Best Approach
Compound Angles
Trigonometry - I
Exercise I
Exercise II
Exercise III
Exercise IV
Trigonometric Ph. 1
KEY
CONCEPTS (COMPOUND ANGLES)
1.
BASIC
(a)
(b)
(c)
TRIGONOMETRIC IDENTITIES :
sin2 + cos2 = 1
;
1  sin   1;
1  cos   1    R
2
2
sec   tan  = 1
;
sec   1    R
cosec2  cot2 = 1 ;
cosec   1    R
2.
IMPORTANT T RATIOS:
(a)
sin n  = 0
;
cos n  = (-1)n ;
tan n  = 0
where n  I
( 2n  1)
( 2n  1)
(b)
sin
= (1)n &
cos
= 0 where n  I
2
2
5

31
(c)
sin 15° or sin
=
= cos 75° or cos
;
12
12
2 2
cos 15° or cos
tan 15° =
31

=
= sin 75° or
12
2 2
3 1
= 2  3 = cot 75° ; tan 75° =
3 1
(d)
sin

=
8
(e)
sin

or sin 18° =
10
2 2
;
2
cos

=
8
51
4
5
;
12
3 1
= 2  3 = cot 15°
3 1
sin

2 2
; tan =
8
2
&
21 ;
cos 36° or cos

=
5
tan
3
=
8
2 1
51
4
3.
TRIGONOMETRIC FUNCTIONS OF ALLIED ANGLES :
If  is any angle, then , 90 ± , 180 ± , 270 ± , 360 ±  etc. are called ALLIED ANGLES.
(a)
sin ( ) =  sin 
; cos ( ) = cos 
(b)
sin (90°- ) = cos 
; cos (90°  ) = sin 
(c)
sin (90°+ ) = cos  ; cos (90°+ ) =  sin 
(d)
sin (180° ) = sin  ; cos (180° ) =  cos 
(e)
sin (180°+ ) =  sin  ; cos (180°+ ) =  cos 
(f)
sin (270° ) =  cos ; cos (270° ) =  sin 
(g)
sin (270°+ ) =  cos ; cos (270°+ ) = sin 
4.
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)
(e)
5.
6.
tan (A ± B) = tan A  tan B
1  tan A tan B
(f)
cot (A ± B) =
cot A cot B  1
cot B  cot A
FACTORISATION OF THE SUM OR DIFFERENCE OF TWO SINES OR COSINES :
CD
C D
CD
C D
(a) sinC + sinD = 2 sin
cos
(b) sinC  sinD = 2 cos
sin
2
2
2
2
C D
C D
CD
CD
(c) cosC + cosD = 2 cos
cos
(d) cosC  cosD =  2 sin
sin
2
2
2
2
TRANSFORMATION OF PRODUCTS INTO SUM OR DIFFERENCE OF SINES & COSINES :
(a) 2 sinA cosB = sin(A+B) + sin(AB)
(b) 2 cosA sinB = sin(A+B)  sin(AB)
(c) 2 cosA cosB = cos(A+B) + cos(AB)
(d) 2 sinA sinB = cos(AB)  cos(A+B)
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Trigonometric Ph. 1
7.
MULTIPLE ANGLES AND HALF ANGLES :

2
sin 2A = 2 sinA cosA ; sin  = 2 sin cos
(b)
cos2A = cos2A  sin2A = 2cos2A  1 = 1  2 sin2A ;
cos  = cos2
8.
9.

2
 sin²

2
= 2cos2

2

2
 1 = 1  2sin2 .
1  cos 2A
2 cos2A = 1 + cos 2A , 2sin2A = 1  cos 2A ; tan2A =
1  cos 2A


2
2
2 cos
= 1 + cos  , 2 sin
= 1  cos .
2
2
2tan( 2)
2tanA
(c)
tan 2A =
; tan  =
2
1 tan 2 ( 2)
1 tan A
2tanA
1tan 2 A
(d)
sin 2A =
,
cos
2A
=
(e) sin 3A = 3 sinA  4 sin3A
2
2
1 tan A
1 tan A
3tanAtan 3 A
3
(f)
cos 3A = 4 cos A  3 cosA
(g)
tan 3A =
13tan 2 A
THREE ANGLES:
tanA  tan B tanCtan A tan BtanC
(a)
tan (A+B+C) =
1 tan A tan B tan BtanC tanC tanA
NOTE IF :
(i) A+B+C =  then tanA + tanB + tanC = tanA tanB tanC

(ii) A+B+C =
then tanA tanB + tanB tanC + tanC tanA = 1
2
(b)
If A + B + C =  then :
(i) sin2A + sin2B + sin2C = 4 sinA sinB sinC
C
A
B
(ii)
sinA + sinB + sinC = 4 cos cos cos
2
2
2
MAXIMUM & MINIMUM VALUES OF TRIGONOMETRIC FUNCTIONS:
(a)
Min. value of a2tan2 + b2cot2 = 2ab where  R
(b)
Max. and Min. value of acos + bsin are a 2  b 2 and – a 2  b 2
(c)
If f() = acos() + bcos() where a, b,  and  are known quantities then
(d)
(e)
10.

2
(a)
– a 2  b 2  2ab cos(  ) < f() < a 2  b 2  2ab cos(  )
If A, B, C are the angles of a triangle then maximum value of
sinA + sinB + sinC and sinA sinB sinC occurs when A = B = C = 600
In case a quadratic in sin or cos is given then the maximum or minimum values can be
interpreted by making a perfect square.
Sum of sines or cosines of n angles,


sin  + sin (+) + sin ( + 2 ) + ...... + sin   n  1  =


n
2
sin 2
sin
cos  + cos (+) + cos ( + 2 ) + ...... + cos   n  1  =
 n1 

sin   
2 

n
2

sin 2
sin
 n1 

cos   
2 

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Trigonometric Ph. 1
1.
SOLVED EXAMPLES
Find the radian measures corresponding to the following degree measures –47°30'.

Sol.
 1
 95 
Note that – 47°30' =   47     
 2
 2 


180° =  radian


1° =
radian 
180

Radian measure of an angle of – 47°30' is 

19
  95 
 95 
  
   radian =  72 radian
 2  180  2 
19
72
2.
Find the degree measures corresponding to the following radian measures (Use =
Sol.
 radian = 180°

5
 80 
 1 radian = 
radian =
 
3
  
3.
Sol.
22 5
)
.
7
3

 180 5 
  = 300°

3 
 
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Number of revolutions made in 60 second = 360
( 1 minute = 60 seconds)
360
=6
60
When the wheel makes one revolution, it turns through 360° or 2radians.
Number of radians turned by the wheel in one second = 6 × 2= 12.
 Number of revolutions made in 1 second =
4.
Sol.
Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an
arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm
In this case the radius = 75 cm
If the angle turned is radian, then
(i)
=
 10cm 2


r 75cm 15
Angle through which the pendulum swings = radian =
(ii)
 15cm 1

 
r 75cm 5
Angle through which the pendulum swings =  radian =
(iii)
2
radian.
15
1
radian.
5
 21cm 7

 
r 75cm 25
 Angle through which the pendulum swings =  radian =
7
radian.
25
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Trigonometric Ph. 1
O
r
5.
Sol.
Evaluate : cosec (– 1410°)
cosec (–1410°) = – cosec 1410°
= – cosec (360° × 3 + 330°) = – cosec 30°
= – cosec (360° – 30°) = – (cosec 30°)
= cosec 30° = 2
(cosec (–) = – cosec )
(cosec (360° n + ) = cosec n  1)
6.
 11 
Evaluate : sin  

 3 
Sol.
11
 11 
sin  
 = –sin
3
 3 
(sin (–) = – sin )
5
= – sin  2  5  = – sin
3
3 

(sin (2+) = sin )




= – sin  2   =    sin 
3
3


(sin (2–) = – sin )
= sin
7.
Sol.

3

3
2
Prove that cot2
L.H.S =
cot2


5
+ cosec + 3tan2 = 6
6
6
6


5
+ coses
+ 3 tan2   =
6
6
6
   1 
3 + cosec      3 
6  3

 
2

+1
6
= 3 + 2 + 1 = 6 = R.H.S
(cosec (–) = cosec 
= 3 + cosec
8.
Find the value of : (i) sin 75° (ii) tan 15°
Sol.
(i) sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin30° =
Also, we can write sin 75° =
2
1
3 1 1
3 1
.

. 
2 2
2 2 2 2
3 1
2
6 2


4
2 2
2
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Trigonometric Ph. 1
1
tan45 – tan 30
3  3 1
(ii) tan 15° = tan (45° – 30°) =
=
1  tan 45 tan 30 1  1. 1
3 1
3
1
2
Also, we can write tan 15° =
3 1
3 1


3 1
3 1
 3  1  3  1  2
 3  1 3 1
3
2
9.


tan   x 
2
4
   1  tan x 
Prove that




1  tan x 
tan   x  
4

Sol.


tan   x  1  tan x
2
4
  1  tan x   1  tan x 
L.H.S.=

 = R.H.S

 1  tan x  1  tan x 
tan   x 
4
 1  tan x

42 3
= 2 3 .
2
10.
Prove that : sin2 6x – sin2 4x = sin 2x sin 10x.
L.H.S. = sin2 6x – sin2 4x = sin (6x + 4x) sin (6x – 4x) ( sin2A – sin2B = sin (A + B) sin (A –B))
= sin 10 x sin 2x = R.H.S.
11.
Prove that :
Sol.
5x  3x
5x  3x
2sin
cos
sin 5x  sin 3x
2
2
L.H.S. =
=
5x  3x
5x  3x
cos5x  cos3x
2cos
cos
2
2
=
sin 5x  sin 3x
= tan4x
cos5x  cos3x
(Using "C, D" formulae)
sin 4x
= tan 4x = R.H.S.
cos 4x
sin x  sin 3x
= 2sin x
sin 2 x  cos 2 x
12.
Prove that :
Sol.
2  sin 3x  sin x  2cos
sin x  sin 3x

L.H.S =
=
sin 2 x  cos 2 x 2  cos 2 x  sin 2 x 
3x  x
3x  x
sin
2
2
cos 2x
(cos2A – sin2A = cos 2A)
= 2 sin x = R.H.S
13.
Sol.
Prove that : cos 4x = 1 – 8 sin2 x cos2x.
L.H.S = cos 4x = 2 cos2 2x – 1
= 2{2cos2 x – 1}2 – 1 = 2{4 cos4 x – 4 cos2 x + 1} – 1
(cos 2 = 2 cos2– 1)
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Trigonometric Ph. 1
= 8 cos4 x – 8 cos2 x + 2 – 1 = 8 cos4 x – 8 cos2x + 1
= 8 cos2 x (cos2 x – 1) + 1
= 8 cos2 x (– sin2 x) + 1
(sin2x + cos2x = 1 cos2x – 1 = – sin2x)
= 1 – 8 sin2 x cos2x
= R.H.S
Alternatively, L.H.S
= cos 4x = cos {2 (2x)}
= 1 – 2 sin2 2x = 1 – 2 (2 sin x cos x)2
= 1 – 8 sin2 x cos2 x
= R.H.S.
14.
Sol.
Prove that (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
L.H.S = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
3x  x
3x  x 
3x  x
3x  x 


cos
sin
=  2sin
 sin x   2sin
 cos x
2
2 
2
2 


(Using : "C, D" formulae)
= 2 sin 2x cos x sin x – 2 sin 2x sin x cos x
= 0 = R.H.S
15.
Prove that
Sol.
L.H.S. =
 sin 7x  sin 5x    sin 9x  sin 3x 
= tan 6x.
 cos 7x  cos5x    cos9x  cos3x 
 sin 7x  sin 5x    sin 9x  sin 3x 
 cos 7x  cos5x    cos9x  cos3x 
7x  5x
7x  5x
9x  3x
9x  3x
cos
 2sin
cos
2
2
2
2
=
7x  5x
7x  5x
9x  3x
9x  3x
2cos
cos
 2cos
cos
2
2
2
2
2sin
16.
Sol.
=
2sin 6x cos x  2sin 6x cos3x
2 cos 6x cos x  2cos 6x cos3x
=
2sin 6x  cos x  cos3x 
2 cos 6x  cos x  cos3x 
=
sin 6x
= tan 6x = R.H.S.
cos 6x
(Using "C, D" formulae)
x
3x
cos
2
2
L.H.S = sin 3x + sin 2x – sin x = (sin 3x – sin x) + sin 2x
Prove that sin 3x + sin 2x – sin x = 4 sin x cos
3x  x
3x  x
sin
+ sin 2x
2
2
= 2 cos 2x sin x + 2 sin x cos x = 2 sin x {cos 2x + cos x)
= 2 cos
2x  x
2x  x 

cos
= 2 sin x 2cos

2
2 

(Using "C, D" formulae)
(Using "C, D" formulae once again)
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= 4 sin x cos
3x
x
cos
2
2
= R.H.S
17.
Sol.
(i) sin 1 > cos 1
(ii) sin1° > cos1°
Which of the above is / are true ?
sin1° = sin 1°
cos 1° = sin 99°
sin1 sin 57° = sin 57°
cos 1 cos 57° = sin33°
In I quadrant sine is increasing.
sin1° < sin 33° < sin 57° < sin99°
sin 1° < cos 1 < sin 1 < cos 1°
If sin 2A = sin 2 B, then the value of
Sol.
tan  A  B  sin  A  B  .cos  A  B 

tan  A  B  sin(A  B).cos(A  B)
1
 sin  2A   sin  2B  
= 2
1
 sin  2A   sin  2B 
2
19.
Sol.
[sin A. cos B =
sin12 sin 48
is :
1  2sin 54
sin12 sin 48
=
1  2sin 54


1
 2sin12.sin 48 
2
1  2sin 54
1
 cos  60  cos  36 
2
1  2cos 36
[–2sin C. sin D = cos (C + D) – cos (C – D)]
[sin= cos (90° –)]
11

   cos36 
2 2
 1
=
=
1

4
2   cos36 
2

20.
Sol.
1
(sin (A + B) + sin (A –B))]
2
 sin 2B  sin 2B   1

 sin 2B  sin 2B   1
The value of
=
(iv) sin1° > cos1
tan  A  B 
is :
tan  A  B 
18.
=
(iii) sin 1 > cos1°
If 13sec – 5tan = 13 then the value of 13tan – 5sec can be
Given 13sec – 5tan = 13 ; Let 13 tan – 5 sec = x
On squaring and subtracting the equations, we get
(13 sec)2 + (5tan)2 – 130 sec . tan = 132
(13tan)2 + (5sec)2 – 130 sec . tan = x2
132 – 52 = 132 – x2
x = 5
EXERCISE–I
1.
Prove that (1 + cot  – cosec )(1 + tan + sec ) = 2.
2.



sec   tan  ; if    

1  sin 
2
2 .
Prove that


3
1  sin  
 sec   tan  ;if   

2
2
3.
4.
Prove that 2
sec2
–
sec4
– 2
cosec2
+
cosec4
1  tan 8 
=
.
tan 4 
3
If cos( – ) + cos( – ) + cos( – ) =  .
2
prove that cos + cos + cos = sin+ sin + sin = 0.
1 1
 .
x y
5.
If tan A – tan B = x and cot A – cot B = y. Prove that cot (A – B) =
6.
Prove that
sin(A  B  C)
= tan A + tan B + tan C – tan A tan B tan C
cos A cos Bcos C
7.
Prove that
sin 5A  sin 3A
= tan A
cos5A  cos 3A
8.
Prove that
cos 4x  cos3x  cos 2x
= cot 3x
sin 4x  sin 3x  sin 2x
9.
Prove that
cos8A cos5A  cos12A cos9A
= tan 4A
sin8A cos5A  cos12A sin 9A
10.
Prove that
sin(A  C)  2sin A  sin(A  C) sin A

sin(B  C)  2sin B  sin(B  C) sin B
11.
Prove that 1 + cos2x + cos4x + cos6x = 4 cosx cos2x cos3x
12.
Prove that cos4
13.
 1 
Prove that cot  7   2  3  4  6
 2
14.
Prove that cos 6A = 32 cos6 A – 48 cos4 A + 18 cos2 A – 1
15.
If cos  + cos  + cos  = 0 ; then

3
5
7 3

+ cos4
+ cos4
+ cos4
8
8
8
8 2
prove that cos3 + cos3 + cos3 = 12 cos cos cos
EXERCISE–II
1.
Prove that: cos2 + cos2(+ )  2 cos  · cos  cos(+ ) = sin2
2.
Prove that: tan  + 2 tan 2 + 4 tan 4 + 8 cot 8 = cot .
3.
Prove that :
(a) tan 20° . tan 40° . tan 60° . tan 80° = 3
4
(b) tan 9°  tan 27°  tan 63° + tan 81° = 4. (c) sin

=
24


3
5
7 3
 sin 4
 sin 4
 sin 4

16
16
16
16 2


4.
Find the positive integers p, q, r, s satisfying tan
5.
If m tan (- 30°) = n tan (+ 120°), show that cos 2 =
6.
If cos (+ ) =
7.
If the value of the expression sin 25° · sin 35° · sin 85° can be expressed as
p q
r s .
mn
.
2( m n )
4
5

; sin (– ) =
& ,  lie between 0 & , then find the value of tan 2.
5
13
4
where a, b, c  N and are in their lowest form, find the value of (a + b + c).
8.
Prove that (4 cos29° – 3)(4 cos227° – 3) = tan 9°.
9.
Prove that 4 cos
10.
2

2
· cos – 1 = 2 cos .
7
7
7
If + = , prove that cos²+ cos² + cos² = 1 + 2 cos  cos cos  .
11.
Calculate without using trigonometric tables:
(a) 4 cos 20° 
(c) cos6
3 cot 20°

3
5
7
+ cos6
+ cos6
+ cos6
16
16
16
16
(b)
2 cos 40  cos20
sin 20
(d) tan 10°  tan 50° + tan 70°
12.
Given that (1 + tan 1°)(1 + tan 2°)......(1 + tan 45°) = 2n, find n.
13.
Let A1, A2, ......, An be the vertices of an n-sided regular polygon such that;
1
1
1
. Find the value of n.


A1 A 2 A1 A 3 A1 A 4
14.
In a right angled triangle, acute angles A and B satisfy
tan A + tan B + tan2A + tan2B + tan3A + tan3B = 70
find the angle A and B in radians.
15.
If the product (sin 1°)(sin 3°)(sin 5°)(sin 7°) ........(sin 89°) =
16.
(a)
(b)
(c)
1
2n
, then find the value of n.
If y = 10 cos2x  6 sin x cos x + 2 sin2x, then find the greatest & least value of y.
If y = 1 + 2 sin x + 3 cos2 x , find the maximum & minimum values of y  x  R.
If y = 9 sec2x + 16 cosec2x, find the minimum value of y  x  R.
a b
c
(d)
(e)
(f)


If a  3 cos     + 5 cos  + 3  b, find a and b.
3

If 4 sin x · cos y + 2 sin x + 2 cos y + 1 = 0 where x, y  [0, 2] find the largest possible value
of the sum (x + y).
Let x > 1, y > 1 and (ln x)2 + (ln y)2 = ln x2 + ln y2, then find the maximum value of x ln y .
17.
If tan  = p/q where  = 6,  being an acute angle, prove that :
1
(p cosec 2  q sec 2 ) = p 2  q 2 .
2
18.
Let A1 , A2 , A3 ............ An are the vertices of a regular n sided polygon inscribed in a circle of radius R.
If (A1 A2)2 + (A1 A3)2 + ......... + (A1 An)2 = 14 R2 , find the number of sides in the polygon.
19.
Let k = 1°, then prove that
88
1
 cos nk ·cos(n  1)k
n 0
20.
Find the exact value of tan2
=
cos k
sin 2 k

3
5
7
+ tan2
+ tan2
+ tan2
16
16
16
16
EXERCISE–III
1.
(a) If y = 10 cos2x – 6 sin x cos x + 2 sin2x, then find the greatest & least value of y.
(b) If y = 1 + 2 sin x + 3 cos2 x , find the maximum & minimum values of y   R.
(c) If y = 9 sec2x + 16 cosec2x, find the minimum value of y   R.



(d) If a  3 cos    3  + 5 cos  + 3  b, find a and b.


(e) If 4 sin x  cos y + 2 sin x + 2 cos y + 1 = 0 where x, y  [0, 2] find the largest possible value of the
sum (x + y).
(f) Let x > 1, y > 1 and (ln x)2 + (ln y)2 = ln x2 + ln y2, then find the maximum value of xln y .
2.
If tan  = p/q where  = 6, being an acute angle, prove that:
1
(p cosec 2  – q sec 2 ) =
2
p2  q 2
3.
Determine the smallest positive value of x (in degrees) for which
tan(x + 100°) = tan(x + 50°) tan x tan (x – 50°).
4.
 x  R, find the range of the function, f (x) = cos x (sin x + sin 2 x  sin 2  );   [0, ]
5.
Prove that : 4 sin 27° = 5  5
6.
Prove that :
7.
Let x1 =
1/2



 3 5
1/2

.
cos 3  cos 3
= (cos + cos) cos() – (sin + sin) sin()
2 cos(  )  1
5
 cos
r 1
x1 · x2 =
r
and x2 =
11
5
r
 cos 11 , then show that
r 1
1 


 cos ec  1 , where  denotes the continued product.
64 
22 
5
. Find the value of (1 – sin t)(1 – cos t).
4
8.
If (1 + sin t)(1 + cos t) =
11.
If cosA = tanB, cosB = tanC and cosC = tanA , then prove that sinA = sinB = sinC = 2 sin18°.
12.
Let R and



 
sin 2k  = a. Find the value
k 2
x2

cot 2  cot2 sin 2  in terms of 'a'.
3
k
k
4
k
k 0
+ 2xy – y2 = 6, find the minimum value of (x2 + y2)2.
13.
If x and y are real number such that
15.
If '' is eliminated from the equations cos  – sin  = b and cos 3 + sin 3 = a, find the eliminant.
16.
Show that elliminating x & y from the equations , sin x + sin y = a ;
8ab
cos x + cos y = b & tan x + tan y = c gives 2 2 2
= c.
a b  4a 2


17.
Given that 3 sin x + 4 cos x = 5 where x  0,  2 . Find the value of 2 sin x + cos x + 4 tan x.
18.
Show that
19.
20.
3  cos x
 x R can not have any value between  2 2 and 2 2 . What inference
sin x
sin x
can you draw about the values of
?
3  cos x
Prove that the triangle ABC is equilateral iff , cot A + cot B + cot C = 3 .
A
C
B
If A+B+C = ; prove that tan2 + tan2 + tan2
 1.
2
2
2
EXERCISE–IV
1.
2.
3.
xy
If cos x + cos y + cos  = 0 and sin x + sin y + sin  = 0, then cot 
[AIEEE : 2002]
=
 2 
(A) sin 
(B) cos 
(C) cot 
(D) 2 sin 
Cos1o cos2o . cos3o ....... cos 179o =
(A) 0
(B) 1
(A) 
(D) 3
3
310
Let cos( + ) =
(B)
3
130
(C)
6
65
(D)
6
65

4

and let sin( – ) =
, where 0 ,   . Then tan2 = [AIEEE : 2010]
5
13
4
25
56
(B)
16
33
2
4
If A = cos  + sin  , then for all value of 
(A) 1 A  2
[AIEEE : 2004]
21
27

& cos + cos = –
, then the value of cos
is :
65
65
2
(A)
5.
(C) 2
Let   be such that  <  –  < 3.
If sin + sin = –
4.
[AIEEE : 2002]
(B)
3
A 1
4
(C)
19
12
(D)
20
7
(C)
13
 A  1
16
(D) None of these
6.
In a PQR , if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then the angle R is equal to
(A)

6
(B)

4
(C)


7.
For 0 < /2, if x =
 cos 2n , y =
n 0
(A) xyz = xz + y
3
4
 sin 2n , z =
n 0
(B) xyz = xy + z
5
6
[AIEEE : 2012]
 sin 2n  , then
[IIT : 1993]
(D)

 cos
2n
n 0
(C) xyz = yz + x
(D) None of these
8.
If K = sin(/18) sin (5/18) sin (7/18), then the numerical value of K is :
[IIT : 1993]
(A) 1/8
(B) 1/16
(C) 1/2
(D) None of these
9.
If A > 0, B > 0 and A + B =  / 3, then the maximum value of tan A tan B is :
(A) 1
10.
11.
12.
(B) 1/3
(C)
[IIT : 1993]
(D) 1/ 3
3
 6




 3

6
The expression 3 sin 4      sin 4 (3  )  – 2 sin      sin (5   )  is equal to
2

 2





(A) 0
(B) 1
(C) 3
(D) sin4 + cos6
[IIT : 1995]
4
2
6
6
3(sinx – cosx) + 6(sin x + cos x) + 4(sin x + cos x) =
[IIT : 1995]
(A) 11
(B) 12
(C) 13
(D) 14
4xy
is true, if and only if :
(x  y)2
(A) x + y  0
(B) x = y, x  0
sec2 =
(D) x  0, y  0
(C) x = y
13.
The number of values of x where the function f(x) = cos x + cos( 2 x) attains its maximum is :
(A) 0
(B) 1
(C) 2
(D) Infinite
[IIT 1998]
14.
Which of the following number(s) is rational :
(A) sin 15o
(B) cos 15o
(C) sin
15o
cos
15o
(D)
sin 15o cos
[IIT 1998]
75o
n
15.
Let n be an odd integer.If sin n =
br sinr, for every value of  then :
[IIT : 1998]
r 0
(A) b0 = 1, b1 = 3
16.

8
In a triangle PQR, R =
(a  0), then :
(A) a + b = c
18.
(C) b0 = –1, b1 = n
(D) b0 = 0, b1 = n2 + 3n + 3
The function f(x) = sin4 x + cos4 x increases if
(A) 0 < x <
17.
(B) b0 = 0, b1 = n
(B)

3
x
4
8
[IIT 1998]
(C)
3
5
x
8
8
(D)
5
3
x
8
4

P
Q
, If tan   and tan   are the roots of the equation ax2 + bx + c = 0
2
2
2
[IIT 1999]
(B) b + c = a
(C) a + c = b
(D) b = c

For a positive integer n, let fn() = tan   (1 + sec) (1 + sec 2 ) (1 + sec 4) ..... ( 1 + sec2n ).
2
Then :
[IIT 1999]

(A) f2   = 2
 16 
 
(B) f3   = 1
 32 
  
(C) f4   = 0
 64 
(D) None of these
19.
Let f () = sin (sin + sin 3 ). Then f ()
(A)  0 only when  0
(C)  0 for all real 
[IIT 2000]
(B)  0 for all real 
(D)  0 only when  0

and , then tan  equals :
2
(A) 2(tan  + tan )
(B) tan  + tan 
(C) tan  + 2 tan 
(D) 2 tan  + tan 
20.
If  +  =
21.
The maximum value of
[IIT 2001]
(cos 1).(cos 2)......(cos n) , under the restrictions 0 1 . 2 ..... n 

and
2
(cot 1).(cot 2).(cot 3)......(cot n) = 1 is
(A)
22.
1
2n/2
(B)
If  & are acute angles such that sin  =
  
(A)  , 
 3 2
23.
25.
cos+ ) =
(D) 1
1
1
and cos  = then  +  lies in
2
3
 2 5 
(C)  , 
 3 6
 
(D)  ,  
6 
1
, cos( – ) = 1 find no. of ordered pair of (, ) – ,  
e
(B) 1
(C) 2
(D) 4
[IIT Scr 2005]
sin 4 x cos 4 x 1

 , then
If
2
3
5
2
3
(B)
1
sin 8 x
cos8 x
+
=
125
8
27
2
sin 8 x
cos8 x
+
=
125
8
27
1
The maximum value of the expression
is
2
sin   3sin  cos   5cos 2 
(C) tan2x =
27.
1
2n

If t = (tan)tan, t2 = (tan)cot, t3 = (cot)tan, t4 = (cot)cot and let   0,  , then [IIT 2006]
 4
(A) t4 < t2 < t1 < t3
(B) t4 < t1 < t3 < t2
(C) t4 < t3 < t2 < t1
(D) t2 < t1 < t3 < t4
(A) tan2x =
26.
(C)
  2 
(B)  , 
2 3 
(A) 0
24.
1
2n
1
3
(D)
[IIT 2010]



Let  [0, 2] be such that 2 cos (1 – sin) = sin2  tan  cot  cos – 1, tan(2 – ) > 0
2
2

and –1 < sin  < –

2
4
3

(C)
3
3
(A) 0 <  < –
3
. Then  cannot satisfy
2
[IIT 2012]

4

2
3
3
   2
(D)
2
(B)
28.
Let f : (–1, 1)  IR be such that f(cos 4) =
2
   
for   0,    ,  . Then the value(s) of
2
2  sec 
 4 4 2
1
f   is(are)
3
(A) 1 –
3
2
(B) 1 +
3
2
(C) 1 –
2
3
(D) 1 +
2
3
tan A
cot A

can be written as :
[JEE Mains 2013]
1  cot A 1  tan A
(A) sec A + cosec A (B) sinA cosA + 1
(C) secA cosecA + 1 (D) tanA + cotA
29.
The expression
30.
Let fk(x) =
(A)
1
(sink x + cosk x) where x  R and k 1. Then f4(x) – f6(x) equals
k
1
12
(B)
1
6
(C)
1
3
(D)
[JEE Mains 2014]
1
4
13
31.
1
is equal to
k 1 sin    (k  1)   sin    k 

 

6  4 6 
4
The value of 
(A) 3  3
32.
(D) 2(2  3)
3
5
(B)
1
3
(C)
2
3
[JEE Mains 2017]
(D) 
7
9
Let a vertical tower AB have its end A on the level ground. Let C be the mid - point of AB and P be a
point on the ground such that AP = 2AB. IfBPC = b then tan b is equal to :
[JEE Mains 2017]
(A)
34.
(C) 2( 3  1)
If 5 (tan2 x – cos2 x) = 2 cos 2x + 9, then the value of cos 4x is
(A) 
33.
(B) 2(3  3)
[JEE Adv. 2016]
6
7
(B)
1
4
(C)
2
9
(D)
4
9
Let  and  be non zero real numbers such that 2(cos  – cos ) + cos  cos  = 1. Then which of the
following is/are true ?
[JEE Adv. 2017]


(A) tan    3 tan    0
2
2


(B) tan    3 tan    0
2
2




3 tan    tan    0
(D) 3 tan    tan    0
2
2
2
2
Let a, b, c be three non-zero real numbers such that the equation
(C)
35.
  
3 a cos x + 2b sin x = c, x    , 
 2 2
has two distinct real roots  and  with  +  =

b
. Then the value of is.
3
a
[JEE Adv. 2018]
36.
 
For any   ,  , the expression 3(sin  – cos )4 + 6(sin  + cos )2 + 4 sin6  equals :
4 2
(A) 13 – 4 cos6 
(B) 13 – 4 cos4  + 2 sin2  cos2 
(C) 13 – 4 cos2  + 6 cos4 
(D) 13 – 4 cos2  + 6 sin2  cos2 
[JEE Main 2019]
37.
The value of cos

2
(A)
38.
Let fk(x) =
(A)
39.
1
256
5
12
2
 cos

2
3
(B)
 ....  cos

10
2
1
2
 sin

210
is :
(C)
[JEE Main 2019]
1
512
(D)
1
1024
1
(sink x + cosk x) for k = 1, 2, 3,..... Then for all x  R, the value of f4(x) – f6(x) is equal to
k
(B)
1
12
(C)
1
4
(D)
1
12

The maximum value of 3 cos  + 5 sin     for any real value of  is :
6

(A) 19
(B)
79
2
(C)
31
(D) 34
[JEE Main 2019]
[JEE Main 2019]
ANSWER
KEY
COMPOUND ANGLES
EXERCISE–II
4.
p = 3, q = 2; r = 2; s = 1
56
33
6.
16. (a) ymax = 11, ymin = 1; (b) ymax =
13
3
(a)
12.
n = 23
15.
89
2
13.
18.
n=7
n=7
20.
24
(c) 49; (d) a = – 4 & b = 10; (e)
3 5
2 3
; (b)
, ymin =  1;
32
16
17.
7.
23
; (f) e4
6
11.
5
(a) 1, (b) 3 , (c) , (d) 3
4
23.
(1,1, 2);
5 2
14.

5
and
12
12
28
EXERCISE –III
3.
12.
x = 30°
a
4
4.
–
13.
18
1  sin 2   y  1  sin 2 
15.
a = 3b – 2b3
17.
8.
13
 10
4
5
1
1 

,
18.  

 2 2 2 2
4.
9.
14.
19.
24.
29.
34.
39.
B
B
C
C
D
C
AB
A
EXERCISE–IV
1.
6.
11.
16.
21.
26.
31.
36.
C
A
C
B
A
2
C
A
2.
7.
12.
17.
22.
27.
32.
37.
A
B
B
A
B
ACD
D
C
3.
8.
13.
18.
23.
28.
33.
38.
A
A
B
B
D
AB
C
D
5.
10.
15.
20.
25.
30.
35.
B
B
B
C
AB
A
0.5