TRIGONOMETRIC Ratio & Identities Theory_E

MATHS

TRIGONOMETRIC RATIO & IDENTITIES The word 'trigonometry' is derived from the Greek words 'trigon' and ' metron' and it means 'measuring the sides and angles of a triangle'.

Angle : Angle is a measure of rotation of a given ray about its initial point. The original ray is called the initial side and the final position of the ray after rotation is called the terminal side of the angle. The point of rotation is called the vertex. If the direction of rotation is anticlockwise, the angle is said to be positive and if the direction of rotation is clockwise, then the angle is negative. B

Vertex O

Initial Side

Vertex O

ide lS ina m Ter

Te rm ina

Initial Side

A

A

lS ide

B (ii) Negative angle (clockwise measurement)

(i) Positive angle (anticlockwise measurement)

Systems For Measurement of Angles : An angle can be measured in the following systems. 1.

Sexagesimal System (British System) : In this system degree (°),

1 1 of a degree is called a minute () and of a minute is called a second (). 60 60

One right angle = 90°, 2.

1° = 60,

1 = 60

Centesimal System (French System) : In this system

grade (g), 

1 of a complete circular turn is called a 360

1 of a grade is called a minute 100

One right angle = 100g; 1g = 100 ;

and

1 of a complete circular turn is called a 400

1 of a minute is called a second 100

.

1 = 100

Note : The minutes and seconds in the Sexagesimal system are different with the minutes and seconds respectively in the Centesimal System. Symbols in both systems are also different. 3.

Circular System (Radian Measurement) The angle subtended by an arc of a circle whose length is equal to the radius of the circle at the centre of the circle is called a radian. In this system the unit of measurement is radian (c) As the circumference of a circle of radius 1 unit is 2 , therefore one complete revolution of the initial side subtends an angle of 2 radian. More generally, in a circle of radius r, an arc of length r will subtend an angle of 1 radian. It is well-known that equal arcs of a circle subtend equal angle at the centre. Since in a circle of radius r, an arc of length r  subtends an angle whose measure is 1 radian, an arc of length  will subtend an angle whose measure is r  radian. Thus, if in a circle of radius r, arc of length  subtends an angle  radian at the centre, we have  = r or = r .

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1

MATHS B

B 1

1

1

 O

1

2

1

O

A

O

 1

1

O

1

A

1

A



1

B  = –1 radian (ii)

 = 1 radian (i)



A

 = 2 radian (iii)

2 B  = –2 radian (iv)

Note : # If no symbol is mentioned while showing measurement of angle, then it is considered to be measured in radians. e.g.  = 15 implies 15 radian # Area of circular sector : Area =

1 2 r  sq. units 2

Relation between radian, degree and grade :  radian = 90° = 100g 2

Trigonometric Ratios for Acute Angles : Let a revolving ray OP starts from OA and revolves into the position OP, thus tracing out the angle AOP. In the revolving ray take any point P and draw PM perpendicular to the initial ray OA. In the right angle triangle MOP, OP is the hypotenuse, PM is the perpendicular, and OM is the base. The trigonometrical ratios, or functions, of the angle AOP are defined as follows : MP Perp. , i.e., , is called the Sine of the angle AOP; OP Hyp.

Base OM , i.e. Hyp. , is called the Cosine of the angle AOP; OP MP Perp. , i.e. , is called the Tangent of the angle AOP; OM Base

Base OM , i.e. , is called the Cotangent of the angle AOP; Perp. MP OP Hyp. , i.e. , is called the Secant of the angle AOP; OM Base

Hyp. OP , i.e. , is called the Cosecant of the angle AOP; Perp. MP The quantity by which the cosine falls short of unity i.e. 1 – cos AOP, is called the Versed Sine of AOP; also the quantity 1 – sin AOP, by which the sine falls short of unity, is called the Coversed Sine of AOP. It can be noted that the trigonometrical ratios are all real numbers. The names of these eight ratios are written, for brevity, sin AOP, cos AOP, tan AOP, cot AOP, cosec AOP, sec AOP, vers AOP, and covers AOP respectively

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MATHS Trigonometric ratios for angle   R : We will now extend the definition of trigonometric ratios to any angle in terms of radian measure and study them as trigonometric functions. (also called circular functions) Consider a unit circle (radius 1 unit) with centre at origin of the coordinate axes. Let at origin of the coordinate axes. Let P(a, b) be any point on the circle with angle AOP = x radian, i.e., length of arc AP = x We define cos x = a and sin x = b Since  OMP is a right triangle, we have OM2 + MP2 = OP2 or a2 + b2 =1 Thus, for every point on the unit circle, we have a2 + b2 = 1 or cos2x + sin2 x = 1 Since one complete revolution subtends an angle of 2 radian at the centre of the circle,  AOB =

 3 ,  AOC =  and AOD = . All angles which 2 2

 are called quadrantal angles. The coordinates of the 2 points A, B, C and D are, respectively, (1, 0), (0, 1), (–1, 0) and (0, –1). Therefore, for quadrantal angles, we have cos 0 = 1 sin 0 = 0, are integral multiples of

 =0 2 cos  = –1

 =1 2 sin  = 0

3 =0 2 cos 2 = 1

3 =–1 2 sin 2 = 0

cos

sin

cos

sin

Now if we take one complete revolution from the position OP, we again come back to same position OP. Thus, we also observe that if x increases (or decreases) by any integral multiple of 2 , the values of sine and cosine functions do not change. Thus, sin (2n + x) = sin x , n  Z, cos (2n + x) = cos x, n  Z Further, sin x = 0, if x = 0, ±  , ± 2 , ± 3 ....., i.e., when x is an integral multiple of  and cos x = 0, if

 3 5  ,± ,± , .....i.e., cos x vanishes when x is an odd multiple of . Thus 2 2 2 2 sin x = 0 implies x = n , where n is any integer x=±

 , where n is any integer.. 2 We now define other trigonometric functions in terms of sine and cosine functions : cos x = 0 implies x = (2n + 1)

cosec x =

sec x =

1 , x  n , sin x

where n is any integer.

1  , x  (2n + 1) , where n is any integer.. cos x 2

tan x =

sin x  , x  (2n + 1) , where n is any integer.. cos x 2

cot x =

cos x , x  n , where n is any integer.. sin x

We have shown that for all real x, sin2x + cos2x = 1 It follows that

1 + tan2x = sec2x

(Think ! )

1 + cot2x = cosec2x

(Think !)

 ; n  Z} 2 {x  n ; n  Z}

{x  (2n + 1)

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MATHS Sign of The Trigonometric Functions (i) (ii) (iii) (iv)

If is in the first quadrant then P(a, b) lies in the first quadrant. Therefore a > 0, b > 0 and hence the values of all the trigonometric functions are positive. If is in the quadrant then P(a, b) lies in the quadrant. Therefore a < 0, b > 0 and hence the values sin, cosec are positive and the remaining are negative. If is in the quadrant then P(a, b) lies in the quadrant. Therefore a < 0, b < 0 and hence the values of tan, cot are positive and the remaining are negative. If is in the V quadrant then P(a, b) lies in the IV quadrant. Therefore a > 0, b < 0 and hence the values of cos, sec are positive and the remaining are negative.

Values of trigonometric functions of certain popular angles are shown in the following table :

 6

0 sin

0 0 4

cos tan

1 0

1 1  4 2 3 2 1 3

 4

 3

2 1  4 2 1 2

3 3  4 2 1 2

1

3

 2 4 1 4 0 N.D.

N.D. implies not defined The values of cosec x, sec x and cot x are the reciprocal of the values of sin x, cosx and tan x, respectively.

Trigonometric Ratios of allied angles If  is any angle, then  

 3 ± ,  ± , ± , 2 ±  etc. are called allied angles. 2 2

Trigonometric Ratios of ( –  ) : Let be an angle in the standard position in the quadrant. Let its terminal side cuts the circle with centre ‘O’ and radius r in P (x, y). Let P(x, y) be the point of intersection of the terminal side of the angle – in the standard position with the circle. Now MOP = MOP (numerically) and P & Phave the same projection M in the x - axis  OMP OMPx = xand y = – y



(x , y) P

–y y sin (– ) = = = – sin . r r

x x cos (– ) = = = cos  r r tan (– ) =

y y =– = – tan  x x

cot (– ) =

x x = = – cot . y –y

sec (– =

r r = = sec. x x

cosec (– ) =

r

y x M

O

P (x, y) POM =  MOP = – 

r r = = – cosec  y –y

Similarly if is in the other quadrants then the above results can also be proved.

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MATHS 

Trigonometric Ratios of –  Let be an angle in the standard position in the quadrant. Let its terminal side cuts the circle with centre ‘O’ and radius r at P (x, y). Let P(x, y) be the point of intersection of the terminal side of the angle – with the circle. Let M and Mbe the projections of P and Prespectively in the x-axis. Since OMPOMP, x= – x, y= y 

sin (– ) =

y y = = sin . r r

cos (– ) =

x x = – = – cos. r r

tan (– ) =

y y = = – tan . x –x

cot (– ) =

x x = – = – cot . y y

sec (– ) =

r r = = – sec . x –x

r r cosec (– ) = y = y = cosec .



  Trigonometric Ratios of  –   : 2 

Similarly we can easily prove the following results.



  sin  –   = cos  , 2 

  tan  –   = cot  2 

  cosec  –   = sec  2 

  cos  –   = sin , 2 

  cot  –   = tan , 2 

  sec  –   = cosec  2 

  Trigonometric Ratios of     2 

Similarly we can easily prove the following results.



  sin     = cos , 2  

  tan     =– cot , 2  

  cosec     = sec  2  

  cos     = – sin , 2  

  cot     = – tan , 2  

  sec     = – cosec  2  

Trigonometric Ratios of ( + ) Similarly we can easily prove the following results. sin (+ ) = – sin , tan (+ ) = tan  cos (+ ) = – cos , cot (+ ) = cot ,



cosec (+ ) = – cosec  sec (+ ) = – sec 

 3  –  Trigonometric Ratios of   2 

Similarly we can easily prove the following results.

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MATHS



 3  –   = – cos , sin   2 

 3  –   = cot , tan   2 

 3  –   = – sec  cosec   2 

 3     = – sin , cos   2 

 3  –   = tan , cot   2 

 3  –   = – cosec , sec   2 

 3    Trigonometric Ratios of  2  

Similarly we can easily prove the following results.  3     = – cos , sin   2 

 3  cos     = sin , 2  

 3     = – cot , tan   2 

 3     = – tan , cot   2 

 3     = cosec , sec   2 

 3     = – sec  cosec   2 

Think, and fill up the blank blocks in following table.

0 sin

0

cos

1

tan

0

 6 1 2 3 2 1

 4 1

2

 3 3 2 1 2

1

3

2 1

3

 2

2 3

5 6

7 6



4 3

3 2

5 3

11 6

2

1 0 N.D.

Trigonometric functions: (i)

Domain : x  R Range : y  [–1, 1]

y = sin x

y 1

3 2

–2

0 –





2

2

3 2



x

2

–1

(ii)

Domain : x  R Range : y  [ – 1, 1]

y = cos x

y 1

–2

3 2

–



0

2

 2



3 2

2

x

–1

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MATHS (iii)

  Domain : x  R – ( 2n  1)  , n  2  Range : y  R

y = tan x

y  –

2

 0

2



– 32

(iv)

3 2

Domain : x  R – {n}, Range : y  R

y = cot x

x 0

n 

y

–

 2

(v)

y = cosec x

(vi)

y = sec x

0

 2



3 2

2

x

Domain : x  R – {n}, n  Range : y  (,  1]  [1, )

  Domain : x R – ( 2n  1)  , n  2 

Range : y  (,  1]  [1, )

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MATHS Trigonometric functions of sum or difference of two angles: (a) sin (A ± B) = sinA cosB ± cosA sinB (b) cos (A ± B) = cosA cosB  sinA sinB (c) sin²A  sin²B = cos²B  cos²A = sin (A+B). sin (A B) (d) cos²A  sin²B = cos²B  sin²A = cos (A+B). cos (A  B) tan A  tan B (e) tan (A ± B) = 1  tan A tan B

cot A cot B  1 (f) cot (A ± B) = cot B  cot A (g) sin (A + B + C) = sin A cos B cos C + sin B cos A cos C + sin C cos A cos B – sin A sin B sin C (h) cos (A + B + C) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C tan A  tan B  tanC tan A tan B tan C (i) tan (A + B + C) = 1  tan A tan B  tan B tan C tan C tan A . (j) tan (1 + 2 + 3 + ....... + n) =

S1  S 3  S 5  ...... 1  S 2  S 4  .......

where Si denotes sum of product of tangent of angles taken i at a time Example # 1 : Prove that

Solution :

(i)

sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B) = cos (A – B)

(ii)

   3     = –1 tan     tan  4 4    

(i)

(ii)

Clearly sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B) = sin (45º + A + 45º – B) = sin (90º + A – B) = cos (A – B)    3  1  tan  1  tan    = tan     × tan  × =–1 4 4 1  tan  1  tan     

Self practice problems : 3 5 , cos  = , then find sin ( + ) 5 13

(1)

If sin  =

(2)

Find the value of sin 105º

(3)

Prove that 1 + tan A tan

Answers :

(1)

–

A A = tan A cot – 1 = sec A 2 2 33 63 , 65 65

(2)

3 1 2 2

Transformation formulae : (i)

sin(A+B) + sin(A  B) = 2 sinA cosB

(a)

sinC + sinD = 2 sin

CD CD cos 2 2

(ii)

sin(A+B)  sin(A  B) = 2 cosA sinB

(b)

sinC  sinD = 2 cos

CD CD sin 2 2

(iii)

cos(A+B) + cos(A  B) = 2 cosA cosB

(c)

cosC + cosD = 2 cos

(iv)

cos(A  B)  cos(A+B) = 2 sinA sinB

(d)

cosC  cosD = 2 sin

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CD CD cos 2 2

CD DC sin 2 2

8

MATHS Example # 2 : Prove that sin 5A + sin 3A = 2sin 4A cos A Solution :

L.H.S. sin 5A + sin 3A = 2sin 4A cos A [ sin C + sin D = 2 sin

= R.H.S.

CD C D cos ] 2 2

Example # 3 : Find the value of 2 sin 3 cos  – sin 4 – sin 2 Solution : 2 sin 3 cos  – sin 4 – sin 2 = 2 sin 3 cos  – [2 sin 3 cos  ] = 0 Example # 4 : Prove that

Solution :

(i)

sin 8 cos   sin 6 cos 3 = tan 2 cos 2 cos   sin 3 sin 4

(ii)

tan 5  tan 3 = 4 cos 2 cos 4 tan 5  tan 3

(i)

2 sin 8 cos   2 sin 6 cos 3 2 cos 2 cos   2 sin 3 sin 4

=

(ii)

sin 9  sin 7  sin 9  sin 3 2 sin 2 cos 5 = cos 3  cos   cos   cos 7 2 cos 5 cos 2

= tan 2

tan 5  tan 3 sin 5 cos 3  sin 3 cos 5 sin 8 = = = 4 cos2 cos 4 tan 5  tan 3 sin 5 cos 3  sin 3 cos 5 sin 2

Self practice problems (4)

Prove that

13 x 3x sin 2 2

(i)

cos 8x – cos 5x = – 2 sin

(iii)

sin A  sin 3 A  sin 5 A  sin 7 A = tan 4A cos A  cos 3 A  cos 5 A  cos 7 A

(iv)

sin A  2 sin 3 A  sin 5 A sin 3 A = sin 3 A  2 sin 5 A  sin 7 A sin 5 A

(v)

sin A  sin 5 A  sin 9 A  sin 13 A = cot 4A cos A  cos 5 A  cos 9 A  cos 13 A

(ii)

sin A  sin 2A A = cot cos A  cos 2A 2

 7 3 11 sin + sin sin = sin 2 sin 5 2 2 2 2

(5)

Prove that sin

(6)

Prove that cos A sin (B – C) + cos B sin (C – A) + cos C sin (A – B) = 0

(7)

Prove that 2 cos

 9 3 5 cos + cos + cos =0 13 13 13 13

Multiple and sub-multiple angles : (a)

sin 2A = 2 sinA cosA Note : sin  = 2 sin

  cos etc. 2 2

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MATHS (b)

cos 2A = cos²A  sin²A = 2cos²A  1 = 1  2 sin²A Note : 2 cos²

 = 1 + cos , 2

2 sin²

 = 1  cos . 2

2 tan A

(c)

tan 2A =

(d)

sin 2A =

(e) (f)

sin 3A = 3 sinA  4 sin3A cos 3A = 4 cos 3A  3 cosA

(g)

tan 3A =

1  tan 2 A 2 tan 2 Note : tan  = 1  tan 2 2 2 tan A 1  tan 2 A

,

cos 2A =

1tan 2 A 1 tan 2 A

3 tan A  tan 3 A 1  3 tan 2 A

Example # 5 : Prove that

Solution :

(i)

sin 2A = tan A 1  cos 2A

(ii)

tan A + cot A = 2 cosec 2 A

(iii)

1  cos A  cos B  cos( A  B) A B = tan cot 1  cos A  cos B  cos( A  B) 2 2

(i)

L.H.S.

(ii)

 1  tan 2 A  2 1  tan 2 A   L.H.S. tan A + cot A = = 2  2 tan A  = = 2 cosec 2 A sin 2A tan A  

(iii)

1  cos A  cos B  cos( A  B) L.H.S. = 1  cos A  cos B  cos( A  B)

2 sin A cos A sin 2A = = tan A 1  cos 2A 2 cos 2 A

A A A   2 sin sin  B  2 2 2   A A  2 A 2 cos  2 cos cos  B  2 2 2   2 sin 2

 A B   B  A A  cos   sin  sin  B    2 sin A  A A B 2 2  2  2   = tan   = tan = tan cot 2  2  2 2 A B B  A A   2 sin 2 sin 2    cos 2  cos 2  B         Self practice problems sin   sin 2 = tan  1  cos   cos 2

(8)

Prove that

(9)

Prove that sin 20º sin 40º sin 60º sin 80º =

(10)

Prove that tan 3A tan 2A tan A = tan 3A – tan 2A – tan A

(11)

A  Prove that tan  45 º   = sec A + tan A 2 

3 16

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MATHS Important trigonometric ratios of standard angles : (a)

sin n  = 0

(b)

sin 15° or sin

;

cos n  = (1)n ; tan n  = 0,

3 1  5 = = cos 75° or cos 12 12 2 2

cos 15° or cos

3 1  5 = = sin 75° or sin 12 12 2 2

3 1 tan 15° =

(c)

sin

3 1

where n   ;

;

3 1 = 2 3 = cot 75° ; tan 75° =

3 1

= 2 3 = cot 15°

 5 1 or sin 18° = = cos 72° 10 4

cos 36° or cos

 = 5

5 1 = sin 54° 4

Conditional identities: If A + B + C =  then : (i)

sin2A + sin2B + sin2C = 4 sinA sinB sinC

(ii)

sinA + sinB + sinC = 4 cos

(iii)

cos 2 A + cos 2 B + cos 2 C =  1  4 cos A cos B cos C

(iv)

cos A + cos B + cos C = 1 + 4 sin

(v)

tanA + tanB + tanC = tanA tanB tanC

(vi)

tan

A B B C C A tan + tan tan + tan tan =1 2 2 2 2 2 2

(vii)

cot

A B C A B C + cot + cot = cot . cot . cot 2 2 2 2 2 2

(viii)

cot A cot B + cot B cot C + cot C cot A = 1  A+B+ C= then tan A tan B + tan B tan C + tan C tan A = 1 2

(ix)

A B C cos cos 2 2 2

A B C sin sin 2 2 2

Example # 6 : If A + B + C = 180°, Prove that, sin2A + sin2B + sin2C = 2 + 2cosA cosB cosC. Solution :

 

Let S = sin2A + sin2B + sin2C so that 2S = 2sin2A + 1 – cos2B +1 – cos2C = 2 sin2A + 2 – 2cos(B + C) cos(B – C) = 2 – 2 cos 2A + 2 – 2cos(B + C) cos(B – C) S = 2 + cosA [cos(B – C) + cos(B+ C)] since cosA = – cos(B+C) S = 2 + 2 cos A cos B cos C

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MATHS Example # 7 : If x + y + z = xyz, Prove that Solution :



2x 1 x

2

+

2y 1 y

2

+

Put x = tanA, y = tanB and z = tanC, so that we have tanA + tanB + tanC = tanA tanB tanC  Hence L.H.S. 2y

2x 1 x

2

+

1 y

2

+

2z 1 z

2

=

2 tan A 1  tan A 2

= tan2A + tan2B + tan2C = tan2A tan2B tan2C =

+

2z 1 z

=

2

2x 1 x

.

2

2y 1 y

2

.

2z 1 z2

.

A + B + C = n where n  

2 tan B 1  tan B 2

+

2 tan C 1  tan2 C

.

[ A + B + C = n ] 2y

2x 1 x

2

.

1 y2

.

2z 1 z2

Self practice problem (12)

(13)

If A + B + C = 180°, prove that (i)

sin(B + 2C) + sin(C + 2A) + sin(A + 2B) = 4sin

(ii)

sin 2A  sin 2B  sin 2C A B C = 8 sin sin sin . sin A  sin B  sin C 2 2 2

BC CA A B sin sin 2 2 2

If A + B + C = 2S, prove that (i) sin(S – A) sin(S – B) + sinS sin (S – C) = sinA sinB. (ii)

sin(S – A) + sin (S – B) + sin(S – C) – sin S = 4sin

A B C sin sin . 2 2 2

Sine and Cosine series: n

(i)

sin 2 n1      sin  + sin (+) + sin ( + 2 ) +...... + sin   (n  1)  = sin  2   sin 2

(ii)

sin 2 n 1      cos  + cos (+) + cos ( + 2 ) +.... + cos   (n  1)  =  cos  2  sin 2

n

where :   2m, m   Example # 8 : Find the summation of the following series

Solution :

(i)

cos

2 4 6 + cos + cos 7 7 7

(ii)

cos

 2 3 4 5 6 + cos + cos + cos + cos + cos 7 7 7 7 7 7

(iii)

cos

 3 5 7 9 + cos + cos + cos + cos 11 11 11 11 11

(i)

 2 6      3 7 7   cos sin 2 4 6 2 7 cos + cos + cos =  7 7 7 sin 7

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MATHS cos =

(ii)

cos

4 3 3 3 6 sin  cos sin sin 7 7 7 7 7 =– 1 = =–    2 sin sin 2 sin 7 7 7

 2 3 4 5 6 + cos + cos + cos + cos + cos 7 7 7 7 7 7

  6     6 cos  7 7  sin 14  2   6   cos sin   2 14 = = =0   sin sin 14 14

(iii)

cos

 3 5 7 9 + cos + cos + cos + cos 11 11 11 11 11

cos =

10  5 10  sin sin 22 11 11 = 1 =   2 sin 2 sin 11 11

Self practice problem Find sum of the following series :

 3 5 + cos + cos + ...... up to n terms. 2n  1 2n  1 2n  1

(14)

cos

(15)

sin2 + sin3 + sin4 + ..... + sin n, where (n + 2) = 2

Answers :

1 2

(14)

(15)

0.

Product series of cosine angles cos  . cos 2 . cos22 . cos23 ...... cos2n–1 =

sin 2n  2n sin 

Range of trigonometric expression: E = a sin  + b cos  

E=

a b 2

2

b Let



a b 2

2

  a b sin   cos    a 2  b 2  a2  b2

= sin  &

a a2  b2

= cos 

E = a 2  b 2 sin ( + ), where tan  =

b a

Hence for any real value of ,

 a2  b2  E 

a2  b2

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MATHS Example # 9 : Find maximum and minimum values of following : (i) 3sinx + 4cosx (ii) 1 + 2sinx + 3cos 2x Solution : (i) We know – (ii)

3 2  4 2  3sinx + 4cosx 

32  42

– 5  3sinx + 4cosx  5 1 + 2sinx + 3cos 2x = – 3sin2x + 2sinx + 4 2 sin x   2 1   + 4 = – 3  sin x   = – 3  sin x  3  3  

2

+

13 3

2

Now

1  16 0   sin x    3 9 



1 16  –  – 3  sin x    0 3 3 

2

2

1  13 13 – 1  – 3  sin x   +  3 3 3  Self practice problems (16)

Find maximum and minimum values of following (i)

3 + (sinx – 2)2

(ii)

10cos 2x – 6sinx cosx + 2sin2x

(iii)

  cos + 3 2 sin     + 6 4 

Answers :

(i) (iii)

max = 12, min = 4. max = 11, min = 1

(ii)

max = 11, min = 1.

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