ASSIGNMENT: CLASS XI, CH: TRIGONOMETRIC FUNCTIONS
BASED ON RADIAN MEASURE OF AN ANGLE
1. Find the length of an arc of a circle of radius 5 cm subtending a central angle
measuring 15 . (Ans: 5 )
12
2. Find in degrees the angle subtended at the centre of a circle of diameter 50cm by
an arc of length 11 cm. ( Ans: 2512/ )
3. A horse is tied to a post by a rope. If the horse moves along a circular path
always keeping the rope tight and describes 88 meters when it has traced out
72 at the centre, find the length of the rope. ( Ans: 70 meters)
4. A circular wire of radius 7.5cm is cut and bent so as to lie along the
circumference of a hoop whose radius is 125 cm. Find in degrees the angle which
is subtended at the centre of the hoop. (Ans: 2230/ )
5. The moon’s distance from the earth is 360000kms and its diameter subtends an
angle of 31/ at the eye of the observer. Find the diameter of the moon. (Ans:
3247.62km)
6. If the angular diameter of the moon be 30/ , how far from the eye a coin of
diameter 2.2cm be kept to hide the moon?(Ans: 252cm)
7. Assuming that a person of normal sight can read at such a distance that the
letters subtended an angle of 5/ at his eye, find what is the height of the letters
that he can read at a distance of 12 meters.(Ans: 1.7cm)
8. Find the angle between the minute hand of a clock and the hour hand when the
time is 7:20 AM (Ans: 100 )
9. Find in degrees and radians the angle between the hour hand and minute hand of
a clock at half past three.(Ans: 5 )
12
10. A railway train is travelling on a circular curve of 1500 meters radius at the rate
c
 11 
of 66 km/hr. Through what angle has it turned in 10 seconds? ( Ans:   )
 90 
11. Find the diameter of the sun in km supposing that it subtends an angle of 32/ at
the eye of an observer. Given that the distance of the sun is 91106 km.(Ans:
847407.4 km)
BASED ON ALLIED ANGLES
&
EVALUATION OF VALUES AT VARIOUS ANGLES
12. Find the value of the following:
(i) sin 315 (ii) cos 210 (iii) cos  480  (iv) sin  1125  (v) cos ec390 (vi) cot 570
(vii) cos ec  1200  (viii) cos855 (ix) sin1845 (x) cos1755 (xi) sin 4530
(Ans:  1
2
, 3
2
, 1 , 1 , 2, 3 , 2 , 1 , 1 , 1 , 1 )
2
2
2
3
2
2
2
Page 1 of 4
11
4 3

17 3  4 3
 2sin
 cos ec 2  4 cos 2

.
3
6 4
4
6
2
cos ec  90     cot  450    tan 180     sec 180   
14. Prove that

2
cos ec  90     tan 180   
tan  360     sec   
13. Prove that tan


 


15. Prove that 1  cot   sec      1  cot   sec       2 cot  .
2
 
2


cos  2    cos ec  2    tan  2   
16. Prove that
1
sec  2    cos  cot    
BASED ON ANGLE’S SUM FORMULAE
17. If cos     
4
5

, sin      and  ,  lie between 0 and , prove that
4
5
13
56
.
33
18. Prove that tan 70  tan 20  2 tan 50 .
19. If tan      n tan     , show that  n  1 sin 2   n  1 sin 2 .
tan 2 
20. If sin   sin   a and cos   cos   b , show that
b2  a 2
2ab
(ii) sin      2
2
2
a  b2
b a
21. If  and  are the solutions of the equation a tan   b sec   c ,then show that
2ac
tan      2 2 .
a c
cos 9  sin 9
22. Prove that
 tan 56 .
cos 9  sin 9
5
1

23. If tan A  and tan B  , prove that A  B  .
4
6
11
24. Prove the following
sin  A  B  sin  B  C  sin  C  A 
(i)


0
sin A sin B sin B sin C sin C sin A
tan 8  tan 6  tan 2  tan 8 tan 6 tan 2
(ii)
(iii)
tan15  tan 30  tan15 tan 30  1
tan 2 2  tan 2 
 tan 3 tan
(iv)
1  tan 2 2 tan 2 
sin  A  B  x  1
.

25. If tan A  x tan B , prove that
sin  A  B  x  1
(i) cos     

2


26. If tan x  tan  x    tan  x 
3
3


3 tan x  tan 3 x

1
  3 , then prove that
1  3 tan 2 x

Page 2 of 4
BASED ON PRODUCT FORMULAE
27. Prove the following:
1
16
1
(ii)
sin10 sin 30 sin 50 sin 70 
16
3
(iii)
sin 20 sin 40 sin 60 sin 80 
16




(iv)
tan 20 tan 40 tan 80  tan 60  3
3
cos10 cos 30 cos 50 cos 70 
(v)
16
28. Without calculating the values of cos75Oand cos15O, find the value of
cos 75 cos15 .
5

32
29. Prove that 2sin
cos 
.
12
12
2
cos 20 cos 40 cos 60 cos80 
(i)
BASED ON SUM FORMULAE
30. Prove the following:
(i)
sin   sin   2 3  sin   4 3  0
(ii)
(iii)
(iv)
(v)
cos   cos   cos   cos        4 cos
 
2
cos
 
2
cos
 
cos 2 A cos 3 A  cos 2 A cos 7 A  cos A cos10 A
 cot 6 A cot 5 A
sin 4 A sin 3 A  sin 2 A sin 5 A  sin 4 A sin 7 A
sin  A  C   2sin A  sin  A  C  sin A

sin  B  C   2sin B  sin  B  C  sin B
sin A  sin 2 A  sin 4 A  sin 5 A  4 cos
A
3A
cos
sin 3 A
2
2

7
3
11
sin sin
 sin sin
 sin 2 sin 5
2
2
2
2
tan  A  B    1
31. If sin 2 A   sin 2 B , prove that:

tan  A  B    1
(vi)
32. If
sin     1  m





, prove that: tan     tan      m .
cos     1  m
4

4

BASED ON DOUBLE, TRIPPLE, HALF ANGLE FORMULA
33. Prove the following:
sin 2
 cot 
(i)
1  cos 2
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2
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
1  sin   cos 

 tan
1  sin   cos 
2
cos 




 tan   
1  sin 
 4 2
2  2  2  2 cos8  2 cos 
sec8  1 tan 8

sec 4  1 tan 2
 
  
5 
7  1

1  cos  1  cos  1  cos
1  cos

8 
8 
8 
8  8

cos 5 A  16 cos5 A  20 cos3 A  5cos A
3
cos3 A  cos3 120  A   cos3  240  A   cos 3 A
4





 1
 1
 1
 1
 1
34. Find the value of cos  22  , sin  22  , tan  22  , sin  7  , cos  7 
 2
 2
 2
 2
 2
(Ans:
2 1
,
2 2
2 1
4 6  2
4 6  2
, 2 1 ,
,
)
2 2
2 2
2 2
BASED ON GENERAL SOLUTION OF TRIGONOMETRICAL
EQUATIONS
35. Solve the following trigonometric equations:
n

sin   sin 3  sin 5  0 (Ans:  
or   m  ,where, m, n  Z )
(i)
3
3
 2s  1  ,where, r , s  Z )
2r
sin m  sin n  0 (Ans:  
or  
(ii)
mn
mn
(iii)
2 tan   cot   1 (Ans:   n 

or   m   ,where m, n  Z and
4
1
)
2
3
n 1 
m 1 
cot 2  
 3  0 (Ans:   n   1
or,   m   1
,
sin 
6
2
m, n  Z )
n
tan   tan 2  tan 3  tan  tan 2 tan 3 (Ans:  
,nZ )
3
n 
tan   tan 2  3 tan  tan 2  3 (Ans:  
 ,nZ )
3 9
tan  
(iv )
(v)
(vi)

(vii)
2sin 2 x  sin 2 2 x  2 (Ans: x  n 
(viii)
cot   cos ec  3 (Ans:   2n 
it makes sin   0 )
Page 4 of 4
2
or x  m 

3

4
,where m, n  Z )
, n  Z and    2n  1  , n  Z as