 2 cot i 2 1 2 = i tan i tan 2 i 1 2 where t = tan = t 2t 2 1 - t2 2 1 2^1 - t h = t 2t 1 1 - t2 = t t 1 - 1 + t2 = t t2 = t =t

LHS = cot

i 2 = RHS = tan

` cot

i i - 2 cot i = tan 2 2

There is also another expression that will help solve some further trigonometric equations.

a sin i + b cos i = r sin ] i + a g where b r = a 2 + b 2 and tan a = a


377

Proof If tan a =

b , then the a

hypotenuse is a 2 + b 2 by Pythagoras’ theorem.

RHS = r sin (i + a ) =

a 2 + b 2 (sin i cos a + cos i sin a ) a = a 2 + b 2 e sin i # + cos i # 2 a + b2 = a sin i + b cos i = LHS ` a sin i + b cos i = r sin ] i + a g

b a + b2 2

o

b a 2 + b 2 and tan a = a

where r =

EXAMPLES 1. Write

3 sin x + cos x in the form r sin ] x + a g.

Solution a sin i + b cos i = r sin ] i + a g where r =


For 3 sin x + cos x: a = 3, b = 1 r=

a2 + b2 2

3 + 12 = = 3+1 = 4 =2 b tan a = a 1 = 3 a = tan - 1 e

1 o 3

= 30c So

3 sin x + cos x = 2 sin(x + 30c) CONTINUED

378


2. Write 3 sin i + 2 cos i in the form r sin ] i + a g.

Solution a sin i + b cos i = r sin (i + a ) where r =


For 3 sin i + 2 cos i: a = 3, b = 2 r= = = =

a2 + b2 32 + 22 9+4 13 b tan a = a 2 = 3 2 a = tan - 1 c m 3 = 33° 41l

So 3 sin i + 2 cos i =

13 sin (i + 33c 41l)

Class Investigation Can you find similar results for these? • a sin i - b cos i • a cos i + b cos i • a cos i - b sin i

6.16 1.

Exercises

Simplify 2t (a) 1 - t2

(e)

1 - t2 (b) 1 + t2 2 tan 10c (c) 1 - tan 2 10c 1 - tan 25c 1 + tan 2 25c 2

(d)

2 tan i 1 + tan 2 i

i 2 (f) i 1 + tan 2 2 1 - tan 2

2.

Find the exact value of 2 tan 30c (a) 1 + tan 2 30c


3.

(b)

1 - tan 2 22.5c 1 + tan 2 22.5c

(c)

1 - tan 2 30c 1 + tan 2 30c

(d)

2 tan 90c 1 - tan 2 90c

Write each expression in terms of i t where t = tan . 2 (a) cosec i (b) sec i (c) cot i (d) sin i + cos i (e) 1 + tan i

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) 7.

Write each expression in the form r sin ] i - a g. (a) sin i - cos i (b) sin i - 2 cos i (c) sin i - 3 cos i (d) 3 sin i - cos i (e) 5 sin i - 2 cos i

8.

Write the expression 3 cos i + sin i in the form r cos ] i - a g.

9.

Write the expression cos i - 3 sin i in the form r cos ] i + a g.

i 2 (g) 3 cos i + 4 sin i (f) 1 + tan i tan

(h)

1 + sin i + cos i 1 + sin i - cos i

2 sin i + cos i sin i + 3 cos i sin i + cos i 5 sin i + 2 cos i 4 sin i + cos i 3 sin i + cos i 2 sin i + 3 cos i 4 sin i + 7 cos i 5 sin i + 4 cos i 3 sin i + 5 cos i

(i) tan i + sec i (j) sin 2i 1 + sin i - cos i = t. 1 + sin i + cos i

4.

Prove

5.

Find an expression for sin 2i - cos 2i in terms of t.

6.

Write each expression in the form r sin ] i + a g.

10. Write the expression 9 sin i + 2 cos i in the form. (a) r sin ] i + a g (b) r cos ] i - a g

We can use these results to help solve some trigonometric equations.

EXAMPLES 1. Solve 2 sin i = cos i for 0c # i # 360c .

Solution 2 sin i = cos i Dividing both sides by cos i:

(check cos i = 0 does not give a solution)

2 sin i cos i = cos i cos i 2 tan i = 1 tan i = 0.5 CONTINUED

379

380


Since tan i is positive in the first and third quadrants: i = 26c 34l, 180c+ 26c 34l = 26c 34l, 206c 34l 2. Solve cos 2i = cos i for 0c # i # 360c.

Solution cos 2i = cos i 2 cos i - 1 = cos i 2 cos i - cos i - 1 = 0 ] 2 cos i + 1 g ] cos i - 1 g = 0 2

2

` 2 cos i + 1 = 0 or cos i - 1 = 0 2 cos i = -1 cos i = 1 1 i = 0c, 360c cos i = 2 i = 120c, 240c ` solutions are i = 0c, 120c, 240c, 360c 3 sin x + cos x = 1 for 0c # x # 360c.

3. Solve

Solution (Method 1) Use the result for a sin x + b cos x. For 3 sin x + cos x, a = 3 and b = 1 r=

a2 + b2

= ^ 3 h + 12 =2 b tan a = a 1 = 3 ` a = 30c 2

`

3 sin x + cos x = 2 sin ] x + 30c g

Solving:

Sine is positive in the first and second quadrants.

3 sin x + cos x = 1 2 sin ] x + 30c g = 1 1 sin ] x + 30c g = 2 `

for 0c # x # 360c for 30c # x + 30c # 390c

x + 30c = 30c , 180c - 30c , 360c + 30c = 30c , 150c , 390c x = 0c , 120c , 360c


Solution (Method 2) Use the results for t = tan

i . 2

3 sin x + cos x = 1 2t 1 - t2 3d + =1 n 1 + t2 1 + t2 2 3 t + 1 - t2 =1 1 + t2 2 3 t + 1 - t2 = 1 + t2 0 = 2t 2 - 2 3 t = 2t ^ t - 3 h ` 2t = 0 or t - 3 t=0 t x x tan = 0 tan 2 2 x x = 0c, 180c 2 2 ` x = 0c, 120c, 360c

=0 = 3 =

3 for 0c #

x # 180c 2

= 60c

Test x = 180c separately: 3 sin 180c + cos 180c = 0 + (- 1) = -1 !1 ` x = 180c is not a solution Solutions are x = 0c , 120c , 360c .

General solutions of trigonometric equations Often the solutions of trigonometric equations are restricted, for example, to 0c # i # 360c . If the solutions are not restricted, then they can be described by a general formula.

EXAMPLE Find all solutions for sin i =

3 . 2

Solution

CONTINUED

381

382


Sin is positive in the 1st and 2nd quadrants.

sin i = sin 60c i = 60c, 180c - 60c, 360c + 60c, 360c + 180c - 60c, 360c + 360c + 60cg = 60c, 180c - 60c, 360c + 60c, 540c - 60c, 720c + 60c, g If i can also be negative, i = - (180c + 60c), - (360c - 60c), - [360c - (180c - 60c)], g = - 180c - 60c, - 360c + 60c, - 540c - 60c, g So the general solution for sin i = sin 60c is i = 180c # n + ] - 1 g n 60c where n is an integer.

In general, the solution for sin i = sin a is given by i = 180n + (-1) n a where n is an integer.

EXAMPLE Find all solutions for cos i =

Solution

Cos is positive in the 1st and 4th quadrants.

1 . 2

cos i = cos 45c i = 45c, 360c - 45c, 360c + 45c, 360c + 360c - 45c, 360c + 360c + 45c, f = 45c, 360c - 45c, 360c + 45c, 720c - 45c, 720c + 45c, f If i can also be negative, i = - 45c, - (360c - 45c), - (360c + 45c), - [360c + (360c - 45c)], f = - 45c, - 360c + 45c, - 360c - 45c, - 720c + 45c, f So the general solution for cos i = cos 45c is i = 360c # n ! 45c where n is an integer.

In general, the solution for cos i = cos a is given by i = 360n ! a where n is an integer.


383

EXAMPLE Find all solutions for tan i = 1.

Solution

tan i = tan 45c i = 45c, 180c + 45c, 360c + 45c, 360c + 180c + 45c, 360c + 360c + 45c, f

Tan is positive in the 1st and 3rd quadrants.

= 45c, 180c + 45c, 360c + 45c, 540c + 45c, 720c + 45c, f If i can also be negative, then i = - (180c - 45c), - (360c - 45c), - [360c + (180c - 45c)], f = -180c + 45c, - 360c + 45c, - 540c + 45c, f The general solution for tan i = tan 45c is i = 180c # n + 45c where n is an integer.

In general, the solution for tan i = tan a is given by i = 180n + a where n is an integer.

6.17 1.

Exercises

Solve for 0c # x # 360c .

2.

Solve for 0c # i # 360c.

(a) sin x = cos x

(a) 3 sin i + 4 cos i = 0

(b) cos x =

(b) 5 cos i - 12 sin i = -3

3 sin x

(c) sin 2x = sin x

(c) sin i - 3 cos i = 0

(d) tan x - tan x = 0

(d) sin i + cos i = -1

(e) 2 sin x - sin x - 1 = 0

(e) 4 sin i - cos i + 3 = 0

(f) 2 sin 2 x + 3 cos x - 3 = 0

(f) sin i - cos i = 1

(g) sin x cot x - sin x = 0

(g)

(h) cos x - 1 = 0 (i) 2 sin x tan x - tan x + 2 sin x - 1 = 0 (j) 3 cos 2 x - 7 cos x + 4 = 0

(h) 2 sin i - cos i =

2

2

2

2 cos i + sin i = 1 5 2

(i) 3 cos i - 5 sin i + 2 = 0 (j)

2 cos i + sin i + 1 = 0

384


3.

Find the general solution for 1 (a) sin i = 2 (b) tan a = 3 3 2 (d) 2 sin x = -1

7.

Find the general solutions of (a) sin i = 0 (b) cos x = 1 (c) tan x = 0 (d) sin i = -1 (e) cos a = 0

8.

For each question (i) solve for 0c # x # 360cand (ii) find the general solutions (a) 2 sin x - 1 = 0 (b) 4 cos x - 3 = 0 (c) sin x = 3 cos x (d) 3 sin x + cos x = 0 (e) sin x + cos x = 2

9.

Find the general solutions of 2 sin 2 x + sin x – 1 = 0.

(c) cos i =

(e) tan i + 1 = 0 (f) 2 cos 2 b = 1 (g) 4 sin 2 c = 3 1 (h) tan i = 3 (i) cos i = 0.245 (j) sin a = 0.399 4.

Solve sin ] 2x - 45c g =

3 for 2

-180c # x # 180c . 5.

Find the general solutions of sin 2x = cos x.

6.

Solve sin 2 x = sin x for -180c # x # 180c .

10. (a) Solve cos 2x = cos x for 0c # x # 360c. (b) Find the general solutions of cos 2x = cos x.


Test Yourself 6 1.

Find the exact value of cos i and sin i if 3 tan i = . 5

2.

Simplify

11. Find the length of AB as a surd.

(a) sin x cot x cos 40c + sin 50c (b) cos 40c 12. Evaluate x, correct to 2 significant figures. (a)

1 + cot 2 A i 1 - t2 (d) where t = tan 2 2 1+t (e) 1 - 2 sin 2 10i (c)

3.

4.

Evaluate to 2 decimal places. (a) sin 39c 54l (b) tan 61c 30l (c) cos 19c 2l Find i to the nearest minute if (a) sin i = 0.72 (b) cos i = 0.286 5 (c) tan i = 7 2 cos i = 2 + 2 sin i. 1 - sin i 2

5.

Prove that

6.

Find the value of b if sin b = cos ] 2b - 30 g c .

7.

Find the exact value of (a) cos 315c (b) sin ] - 60c g (c) tan 120c (d) 2 sin 105c cos 105c (e) sin ^ x - y h when sin x = cos y =

(b)

13. Evaluate i to the nearest minute. (a)

(b)

8 and 17

5 13

8.

Solve 2 cos x = -1 for 0c # x # 360c.

9.

Sketch the graph of y = cos x, and hence solve cos x = 0 for 0c # x # 360c .

10. A ship sails on a bearing of 215c from port until it is 100 km due south of port. How far does it sail, to the nearest km?

(c)

385

386


14. Find the area of triangle MNO.

15. Solve for -180c # x # 180c . 3 (a) sin 2 x = 4 1 (b) tan 2x = 3 (c) 3 tan 2 x = tan x 5 16. If sec i = - and tan i 2 0, find sin i 4 and cot i. 17. Jacquie walks south from home for 3.2 km, then turns and walks west for 1.8 km. What is the bearing, to the nearest degree, of (a) Jacquie from her home? (b) her home from where Jacquie is now? 18. Find the general solution of 6 sin i - 8 cos i = 5.

(a) Find an expression for the length of AD. (b) Find the height of the pole, to 1 decimal place. 20. A plane flies from Orange for 1800 km on a bearing of 300c . It then turns and flies for 2500 km on a bearing of 205c . How far is the plane from Orange, to the nearest km? 21. Find the exact value of (a) sin 75c (b) cos 105c (c) sin 22c 30l cos 22c 30l. 22. Find the general solutions of (a) 2 cos x – 1 = 0 (b) tan x = 1 3 (c) sin x = . 2 23. Solve 3 sin i + cos i = 1 for 0c # i # 360c . 24. Evaluate a in the figure below. a

19. The angle of elevation from point B to the top of a pole is 39c , and the angle of elevation from D, on the other side of the pole, is 42c. B and D are 20 m apart.

10 mm

4 mm 12 mm

25. (a) Simplify cos x cos y – sin x sin y. (b) Show that cos 2x = 1 – sin2 x.


Challenge Exercise 6 1.

2.

3.

Two cars leave an intersection at the same time, one travelling at 70 km/h along one road and the other car travelling at 80 km/h along the other road. After 2 hours they are 218 km apart. At what angle, to the nearest minute, do the roads meet at the intersection? A ship sails from port on a bearing of 055c , then turns and sails on a bearing of 153c for 29.1 km, when it is due east of port. How far, to 1 decimal place, is the ship from its starting point? Evaluate x correct to 3 significant figures.

6.

Simplify sin ] 360c - x g $ tan ] 90c- x g .

7.

Find the exact area of D ABC.

8.

Find the exact value of cos (-315c) .

9.

Solve tan 2x - 1 = 0 for 0c # x # 360c .

10. Find i to the nearest minute.

4.

(a) Find an exact expression for the length of AC. (b) Hence, or otherwise, find the value of h correct to 1 decimal place. 11. The angle of depression from the top of a 4.5 m mast of a boat down to a fish is 56c 28l . How far down, to 1 decimal place, does a pelican sitting at the top of the mast need to fly to catch the fish? 12. Solve 2 cos (i + 10c ) = - 1 for 0c # i # 360c.

5.

A man walks 3.8 km on a bearing of 134c from a house. He then walks 2.9 km on a bearing of 029c . How far is he from the house, to 1 decimal place?

13. Two roads meet at an angle of 74c . Find the distance, correct to 3 significant figures, between two cars, one 6.3 km from the intersection along one road and the other 3.9 km along the other road.

387

388


14. Find the exact value of cos i, given 5 sin i = and cos i 1 0. 9 15. From the top of a vertical pole the angle of depression to a man standing at the foot of the pole is 43c . On the other side of the pole is another man, and the angle of depression from the top of the pole to this man is 52c . The men are standing 58 m apart. Find the height of the pole, to the nearest metre. 16. Show that cos i ] sin i + cos i g = 1 + tan i. ] 1 + sin i g ] 1 - sin i g 17. If x = 3 sin i and y = 3 cos i - 2, eliminate i to find an equation relating x and y. 18. From point A, 93 m due south of the base of a tower, the angle of elevation is 35c . Point B is 124 m due east of the tower. Find (a) the height of the tower, to the nearest metre (b) the angle of elevation of the tower from point B.

20. A cone has a base diameter of 14 cm and a perpendicular height of 26 cm. Find the vertical angle at the top of the cone. 21. Show that cos 6i cos 4i - sin 6i sin 4i = 2 cos 2 5i - 1. 22. A cable car 100 m above the ground is seen to have an angle of elevation of 65c when it is on a bearing of 345c . After a minute, it has an angle of elevation of 69c and is on a bearing of 025c . Find how far it travels in that minute, and its speed in ms - 1 . 23. Solve cos 2i - sin i = 0 for 0c # i # 360c .

19. ABCD is a triangular pyramid with 24. Find the general solutions of sin i = -1. BC = 7 m, CD = 10 m, BD = 8 m, AB = AC and +ACB = 67c . Calculate 25. Simplify cosec i ] cos i - 1 g by expressing (a) +BCD i it in terms of t c tan m . (b) length AB, to the nearest metre. 2

Ch 6 Trigonometry

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