ch3.pdf

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C H A P T E R

3 Circular functions Objectives To understand the reciprocal circular functions cosec, sec and cot To understand and apply the identities sec2 = 1 + tan2 and cosec2 = 1 + cot2 To understand and apply the compound angle formulas To understand and apply the double angle formulas To understand the restricted circular functions and their inverses sin−1 , cos−1 and tan−1 To understand the graphs of the inverse functions sin−1 , cos−1 and tan−1 To solve equations involving circular functions

The sine, cosine and tangent functions are discussed in some detail in section 1.1. Several new circular functions are introduced in this chapter.

3.1

The reciprocal circular functions The cosecant function, y = cosec The cosecant function is defined as: 1 provided sin = 0 cosec = sin Since sin = 0 when = n, n ∈ Z, the domain of cosec = R \ { : = n, n ∈ Z}. The graph of cosec is derived from the graph of sin . The range of sin is [−1, 1] therefore the range of cosec is R \ (−1, 1). The graph of y = sin has turning points at = + n, n ∈ Z as does the graph of 2 y = cosec . sin = 0 at = n, n ∈ Z. These values of will be vertical asymptotes for y = cosec . A sketch of the graph of f : R \ { : = n, n ∈ Z} → R, f () = cosec is shown below. The graph of y = sin is shown on the same set of axes.

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Essential Specialist Mathematics

y y = cosec θ 1

y = sin θ

0 –π

π π 2

–π 2 –1

3π 2

2π

θ

Using a graphics calculator 1 1 , the function y = can be entered into the Y= window. sin x sin x

Since cosec x =

Check that the calculator is in RADIAN mode and press y = cosec x.

ZOOM

7 to display the graph of

The secant function, y = sec The secant function is defined as: 1 provided cos = 0 sec = cos As the graph of y = cos is a translation of the graph of y = sin , the graph of y = sec must be a translation of the graph of y = cosec , units in the negative direction of the x axis. 2 The domain of sec = R \ : = + n, n ∈ Z . 2 A sketch of the graph of f : R \ : = + n, n ∈ Z → R, f () = sec is shown. 2 The graph of y = cos is shown on the same set of axes. y y = sec θ 1 0 –π

–π 2 –1

y = cos θ π 2

π

3π 2π 2

θ

The range of sec is R \ (−1, 1). sec has turning points at = n, n ∈ Z, and has asymptotes at = + n, n ∈ Z. 2

The cotangent function, y = cot The cotangent function is defined as: cot =

cos sin

provided sin = 0

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Chapter 3 — Circular functions

∴

cot = tan

2

− (the complementary properties of cot and tan are listed

later in this section) − = −tan − 2 = −tan + 2 The graph of cot is a translation of tan , units in the negative direction 2 of the x axis, and a reflection in the –π –π axis. A sketch of the graph of f : R \ { : = n, n ∈ Z} → R, f () = cot is shown opposite.

y

0

π 2

2

π 3π 2

θ

2π

Using a graphics calculator cos x cos x , the function y = can be entered into the Y = window. sin x sin x Check that the calculator is in RADIAN mode and press ZOOM 7 to display the graph of y = cot x. Since cot x =

Example 1 Sketch the graph of each of the following over the interval [0, 2]. b y = sec x + c y = cot x − a y = cosec(2x) 3 4 y

Solution a The graph of y = cosec(2x) is obtained from the graph of y = cosec x by a dilation of factor 12 from the y axis. The graph of y = sin(2x) is also shown. is b The graph of y = sec x + 3 a translation of the graph of y = sec x, units in the negative 3 direction of the x axis. The y axis = 2. The intercept is sec 3 7 . asymptotes are at x = and x = 6 6

y = cosec 2x y = sin 2x

1 0 –1

π 2

π

2π

3π 2

x

y

(2π, 2)

2 1 0 –1 –2

π 6

π 7π 6

x 2π

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c The graph of y = cot x − is a 4 translation of the graph of y = cot x, units in the positive 4 direction of the x axis. The y axis = −1 intercept is cot − 4 5 . The asymptotes are at x = and x = 4 4 3 7 The x-axis intercepts are at and . 4 4

y

1 0 –1

π 3π π 4 4

5π 7π 2π 4 4 (2π, –1)

For right-angled triangles, the reciprocal functions can be defined through ratios. hyp opp hyp sec x ◦ = adj adj cot x ◦ = opp

x

C

cosec x ◦ =

and

hyp opp x° A

adj

Example 2

B

C

In triangle ABC, ∠ABC = 90◦ and ∠CAB = x◦ . AB = 6 cm and BC = 5 cm. Find: b the trigonometric ratios related to x◦ a AC

5

Solution a By Pythagoras’ theorem, AC2 = 52 + 62 = 61 √ ∴ AC = 61 cm 5 6 cos x ◦ = √ b sin x ◦ = √ 61 61 √ √ 61 61 cosec x ◦ = sec x ◦ = 5 6

x° A

6 tan x ◦ =

5 6

cot x ◦ =

6 5

B

Useful properties The symmetry properties established for sine, cosine and tangent can be used to establish the following results: sec( − x) = −sec x cosec( − x) = cosec x cot( − x) = −cot x sec(2 − x) = sec x cosec(2 − x) = −cosec x cot(2 − x) = –cot x

sec( + x) = −sec x cosec( + x) = −cosec x cot( + x) = cot x sec(−x) = sec x cosec(−x) = −cosec x cot(−x) = −cot x

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The complementary properties are also useful. − x = cosec x − x = tan x sec cot 2 2 − x = cot x − x = sec x tan cosec 2 2 Example 3 Find the exact value of each of the following: 11 −23 a sec b cosec 4 4 Solution 3 11 = sec 2 + a sec 4 4 3 = sec 4 1 = 3 cos 4 1 = − √12 √ =− 2 11 c cot = cot 4 − 3 3 = cot − 3 = − cot 3 1 =− tan 3 1 = −√ 3

c cot b cosec

11 3

−23 4

= cosec −6 + 4 = cosec 4 1 = 1 √ 2 √ = 2

Two new identities It was shown earlier that, for all values of x, sin2 x + cos2 x = 1. From this identity the following identities can be derived: 1 + cot2 x = cosec2 x 1 + tan2 x = sec2 x

provided sin x = 0 provided cos x = 0

The first of these identities is obtained by dividing each term of the original identity by sin2 x: sin2 x cos2 x 1 i.e. + = 2 2 sin x sin x sin2 x 2 which implies 1 + cot x = cosec2 x The derivation of the second identity is left as an exercise for the reader.

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Example 4 Simplify the expression

cos x − cos3 x . cot x

Solution cos x(1 − cos2 x) cos x − cos3 x = cot x cot x sin x = cos x × sin2 x × cos x = sin3 x

Using a CAS calculator Key the expression (cos(x) − cos(x)ˆ3)/(cos(x)/sin(x)) into the entry line and press ENTER to display the answer of (sin(x))3 on the screen.

Example 5 , find: If tan x = 2, x ∈ 0, 2 b cos x a sec x

c sin x

Solution sec x = tan x + 1 =4+1 √ ∴ sec x = ± 5 √ As x ∈ 0, , sec x = 5 2 √ 2 5 c sin x = tan x × cos x = 5

a

2

2

d cosec x √ 5 1 = b cos x = sec x 5

√ 5 1 = d cosec x = sin x 2

Using a CAS calculator Check the calculator is in RADIAN and EXACT modes. Key the following expression into the entry line using solve from the Algebra (F2) menu: solve(tan(x) = 2, x)| x > 0 and x < /2 Press STO➢ ALPHA A and then ENTER . To find sec x, key 1/cos(A) into the entry line and then press process to find cos x, sin x and cosec x.

ENTER

. Repeat the

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Exercise 3A 1 Sketch the graph of each of the following, over the interval [0, 2]: a y = cosec x + b y = sec x − c y = cot x + 4 6 3 3 2 e y = cosec x − f y = cot x − d y = sec x + 2 4 3 2 Sketch the graph of each of the following, over the interval [0, ]: c y = cot(4x) f y = cot 2x − e y = sec(2x + ) d y = cosec 2x + 3 2 3 Sketch the graph of each of the following, over the interval [−, ]: 2 a y = sec 2x − b y = cosec 2x + c y = cot 2x − 2 3 3 a y = sec 2x

b y = cosec(3x)

4 Find the trigonometric ratios cot x◦ , sec x◦ and cosec x◦ for each of the following triangles: c b a x°

5 5

9

x° 7

x°

8

7

5 Find the exact value of each of the following: − 2 3 c tan a sin b cos 4 3 4 − 5 f cot g sin e sec 6 4 4 − 3 9 i sec j cosec k cot 3 4 4 6 Simplify each of the following expressions: a sec2 x − tan2 x

sin2 x e sin4 x − cos4 x + cos x cos x

− If tan x = −4, x ∈ , 0 , find: 2 b cos x a sec x

3 , find: If cot x = 3, x ∈ , 2 b sin x a cosec x

− , 0 , find: If sec x = 10, x ∈ 2 b sin x a tan x

3 , 2 , find: If cosec x = −6, x ∈ 2 b cos x a cot x

d 7

8

9

10

b cot2 x − cosec2 x

c

6 5 h tan 6 −7 l cos 3

d cosec

tan2 x + 1 tan2 x

f tan3 x + tan x

c cosec x

c sec x

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11 sin x◦ = 0.5, 90 < x < 180. Find: a cos x◦

c cosec x◦

b cot x◦

12 cosec x◦ = −3, 180 < x < 270. Find: a sin x◦

c sec x◦

b cos x◦

13 cos x◦ = −0.7, 0 < x < 180. Find: a sin x◦

c cot x◦

b tan x◦

14 sec x◦ = 5, 180 < x < 360. Find: a cos x◦

c cot x◦

b sin x◦

15 Simplify each of the following expressions: a sec2 + cosec2 − sec2 cosec2

b (sec − cos ) (cosec − sin )

c (1 − cos2 ) (1 + cot2 )

d

sec2 − cosec2 tan2 − cot2

1 16 If x = sec − tan , prove that x + = 2 sec and also find a simple expression for x 1 x − in terms of . x

3.2

Compound and double angle formulas Compound angle formulas The following identities are called the compound angle formulas. 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) = sin x cos y − cos x sin y sin(x + y) = sin x cos y + cos x sin y tan x − tan y tan(x − y) = 1 + tan x tan y tan x + tan y tan(x + y) = 1 − tan x tan y A proof of the first identity is given below and the other identities can be derived from that result.

y

Proof cos(x − y) = cos x cos y + sin x sin y Consider angles x and y, x > y, measured counter-clockwise, and the corresponding points P (cos x, sin x) and Q (cos y, sin y) respectively. Let be the angle measured anticlockwise from OQ to OP. Then x − y = + 2k for some integer constant k. Two cases need to be considered.

Q(cos y, sin y)

1 O

–1

x

α

1

–1

P(cos x, sin x)

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Case 1 0<≤ Using vectors, − → OP = cos xi + sin xj − → OQ = cos yi + sin yj

− → and |OP| = 1 − → and |OQ| = 1

We apply a.b = |a| |b| cos − → − → to obtain OP.OQ = cos

Case 2

y < ≤ 2

In the diagram opposite: → − → − OP.OQ = cos(2 − )

Q

O

= cos(−)

–1

= cos

∴ in all cases: − → − → OP.OQ = cos − → − → Now OP.OQ = cos x cos y + sin x sin y

α –1

(using dot product)

and cos(x − y) = cos( + 2k) where k is an integer = cos

∴

1

cos(x − y) = cos x cos y + sin x sin y

The derivation of other identities cos(x + y) = cos[x − (−y)] = cos x cos(−y) + sin x sin(−y) = cos x cos y − sin x sin y −x+y sin(x − y) = cos 2 = cos − x cos y − sin − x sin y 2 2 = sin x cos y − cos x sin y sin(x − y) tan(x − y) = cos(x − y) sin x cos y − cos x sin y = cos x cos y + sin x sin y

x 1 P

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Dividing top and bottom by cos x cos y we have: sin x cos y cos x sin y − cos x cos y cos x cos y tan(x − y) = sin x sin y 1+ cos x cos y tan x − tan y = 1 + tan x tan y The derivation of the remaining two identities is left as an exercise for the reader. Example 6 a Use

5 5 = + to evaluate sin . 12 6 4 12

b Use

= − to evaluate cos . 12 3 4 12

Solution 5 = sin + a sin 12 6 4 = sin cos + cos sin 6 4 6 4 √ √ √ 1 2 3 2 = × + × 2 2 2 2 √ √ 2 = (1 + 3) 4

Using a CAS calculator Check the calculator is in RADIAN and EXACT modes. Key the expression sin(5/12) into the entry line and then press ENTER . The exact answer is given. b cos

= cos − 12 3 4 = cos cos + sin sin 3 4 3 4 √ √ √ 2 3 2 1 = × + × 2 2 2 2 √ √ 2 = (1 + 3) 4

Example 7

3 and cos y = −0.4, y ∈ , . Find sin(x + y). sin x = 0.2, x ∈ 0, 2 2 Solution sin x = 0.2

∴

√ cos x = ± 1 − 0.22 = ± 0.96

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√ √ 2 6 , cos x = 0.96 = As x ∈ 0, 2 5 √ cos y = −0.4 ∴ sin y = ± 1 − (−0.4)2 = ± 0.84 √

√ 3 21 As y ∈ , , sin y = − 0.84 = − 2 5 sin(x + y) = sin x cos y + cos x sin y √ √ 2 6 21 ×− = 0.2 × −0.4 + 5 5 √ 2 = −0.08 − × 3 14 25 √ 2 = − [1 + 3 14] 25

Using a CAS calculator Check the calculator is in RADIAN and EXACT modes. Key the following expression into the entry line using solve from the Algebra (F2) menu: solve(sin(x) = 0.2, x)| x > 0 and x < /2 Press STO➢ ALPHA A and then ENTER . Repeat for the following expression: solve(cos(y) = −0.4, y)| y > and y < 3/2 Press

STO➢ ALPHA B

and then

ENTER

.

To find sin(x + y), use Trig Expand from the Algebra (F2) menu and key the following expression into the entry line: tExpand(sin(A + B)) Press

ENTER

and the answer is displayed.

Double angle formulas The double angle formulas are: sin 2x = 2 sin x cos x cos 2x = cos2 x − sin2 x = 1 − 2 sin2 x = 2 cos2 x − 1 2 tan x tan 2x = 1 − tan2 x

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These formulas can be derived from the compound angle formulas by substituting y for x: cos(x + y) = cos x cos y − sin x sin y cos(x + x) = cos x cos x − sin x sin x cos 2x = cos2 x − sin2 x

∴

The two other equivalent forms of cos 2x can be obtained by applying the Pythagorean identity sin2 x + cos2 x = 1 e.g.

cos2 x − sin2 x = (1 − sin2 x) − sin2 x = 1 − 2 sin2 x = cos2 x − (1 − cos2 x)

also

= 2 cos2 x − 1 Example 8 If sin = 0.6, ∈

, , find sin 2. 2

Solution sin = 0.6 As ∈

2

∴ cos = ± 1 − 0.62

= ±0.8

, , cos = −0.8

sin 2 = 2 sin cos = 2 × 0.6 × −0.8 = −0.96 Example 9

3 If cos = 0.7, ∈ , 2 , find sin . 2 2 Solution cos 2x = 1 − 2 sin2 x ∴ cos = 1 − 2 sin2 2 2 2 sin = 1 − 0.7 2 = 0.3 √ sin = ± 0.15 2

3 3 ∈ , 2 , ∈ , 2 2 4 ∴ sin is positive 2 √ √ 15 sin = 0.15 = 2 10

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Exercise 3B 1 Use the compound angle formulas to expand each of the following: b cos(x2 + y)

a sin(2x − 5y)

c tan(x + (y + z))

2 Simplify each of the following: a sin x cos 2y − cos x sin 2y b cos 3x cos 2x + sin 3x sin 2x tan A − tan(A − B) d sin(A + B) cos(A − B) + cos(A + B) sin(A − B) c 1 + tan A tan(A − B) e cos y cos(−2y) − sin y sin(−2y) 3 a Expand sin(x + 2x).

b Hence, express sin 3x in terms of sin x.

4 a Expand cos(x + 2x).

b Hence, express cos 3x in terms of cos x.

5 Use the compound angle formulas and appropriate angles to find the exact value of each of the following: 5 7 a sin b tan c cos d tan 12 12 12 12 , and tan y = 2.4, y ∈ 0, . Find the exact value of each 6 Let sin x = 0.6, x ∈ 2 2 of the following: c b sec y f e tan x i h tan(x + y)

3 and sin y = 0.4, y ∈ 0, 7 Let cos x = −0.7, x ∈ , 2 of the following, correct to two decimal places: a cos x d sin y g sin(x − y)

a sin x

c tan(x − y)

b cos y

cos y cos(x − y) tan(x + 2y) , Find the value of each 2

d cos(x + y)

8 Simplify each of the following: a

1 2

sin x cos x

b sin2 x − cos2 x

4 sin3 x − 2 sin x cos x cos 2x

3 9 Let sin x = −0.8, x ∈ , . Find: 2 c tan 2x b cos 2x a sin 2x . Find: 10 Let tan x = 3, x ∈ 0, 2 b tan 3x a tan 2x d

sin4 x − cos4 x cos 2x

e

c

tan x 1 − tan2 x

f

4 sin2 x − 4 sin4 x sin 2x

11 Use the double angle formula for tan 2x, and the fact that tan = 1, to find the exact 4 value of tan . 8

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3 . Find correct to two decimal places: 12 Let sin x = −0.75, x ∈ , 2 b sin 12 x a cos x 13 If cos x = 0.9, x ∈ 0, , find cos 12 x correct to two decimal places. 2 14 In a right-angled triangle GAP, AP = 12 m and GA = 5 m. T is a point on AP, such that ∠AGT = ∠TGP = xc . Without using a calculator, find the exact values of the following:

P

a tan 2x b tan x, by using the double angle formula c AT

12 m T

G

3.3

5m

A

Inverses of circular functions All six circular functions discussed earlier are periodic and are, therefore, many-to-one functions. The inverse of these functions cannot, therefore, be functions. However, by restricting the domain so that the circular functions are one-to-one functions, the inverse circular functions can be defined.

The inverse sine function, y = sin-- 1 x

When the domain for the function sin is restricted to the interval − , , it is a one-to-one 2 2 function and an inverse function exists. defined through consecutive turning

Other intervals 3 5 3 , or , ) could have been used for the domain, but points of the graph (e.g. 2 2 2 2 is the standard convention. − , 2 2 The inverse of the restricted sine function is denoted by sin−1 and is defined by: sin−1 : [−1, 1] → R, sin−1 x = y where sin y = x, y ∈ − , 2 2 The domain of sin−1 = range of the restricted sine function = [−1, 1]. . The range of sin−1 = domain of the restricted sine function = − , 2 2 , By the property of inverse functions, sin(sin−1 x) = x and, for x ∈ − , 2 2 −1 sin (sin x) = x. , through The graph of y = sin−1 (x) is obtained from the graph of y = sin (x), x ∈ − , 2 2 a reflection in the line y = x. Note:

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y y

1

π –1 2 y = sin x y = sin x 1

0 –π 2

y

–1

y=x

–π –1 2

y = sin–1 x

y = sin x

0

x

π 2

π 2

1 –1

0

x

π 2

x

–1

–π 2 sin−1 is also denoted by arsin, arcsin or asin.

1

–π 2

The inverse cosine function, y = cos--1 x

cos−1 x is defined in a similar way to the inverse sine function. The standard domain of the restricted cosine function is [0, ], although other intervals would also produce an inverse function. cos−1 x, the inverse function of the restricted cosine function, is defined as follows: cos−1 : [−1, 1] → R, cos−1 x = y where cos y = x, y ∈ [0, ] The graph of the restricted cosine function, its reflection in the line y = x to produce y = cos−1 x, and the graph of y = cos−1 x are shown in the three figures below.

y

y

y

π y = cos–1 x

π

y=x

π

1 0 –1

Note:

1 0

y = cos x π 2

π

x

–1 –1

2

y = cos x π 2

π

x

y = cos–1 x

0 –1

x 1

The domain of cos−1 x is [−1, 1] and the range is [0, ]. Also, cos(cos−1 x) = x and, for x ∈ [0, ], cos−1 (cos x) = x. cos−1 is also denoted by arcos or acos.

--1 x The inverse tangent function, y = tan

, a one-to-one function is formed If the domain of the tangent function is restricted to − , 2 2 and the inverse function exists. tan−1 : R → R, tan−1 x = y where tan y = x, y ∈ − , 2 2

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The graph of the restricted tangent function, its reflection in the line y = x to produce y = tan−1 x, and the graph of y = tan−1 x are shown in the three figures below.

y

y

y y = tan x

0

π 2

–π 2

π 2 x

0 –π 2

y = tan x

π 2

y = tan–1 x x

–π 2 Note:

π 2

y=x

y=

π 2

y = tan–1 x x

0 –π 2

y=–

π 2

The domain of tan−1 x is R and its range is − , . 2 2 Also tan(tan−1 x) = x and, for x ∈ − , , tan−1 (tan x) = x. 2 2 tan−1 is also denoted by artan, arctan or atan.

Example 10 Sketch the graph of each of the following for their maximal domain: b y = tan−1 (x + 2) + a y = cos−1 (2 − 3x) 2 y Solution a For the function to be defined

(1, π)

−1 ≤ 2 − 3x ≤ 1 ⇔ ⇔

−3 ≤ −3x ≤ −1 1 3

≤x ≤1

x i.e. the implied domain is 13 , 1 0 1 1

3 Note: y = cos−1 −3 x − 23 The graph is obtained from the graph of y = cos−1 (x) by the following sequence of transformations: a dilation from the y axis of factor 13 a reflection in the y axis a translation of 23 units in the positive direction of the x axis b The domain of tan−1 is R. The graph of y = tan−1 (x + 2) + is obtained from the 2 graph of y = tan−1 (x) by a translation of 2 units in the negative direction of the x axis and units in the positive direction of the y axis. 2

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y

y=π –2,

π 2 x

0

Example 11 √ 3 a Evaluate sin−1 − 2 b Simplify: i sin−1 sin 6 iii sin−1 cos 3

5 ii sin sin 6 √ 2 iv sin cos−1 2 −1

Solution

√ √ − 3 3 −1 − is equivalent to solving the equation sin x = a Evaluating sin 2 2

∴ ∴

b

√ 3 sin = 3 2 √ 3 =− sin − 3 2 √ 3 =− sin−1 − 2 3

∈ − , , by definition sin−1 sin = 6 2 2 6 6

5 5 ii sin−1 sin = sin−1 sin − 6 6 = sin−1 sin 6 = 6 −1 = sin−1 sin − cos iii sin 3 2 3 −1 sin = sin 6 = 6 i Since

119

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−1

iv sin cos

√ 2 = sin 2 4 √ 2 = 2

Example 12 Find the implied domain and range of: b y = 3 cos−1 (2 − 2x) a y = sin−1 (2x − 1) Solution a For sin−1 (2x − 1) to be defined

b For 3 cos−1 (2 − 2x) to be defined −1 ≤ 2 − 2x ≤ 1 ⇔ −3 ≤ −2x ≤ −1 3 1 ≤x≤ ⇔ 2 2

−1 ≤ 2x − 1 ≤ 1 ⇔ 0 ≤ 2x ≤ 2 ⇔ 0≤x ≤1

∴ the implied domain is

∴ the implied domain is [0,1] . The range is − , 2 2

1 3 , 2 2

The range is [0, 3].

Example 13 Find the implied domain and range of y = cos(−sin−1 x), where cos has restricted domain [0, ]. Solution where u = −sin−1 x u

Let y = cos u, u ∈ [0, ]

y

π 2

1 0 –1

π 2

π

u

x

0

–1 –

1

π 2

− , , but for the composite 2 2 function to exist the values of u must be a subset of [0, ], the domain of cos u. Hence the values of u for this composite function to exist, and hence the domain of cos, . must lie in the interval 0, 2 From the graphs it can be seen that the range of u =

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2 0 ≤ − sin−1 x ≤ 2 −1 − ≤ sin x ≤ 0 2 −1 ≤ x ≤ 0

121

0≤u≤

i.e.

∴ ∴ ∴

since u = − sin−1 x

Hence, the domain of cos(−sin−1 x) is [−1, 0]. The range of cos(−sin−1 x) is [0, 1].

Exercise 3C 1 Sketch the graphs of each of the following, stating clearly the implied domain and range each time: − 1) a y = tan−1 (x c y = 2 sin−1 x +

1 2

b y = cos−1 (x + 1)

e y = cos−1 (2x)

f y=

1 2

2 Evaluate each of the following: −1

1 √ − 3 −1 d cos 2 √ g tan−1 (− 3) a sin

−1

b sin

a sin(cos

0.5)

√ − 2 2

f −1 (−1)

f tan−1 1

b

v

i cos−1 (−1)

5 cos sin 6 5 tan−1 sin

2 2 −1 sin − sin 3 cos−1 tan − 4

−1

−1

e d cos(tan−1 1) 7 g cos−1 cos h 3 j cos−1 sin − k 3

3 , → R, f (x) = sin x. 4 Let f : 2 2 a Define f −1 , clearly stating its domain and its range. b Evaluate: 3 ii f i f 4 2 iv

c sin−1 0.5

e cos−1 0.5 √ 3 −1 h tan 3

3 Simplify: −1

2 sin−1 (3x) + 4

d y = 2 tan−1 (x) +

f −1 (0)

c tan sin

√ 2 − 2

f tan(cos−1 0.5) 11 −1 tan i tan 4

3 −1 l sin cos − 4

iii vi

7 6 −1 f (0.5)

f

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5 Given that the domain of sin x, cos x and tan x are restricted to − , , [0, ] and 2 2 − , respectively, define the implied domain and range of each of the following 2 2 where y is equal to: b sin x + c sin−1 (2x + 4) a sin−1 (2 − x) 4 e cos x − d sin 3x − f cos−1 (x + 1) 6 3 2 g cos−1 (x2 ) i tan−1 (x2 ) h cos 2x + 3 j tan 2x − l tan(x2 ) k tan−1 (2x + 1) 2 6 Simplify each of the following expressions, in an exact form: 5 7 c cos tan−1 24 b tan cos−1 13 a cos sin−1 45 40 f sin cos−1 32 e tan cos−1 12 d tan sin−1 41 i sin(tan−1 0.7) h cos sin−1 73 g sin(tan−1 (−2)) 5 , ∈ 0, and ∈ 0, 7 Let sin = 35 and sin = 13 2 2 a Find: i cos ii cos b Use a compound angle formula to show that: i sin−1

3 5

− sin−1

5 13

= sin−1

16 65

ii sin−1

3 5

5 + sin−1 13 = cos−1

33 65

8 Given that the domain of sin x and cos x are restricted to − , and [0, ] respectively, 2 2 define the implied domain and range of each of the following where y is equal to: a sin−1 (cos x) d sin(−cos−1 x) g cos(tan−1 x)

b cos(sin−1 x) e cos(2 sin−1 x) h sin(tan−1 x)

c cos−1 (sin 2x) f tan−1 (cos x)

1 9 a Use a compound angle formula to show that tan−1 3 − tan−1 = . 2 4 −1 −1 x − 1 = , x > –1. b Hence, show that tan x − tan x +1 4 and [0, ] 10 Given that the domain of sin x and cos x are restricted to − , 2 2 respectively, explain why each of these expressions cannot be evaluated: a cos[sin−1 (−0.5)]

3.4

b sin[cos−1 (−0.2)]

c cos[tan−1 (−1)]

Solution of equations In section 1.1, the solution of equations involving sine, cosine and tangent was discussed. In this section, equations that involve the reciprocal circular functions and the use of the double angle formulas are introduced. Equations that are not able to be solved by analytic methods are also considered.

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Example 14 Find x, such that sec x = 2, in the interval [0, 2].

y

Solution sec x = 2

∴

cos x =

1

1 2

0.5

The values of x which exist in [0, 2] are x = and 2 − 3 3 5 x = and 3 3

2π y = cos x

0

x

–1

Example 15

y

Find all the values of x for which cot x = −1. Solution The period of cot x is . In the interval [0, ] 3 the solution of cot x = −1 is x = 4 ∴ the solutions of the equation are x=

3 + n, n ∈ Z 4

Example 16 √ −2 3 = , for x ∈ [0, 2]. Find x, such that cosec 2x − 3 3 Solution

implies Let then

∴

= cosec 2x − 3 sin 2x − = 3

= 2x −

3

√ −2 3 3 √ −3 − 3 √ = 2 2 3

11 where ∈ − , 3 3

√ − 3 sin() = 2 4 5 10 11 , , , =− , 3 3 3 3 3

3π 4 0 –1

π 2

π

x

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∴

2x −

∴ ∴

4 5 10 11 =− , , , , 3 3 3 3 3 3 5 11 , 2, , 4 2x = 0, 3 3 5 11 x = 0, , , , 2 6 6

Example 17 Solve each of the following for x ∈ [0, 2]. x b cos x = sin a sin(4x) = sin(2x) 2 Solution sin(4x) = sin(2x)

a

∴

2 sin(2x) cos(2x) = sin(2x)

where

2x ∈ [0, 4]

implies sin(2x)(2 cos(2x) − 1) = 0

∴

sin(2x) = 0

∴

cos(2x) =

∴ ∴ i.e. b

or

2 cos(2x) − 1 = 0

1 2

5 7 11 , , , 3 3 3 3 3 5 7 11 , 2 or x = , , , x = 0, , , 2 2 6 6 6 6 5 7 3 11 x = 0, , , , , , , , 2 6 2 6 6 2 6

2x = 0, , 2, 3, 4 or 2x =

x 2 x x 2 x ∴ 1 − 2 sin = sin where ∈ [0, ] 2 2 2 x 2 x ∴ 2 sin + sin − 1 = 0 2 2 x Let a = sin 2 2 2a + a − 1 = 0 cos x = sin

∴

(2a − 1)(a + 1) = 0

∴

2a − 1 = 0

∴ ∴ ∴ ∴ ∴

a=

1 2

or a + 1 = 0 or a = −1

a= since a ∈ [0, 1] x sin = 12 2 5 x = , 2 6 6 5 x= , 3 3 1 2

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Example 18 Find the maximum and minimum values of: a sin2 (2x) + 2 sin(2x) + 2

b

1 sin (2x) + 2 sin(2x) + 2 2

Solution a Let

a = sin(2x)

Then sin2 (2x) + 2 sin(2x) + 2 = a 2 + 2a + 2 = (a + 1)2 + 1 = (sin(2x) + 1)2 + 1 y

Now −1 ≤ sin(2x) ≤ 1 and therefore the maximum value is 5 and the minimum value 1.

y = (sin (2x) + 1)2 + 1 5

2 –3π 2

–π

–π 2

π 2

0 y=

π

x 3π 2

1 (sin (2x) + 1)2 + 1

Using a graphics calculator Enter the function y = sin(2x)2 + 2sin(2x) + 2 into the Y= window. Check that the calculator is in RADIAN mode and press ZOOM 7 and then ZOOM 0 to display the graph. Use 3:minimum and 4:maximum from the CALCULATE menu to find the minimum and maximum values. b Note that sin2 (2x) + 2 sin(2x) + 2 ≥ 0 for all x. Thus its reciprocal also has this property. The local maximum for the original function yields a local minimum for the reciprocal. The local minimum for the original function yields a local maximum for the reciprocal. ∴ maximum value is 1 and the minimum value is 15

Using a graphics calculator In the Y= window, deselect but do not clear the function y = sin(2x)2 + 2 sin(2x) + 2. Enter the reciprocal function as Y2 = 1/ Y1 . Press ZOOM 0 to display the graph. Use 3:minimum and 4:maximum from the CALCULATE menu to find the minimum and maximum values.

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There are many equations involving circular functions which are not solvable by analytic techniques. A graphics calculator can be used for the solution of such equations. Example 19 Find the solution of the equation 2 sin(3x) = x correct to three decimal places. y

Solution

y=x

The graphs of y = 2 sin(3x) and y = x are plotted with a graphics calculator. The solutions are x = 0, x ≈ 0.893 and x ≈ −0.893

2 (0.8929..., 0.8929...) 0

–2 y = 2 sin (3x) (–0.8929..., –0.8929...)

Exercise 3D 1 Solve each of the following equations for x ∈ [0, 2]: = −2 b cosec x − a cosec x = −2 4 √ d cosec(2x) + 1 = 2 c 3 sec x = 2 3 √ = −1 f cot 2x − e cot x = − 3 3 2 Solve each of the following equations, giving solutions in the interval [0, 2]: √ √ − 3 a sin x = 0.5 b cos x = c tan x = 3 2 √ f cosec x = − 2 e sec x = 2 d cot x = −1 3 Find all the solutions of each of the following equations: √ √ 2 c cot x = 3 b sec x = 1 a sin x = 2 4 Solve each of the following, in the interval [−, ], giving the answers correct to two decimal places: a sec x = 2.5

b cosec x = −5

c cot x = 0.6

5 Solve each of the following equations for x ∈ [0, 2]: a c e g i

cos2 x − cos x sin x = 0 sin 2x = cos x cos 2x = cos x sec2 x + tan x = 1 cot x + 3 tan x = 5 cosec x

b d f h j

sin 2x = sin x sin 8x = cos 4x cos 2x = sin x tan x (1 + cot x) = 0 sin x + cos x = 1

x

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127

6 Find the maximum and minimum values of each of the following: 1 a 2 + sin b c sin2 + 4 2 + sin 1 d f cos2 + 2 cos + 6 e cos2 + 2 cos 2 sin + 4 7 Using a graphics calculator, find the coordinates of the points of intersection for the graphs of the following pairs of functions. (Give values correct to two decimal places.) a y = 2x c y=3−x

y = 3 sin(2x) y = cos x

b y=x d y=x

y = 2 sin(2x) y = tan x x ∈ [0, 2]

8 Let cos x = a, a = −1, x ∈ [0, 2]. If q is one of the solutions, find, in terms of q, the second solution. . Find, in terms of , two values of x in the range [0, 2] 9 Let sin = a where ∈ 0, 2 which satisfy each of the following equations: a sin x = −a

b cos x = a

, . Find, in terms of , two values of x in the range 2 [−, ] which satisfy each of the following equations:

10 Let sec = b where ∈ a sec x = −b

b cosec x = b

3 . Find in terms of , two values of x in the range 2 [0, 2] which satisfy each of the following equations:

11 Let tan = c where ∈ , a tan x = −c

b cot x = c

12 Solve, correct to two decimal places, the equation sin2 =

for ∈ [0, ].

13 Find the value of x, correct to two decimal places, such that tan−1 x = 4x − 5. 14 A curve on a light rail track is an arc of a circle of length 300 m and the straight line joining the two ends of the curve is 270 m long. a Show that, if the arc subtends an angle of 2 ◦ at the centre of the circle, is a solution ◦ . of the equation sin ◦ = 200 b Solve, correct to two decimal places, the equation for . 15 Solve, correct to two decimal places, the equation tan x =

1 for x ∈ [0, ]. x

16 The area of a segment of a circle is given by the equation A = 12 r 2 ( − sin ), where is the angle subtended at the centre of the circle. If the radius of the circle is 6 cm and the area of the segment is 18 cm2 , find the value of correct to two decimal places.

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17 Two tangents are drawn from a point so that the area of the shaded region is equal to the area of the remaining region of the circle. a Show that satisfies the equation tan = − . b Solve for , giving the answer correct to three decimal places.

A

O

X

2θ

B ∠AOB = 2θ

18 Two particles A and B move in a straight line. At time t their positions relative to a point O are given by xA = 0.5 sin t and xB = 0.25t2 + 0.05t Find the times at which their positions are the same, and give this position (distances are measured in cm and time in seconds). 19 A string is wound around a disc and a horizontal length of the string AB is 20 cm long. The radius of the disc is 10 cm. The string is then moved so that the end of the string, B , is moved to a point at the same level as O, the centre of the circle. BP is a tangent to the circle.

O

O

B'

θ

10 cm

P A

20 cm

B

A

B

− + tan = 2. 2 b Find the value of , correct to two decimal places, which satisfies this equation. a Show that satisfies the equation

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The reciprocal circular functions cosec x, sec x and cot x are defined as follows: 1 sin x = 0 sin x 1 cos x = 0 sec x = cos x cos x cot x = sin x = 0 sin x Useful symmetry properties for the reciprocal circular functions are: cosec x =

sec( − x) = −sec x cosec( − x) = cosec x cot( − x) = −cot x sec(2 − x) = sec x cosec(2 − x) = −cosec x cot(2 − x) = –cot x

sec( + x) = −sec x cosec( + x) = −cosec x cot( + x) = cot x sec(−x) = sec x cosec(−x) = −cosec x cot(−x) = −cot x

Useful complementary properties for the reciprocal circular functions are: − x = cosec x cot − x = tan x sec 2 2 − x = sec x tan − x = cot x cosec 2 2 The Pythagorean identities derived from Pythagoras’ theorem are: sin2 x + cos2 x = 1 cot2 x + 1 = cosec2 x tan2 x + 1 = sec2 x The compound angle formulas express circular functions of sums and differences of two angles (variables) in terms of circular functions of each of the angles: 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) = sin x cos y − cos x sin y sin(x + y) = sin x cos y + cos x sin y tan x − tan y tan(x − y) = 1 + tan x tan y tan x + tan y tan(x + y) = 1 − tan x tan y

Review

Chapter summary

129

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The double angle formulas are derived from the compound angle formulas: cos 2x = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x sin 2x = 2 sin x cos x 2 tan x tan 2x = 1 − tan2 x One-to-one inverse circular functions are defined as follows: cos−1 : [−1, 1] → R, cos−1 x = y sin−1 : [−1, 1] → R, sin−1 x = y where cos y = x, y ∈ [0, ] where sin y = x, y ∈ − , 2 2

y

y π 2

_1

y = sin–1 x

y = cos–1 x

x

0 _

π

1

π 2

_1

tan−1 : R → R, tan−1 x = y where tan y = x, y ∈ − , 2 2

y π 2 0

_π

2

y = tan–1 x

x

0

1

x

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Review

Multiple-choice questions 1 Which of the following is the graph of the function y = cos−1 (x)?

A

B

y

y

C

y

π

π

1 π 2

0 –1

π 2

x

π

D

E

y

1

π 2

0

0

2 If cos x = A 2 +

−2 3 √ 5 3

x

1

1

x

–1

0

x

1

y

π

–1

π 2

0

–1

–1

π

π 2

x

and 2 < x < 3 then the exact value of sin x is: B 2 −

√ 5 3

√ 5 3

D

√ − 5 3

E

5 9

D

√ √ − 2 2 , 3 3

E

−1 1 , 2 2

D

2 0, 5

E

, , the value of cot(x) is: √ √ √ C −3 11 D 3311 E −3311 4 The graph of the function y = 2 + sec(3x), for x ∈ − 6 , 7 , has stationary points at: 6 5 B x = 6, 2, 6 C x = 2 A x = 3, D x = 0, 3 , 2 , E x = 0, 2 3 3 3 Given that cos(x) = −1 and x ∈ 10 √ A 3√1011 B 3 11

C

2

5 If sin x = − 13 , the possible values of cos x are: A

√ √ −2 2 2 2 , 3 3

B

−2 2 , 3 3

C

−8 8 , 9 9

6 The maximum domain of cos−1 (1 − 5x) is given by:

B 1− , 15 C [−1, 1] A 0, 25 5 7 (1 + tan x)2 + (1 − tan x)2 is equal to: A 2 + tan x + 2 tan(2x) B 2 C −4 tan x

D 2 + tan(2x)

8 The number of solutions of cos2 (3x) = 14 , given that 0 ≤ x ≤ , is: A 1 B 2 C 3 D 6 tan(2) equals: 9 1 + sec(2) A tan(2) B tan(2) + 1 C tan + 1 D sin(2) 2

− 15 , 15

E 2 sec2 x E 9

E tan

< A < and 0 < B < 2 with sin A = t and cos B = t, cos(B + A) is equal to: √ √ 1 − t2 C 2t 2 − 1 D 1 − 2t 2 E −2t 1 − t 2 A 0 B

10 For

131

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Short-answer questions (technology-free) 4 1 If is an acute angle and cos = , find: 5 b sin 2 a cos 2 d cosec c tan 2 e cot 2 Solve each of the following equations for − < x ≤ 2: b cos x − 1 = cos 2x a sin 2x = sin x d sin2 x cos3 x = cos x c sin 2x = 2 cos x f 2 cos2 x − 3 cos x + 1 = 0 e sin2 x − 12 sin x − 12 = 0 3 Solve each of the following equations for , 0 ≤ ≤ 2, giving exact answers: b sec 2 = 2 a 2 − sin = cos2 + 7 sin2 1 d sec = 2 cos c 2 (5 cos − 3 sin ) = sin 4 Find the value of each of the following: exact 5 5 a sin b cosec − 3 3 7 5 c sec d cosec 3 6 3 e cot − f cot − 4 6 5 Given that tan = p, where is an acute angle, find each of the following in terms of p: b tan(− ) a tan(−) 3 d tan + c tan − 2 2 e tan(2 − ) √ 3 a sin−1 2

2 c cos−1 cos 3

1 −1 − e cos sin 2

6 Find:

−1 1 b cos cos 2

4 d cos−1 cos 3 f cos[tan−1 (−1)]

7 Sketch the graph of each of the following, stating the maximal domain and range each time: b y = sin−1 (3 − x) a y = 2 tan−1 x d y = −cos−1 (2 − x) c y = 3 cos−1 (2x + 1) −1 e y = 2 tan (1 − x)

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Review

Extended-response questions B2

1 A horizontal rod 1 m long is hinged at A at one end and rests on a support B at the other end. The rod can be rotated about A with the other end taking the two positions B1 and B2 which are x m and 2x m above AB, x < 0.5. Let ∠BAB1 = and ∠BAB2 = .

2x m

B1

a Find, in terms of x, each of the following: xm β ii cos i sin α iv sin iii tan A B vi tan v cos b Using the results of a, find: iii tan( − ) ii cos( − ) i sin( − ) vi cos(2) v sin(2) iv tan(2) c If x = 0.3, find the magnitude of ∠B2 AB1 and 2, correct to two decimal places. 2 a On the one set of axes sketch the graphs of: i y = cosec(x) x ∈ (0, ) ∪ (, 2) ii y = cot(x) x ∈ (0, ) ∪ (, 2) iii y = cosec(x) − cot(x) x ∈ (0, ) ∪ (, 2) b i Show that cosec x − cot x > 0 for all x ∈ (0, ) and hence that cosec x > cot x for all x ∈ (0, ). ii Show that cosec x − cot x < 0 for all x ∈ (, 2) and hence that cosec x < cot x for all x ∈ (, 2). x for x ∈ (0, 2) and c On separate axes sketch the graph of y = cot 2 y = cosec(x) + cot(x) for x ∈ (0, 2). d i Prove that cosec + cot = cot where sin = 0. 2 ii Use this result to find cot and cot . 8 12 = cosec2 to find the exact value of sin . iii Use the result 1 + cot2 8 8 8 e Use the result of d to show that cosec() + cosec(2) + cosec(4) can be expressed as the difference of two cotangents. 3 a ABCD is a rectangle with diagonal AC of length 10 units. i Find the area of the rectangle in terms of . ii Sketch the graph of R against where R is the area ofthe rectangle in square . units, ∈ 0, 2

B

C

10

θ A

D

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iii Find the maximum value of R (do not use calculus). iv Find the value of for which this maximum occurs. b ABCDEFGH is a cuboid. ∠GAC = , ∠CAD = , 2 AC = 10

F E

B

G

θ 2

H

C

i Show that the volume, V, of the cuboid θ is given by V = 1000 cos sin tan . A D 2 ii Find the values of a and b such that V = a sin2 + b sin4 . 2 2 2 iii Let p = sin , and express V as a quadratic in p. 2 iv Find the possible values of p for 0 < < . 2 v Sketch the graphs of V against and V against p with the help of a graphics calculator. vi Find the maximum volume of the cuboid and the values of p and for which this occurs. (Determine the maximum through the quadratic found in b iii.) c If for the cuboid ∠CAD = and ∠GAC = : i find V in terms of ii sketch the graph of V against iii discuss the relationship between V and using the graph of c ii. 4 ABCDE is a pentagon inscribed in a circle.

C AB = BC = CD = DE = 1 and ∠BOA = 2. B O is the centre of the circle. Let AE = p 2θ O sin 4 . a Show p = A sin b Express p √ as a function of cos . Let √ x = cos . 3 c i If p = 3 show √ that 8x − 4x − 3 = 0. 3 ii Show that is a solution to the equation and that it is the only real solution. 2 √ iii Find the value of for which p = 3. iv Find the radius of the circle. . d Using a graphics calculator sketch the graph of p against for ∈ 0, 4 e If A = E, find the value of . = 0. f i If AE = 1, show that 8x3 − 4x −1 1 √ . ii Hence show ( 5 + 1) = cos 4 5

D

E

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135

Review

i Prove that tan x + cot x = 2 cosec(2x) for sin 2x = 0. ii Solve the equation tan x = cot x for x. iii On the one set of axes, sketch the graphs of y = tan x, y = cot x and y = 2 cosec(2x) for x ∈ (0, 2). b i Prove that cot(2x) + tan x = cosec(2x) for sin(2x) = 0. ii Solve the equation cot(2x) = tan x. iii On the one set of axes, sketch the graphs of y = cot(2x), y = tan x and y = cosec(2x) for x ∈ [0, 2]. cos[(m − n)x] m, n ∈ Z . i Prove that cot(mx) + tan(nx) = c sin(mx) cos(nx) ii Hence show that cot(6x) + tan(3x) = cosec(6x).

5 a

6 Triangle ABE is isosceles with AB = BE and triangle ACE is isosceles with AC = AE. AE = 1 a b c d e

B 36°

i Find the magnitude of ∠BAE, ∠AEC and ∠ACE. ii Hence find the magnitude of ∠BAC. Show that BD = 1 + sin 18◦ . 1 + sin 18◦ Use triangle ABD to prove that cos 36◦ = . 1 + 2 sin 18◦ Hence show that 4 sin2 18◦ + 2 sin 18◦ − 1 = 0. Find sin 18◦ in exact form.

C D

7 VABCD is a right pyramid with diagonal length AC = 10. ABCD is a rectangle. a i If ∠CAD = ◦ and ∠VAX = ◦ show that the volume, V, of the pyramid is given by V =

V

500 2 sin 3

ii Sketch the graph of V against for ∈ (0, 90). iii Comment on the graph.

E

A

B

C X

A

θ° D

◦ b If the magnitude of angle CAD is ◦ and the magnitude of angle VAX is : 2 i show that the volume, V, of the pyramid is given by 1000 2 2 sin 1 − 2 sin V = 3 2 2 ii state the maximal domain of the function V(). iii Let a = sin2 and write V as a quadratic function with variable a. 2 iv Hence find the maximum value of V and the value of for which this occurs. v Sketch the graph of V against for the domain established in ii.

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8 VABCD is a right pyramid with diagonal length AC = 10. ABCD is a rectangle. ∠CAD = ◦ AY = BY V a If ∠VYX = ◦ find: i an expression for the volume of the pyramid in terms of B ii the maximum volume and C the value of for which this occurs. Y X ◦ b If ∠VYX = : 2 θ° 500 A D cos2 (1 − cos ) i show that V = 3 ii state the implied domain for the function. 500 2 a (1 − a). Use a graphics calculator to find the maximum c Let a = cos . Then V = 3 value of V and the values of a and for which this maximum occurs. 9 A camera is in a position x m from a point A. A body a metres in length is projected vertically upwards from A. When the body has moved b metres vertically up: θ a+b b − tan−1 a show = tan−1 x x b use the result of a to show ax tan = 2 xm x + ba + b2 √ ii x if a = 2(1 + 2) and b = 1 c If = find: i x in terms of a and b 4 √ d If a = 2(1 + 2), b = 1 and x = 1, find an approximate value of . e Using a graphics calculator, plot the graphs of against b and tan against b for constant values of a and x as indicated below: i a = 1, x = 5 ii a = 1, x = 10 iii a = 1, x = 20

am

bm

A

f Comment on these graphs. 10 Points A, B and C lie on a circle with centre O and radius 1 as shown. y a Give reasons why triangle ACD is similar to triangle ABC. b Give the coordinates of C in terms of circular functions applied to 2. c i Find CA in terms of from triangle ABC. ii Find CB in terms of from triangle ABC. d Use the results of b and c to show sin(2) = 2 sin cos . 2θ θ A O e Use the results of b and c to show 2 cos(2) = 2 cos − 1.

C x2 + y2 = 1

D

B

x

ch3.pdf

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