Download EXERCISE 1. If where and . Show that (i) (ii) 2. show that 3. Prove

EXERCISE
1.
2.
3.
24
3
3
3
, cos   where    
and     2 .
25
5
2
2
3
4
Show that (i) sin     
(ii) cos      
5
5
3

12
3
16
sin   ,0    ,cos    ,    
show that tan     
5
2
13
2
63
If cos   
Prove the following identities:
(1)
cos  A  B  cos B  sin  A  B  sin B  cos A
(2)
sin 3A cos A – cos 3A sin A = sin 2A
(3)
cos(30 + A) cos(30 - A) – sin(30 + A)sin(30 - A) = ½
(4)
sin(60 - A) cos(30 + A) + cos(60 - A)sin(30 + A) = 1
(5)
sin 2 cos  + cos 2 sin  = sin 4 cos  - cos 4 sin 
(6)
cos 2 cos  + sin 2 sin  = cos 
(7)
cos 4 cos  + sin 4 sin  = cos 2 cos  - sin 2 sin 
(8)
cos 4 cos  - sin 4 sin  = cos 3 cos 2 - sin 3 sin 2
(9)
tan      tan 
1  tan     tan 
 tan 
(10)
tan 4 A  tan 3 A
 tan A
1  tan 4 A tan 3 A
(12)
sin(150 + x) + sin(150 - x) = cos x
(15)
cos2 2x – cos2 6x = sin 4x sin 8x
(11)
cot(   ) cot   1
 cot 
cot   cot(   )
(13)
 3

 3

cos 
 x   cos 
 x    2 sin x
 4

 4

(14)




cos   x   cos   x    2 sin x
4

4

(16)
tan 3x tan 2x tan x = tan 3x – tan 2x – tan x
(17)
sin(60 + A) – sin(60 – A) = sin A
(18)
cos  30  A  cos  30  A   3 cos A (19)




cos      cos       2 sin 
4

4

(20)
sin2 5A – sin2 2A = sin 7A sin 3A
cos 2A cos 5A = cos2
(22)
sin     
sin  sin 

sin    
sin  sin 

sin    
sin  sin 
(21)
0
7A
3A
 sin 2
2
2
(23)
sin  B  C 
cos B cos C

sin  C  A 
cos C cos A

sin  A  B 
cos A cos B
0
(24)
tan( - ) + tan( - ) + tan( - ) = tan( - )tan( - )tan( - )
(25)
cos2  + cos2  - 2cos  cos  cos( + ) = sin2 ( + )
(26)
sin2  + sin2  + 2sin  sin  cos( + ) = sin2( + )
(27)
cos ec     
(29)

 


 

cos   x  cos   y   sin   x  sin   y   sin( x  y)
4
 4

4
 4

(30)


tan   x 
2
4

   1  tan x 
 1  tan x 



tan   x  
4

(31)
sin(n  1) x sin(n  2) x  cos(n  1) x cos(n  2) x  cos x
(32)
 3

 3

cos 
 x   cos 
 x    2 sin x
 4

 4

(33)
sin 2 6 x  sin 2 4 x  sin 2 x sin10 x (34)
(35)
cot x cot 2 x  cot 2 x cot 3x  cot 3x cot x 1
(36)
cos2 cos2  sin 2 (   )  sin 2 (   )  cos2(   )
cos ec cos ec
cot   cot 
(28)
cos     
1  tan  tan 
sec  sec 
cos2 2 x  cos2 6 x  sin 4 x sin8 x
4.
If 2 tan   cot   tan  , prove that cot   2 tan(    )
5.
If tan  
6.
If A  B 
7.
If A + B = 225, prove that
8.
Prove that :
9.
If sin(    )  1 and sin(    ) 
n sin  cos 
, show that tan(    )  (1  n) tan 
1  n sin 2 

4
, prove that : (i)
(1  tan A)(1  tan B)  2 (ii) (cot A  1)(cot B  1)  2
cot A
cot B
1


1  cot A 1  cot B 2
tan 2 2 x  tan 2 x
 tan 3x tan x
1  tan 2 2 x tan 2 x
tan(   2 ) and tan( 2   ) .
1

, where 0   ,   , then find the values of
2
2
10.
Prove that
1
cot( x  a)  cot( x  b)

sin( x  a) sin( x  b)
sin( a  b)
11.
If tan  
m
1
, tan  
, show that  +  = /4.
m 1
2m  1
Section –IV Expressing product as sum and sum as products
Product as sum : In the last section we have learnt four important addition and subtraction formulas. If
we add or subtract two such formulas of same kind we get formulas expressing product as a sum and
sum as a product. Indeed by adding sin  A B and sin  A B formulas we easily get
sin  A  B  sin  A  B  2 sin Acos B . If we write it in the reverse order we get a formula expressing
the product of the type 2 sin  cos  as a sum of two sines.
For example, 2 sin 65 cos 25  sin 65  25  sin 65  25  sin 90  sin 30  1  1/ 2  3 / 2 .
Similarly, we can have formulas 2 cos Asin B  sin  A  B  sin  A  B
2 cos Acos B  cos A  B  cos A  B
2 sin Asin B  cos A  B  cos A  B
One has to be careful in applying last formula in which   comes at first place
Sum as product : - If in all above four formulas we put A  B  C , A  B  D then by writing in the
CD
CD
cos
reverse direction we easily get sin C  sin D  2 sin
2
2
CD
CD
sin C  sin D  2 cos
sin
2
2
CD
CD
cos C  cos D  2 cos
cos
2
2
CD
CD
cos C  cos D  2 sin
sin
2
2
Once again we have to be careful in applying formula number 4. The intricacies of these formulas will be
more clear when we do sufficient number of problems. For instance the formula
2 cos Asin B  sin  A  B  sin  A  B may be avoided if we write
2 cos Asin B  2 sin B cos A  sin B  A  sin B  A  sin  A  B  sin  A  B
A   will never appear in cos C  cos D because of cos    cos . Let us illustrate usage of these
formulas by means of examples.
SOLVED EXAMPLES
1.
1
Show that sin140 cos 40 cos10
2
Sol :
sin 140 cos 40 

1
2 sin 140 cos 40
2
1
sin( 140  40)  sin( 140  40)
2
1
sin 180  sin 100  1 sin100  1 cos10
2
2
2
1
Show that sin 10 sin 50 sin 70 
8
1
1
LHS  sin 102 sin 50 sin 70  sin 10cos (50  70)  cos(50  70)
2
2
( on applying 2 sin A sin B  cos( A  B)  cos( A  B)]
1
= sin 10 cos( 20 )  cos120
2
1
1
1

 sin 10cos 20  
(  cos( 20)  cos 20 and cos 120   )
2
2
2

1
1
1
1
 sin 10 cos 20  sin 10
 2 sin 10 cos 20  sin 10
2
4
4
4
1
1
 sin (10  20)  sin( 10  20)  sin 10
4
4
1
1
sin 30  sin 10  sin 10  1 sin 30  1 sin 10  1 sin 10  1
=
4
4
4
4
4
8

2.
Sol :
3.
1
Show that sin A sin(60  A)sin(60  A)  sin 3 A
4
Sol :
LHS 
4.
1
1
1
1

sin A  cos 2 A    sin A cos 2 A  sin 3 A
2
2
2
4

1
1
1
  sin 3 A  sin A  sin A  sin 3 A
4
4
4
Show that sin( B  C ) cos( A  D)  sin( C  A) cos( B  D)  sin( A  B) cos(C  D)  0
1
sin A 2 sin( 60  A) sin( 60  A) 
2

1
sin A  cos( 2 A)  cos120
2

1
[ sin( B  C  A  D)  sin( B  C  A  D)  sin( C  A  B  D)  sin( C  A  B  D)
2
 sin( A  B  C  D)  sin( A  B  C  D)]
= 0 ( sin( B  C  A  D) gets cut with sin( C  A  B  D) etc )
Show that cos 4  cos10  2 cos 7 cos 3
CD
CD
cos
By applying cos C  cos D  2 cos
2
2
4  10
4  10
cos 4  cos10  2 cos
cos
 2 cos 7 cos 3
2
2
Show that cos18  sin 18  2 cos 63
LHS  cos18  sin 18  sin 72  sin 18
72  18
72  18
 2 cos
sin
 2 cos 45 sin 27  2 cos 63
2
2
LHS 
5.
Sol:
6.
Sol:
7.
Sol:
8.
Sol:
sin 5 x  sin 3 x
 tan 4 x
cos 5 x  cos 3x
5 x  3x
5 x  3x
2 sin
cos
2 sin 4 x cos x
2
2
 tan 4 x
=
LHS 
5 x  3x
5 x  3x
2 cos 4 x cos x
2 cos
cos
2
2
Show that
Show that sin x  sin 3x  sin 5x  sin 7 x  4 cos x cos 2 x cos 4 x
We will use sin C  sin D formula twice on correct pairs. The correct pairs are
sin x , sin 7 x ; sin 3 x and sin 5x
LHS   sin x  sin 7 x    sin 3x  sin 5 x 
 2 sin 4 x cos 3x  2 sin 4 x cos x
 2sin 4 x  cos3x  cos x   2 sin 4 x.2 cos 2 x cos x = 4 cos x cos 2x sin 4x
9.
Show that
Sol:
LHS 
sin 8 A cos 5 A  cos12 A sin 9 A
 cot 4 A
cos 8 A cos 5 A  cos12 A cos 9 A
sin 13 A  sin 3 A  sin 21A  sin 3 A
2sin8 A cos5 A  2cos12 A sin 9 A
=
2cos8 A cos5 A  (2cos12 A cos9 A)
cos13 A  cos 3 A  (cos 21A  cos 3 A)
sin 13 A  sin 21A
2 sin 17 A cos 4 A
CD
CD

cos
( Applying sin C  sin D  2 sin
cos13 A  cos 21A 2 sin 17 A sin 4 A
2
2
CD
CD
sin
and cos C  cos D   2 sin
)  cot 4 A
2
2
tan( A  B) k  1
If sin 2 A  k sin 2 B prove that
.

tan( A  B) k  1
sin 2 A k
sin 2 A  sin 2 B k  1


We
(applying componendo & divenendo )

sin 2 B 1
sin 2 A  sin 2 B k  1

10.
Sol:


2 A  2B
2 A  2B
cos
sin( A  B) cos( A  B) k  1
2
2
=

2 A  2B
2 A  2B
cos( A  B) sin( A  B) k  1
2 cos
sin
2
2
2 sin
tan( A  B) k  1
.

tan( A  B) k  1
EXERCISE
5
 1
sin

12
12 2
1.
Show 2 sin
2.
Show that sin 25 cos115 
1
(sin 40  1)
2
1
8
3.
Show that cos 40 cos 80 cos160  
4.
Show that cos
5.
4 cos12 cos 48 cos 72  cos 36
6
cos 3A sin 2A – cos 4A sin A = cos 2A sin A
7.
Prove the following identities:
5

1
cos

12
12 4
(1)
sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x
(2)
sin  + sin 3 + sin 5 + sin 7 = 4cos  cos 2 sin 4
(3)
cos 7x + cos 5x + cos 3x + cos x = 4 cos x cos 2x cos 4x
(4)
cot 4x(sin 5x + sin 3x) = cot x(sin 5x – sin 3x)
(5)
cos9 x  cos5 x
sin 2 x

sin17 x  sin 3x
cos10 x
(6)
sin 5 x  sin 3x
 tan 4 x
cos5 x  cos3x
(7)
sin x  sin y
x y
 tan
cos x  cos y
2
(8)
sin x  sin 3x
 tan 2 x
cos x  cos3x
(9)
sin x  sin y
x y
 tan
cos x  cos y
2
(10)
tan 5  tan 3
 4cos 2 cos 4
tan 5  tan 3
(11)
sin x  sin 3x
 2sin x
sin 2 x  cos 2 x
(12)
sin 5x  2sin 3x  sin x
 tan x
cos5 x  cos x
(13)
(sin 7 x  sin 5 x)  (sin 9 x  sin 3x)
 tan 6 x
(cos7 x  cos5x)  (cos9 x  cos3x)
(14)
cos 4 x  cos3x  cos 2 x
 cot 3x
sin 4 x  sin 3x  sin 2 x
(15)
sin 3x + sin 2x – sin x = 4 sin x cos
(16)
cos  cos3
 tan 2
sin 3  sin 
(17)
sin 2  sin 3

 cot
cos 2  cos3
2
(18)
cos 4  cos
5
 tan
sin   sin 4
2
(19)
cos 2  cos12
 tan 5
sin12  sin 2
(20)
(22)
cos  2  3   cos3
sin  2  3   sin 3
sin      sin 4
cos      cos 4
 cot 
 tan
  3
2
x
cos
2
(21)
cos   3   cos  3   
sin  3     sin   3 
 tan   2 
8.
(23)
cos 3A + sin 2A – sin 4A = cos 3A(1 – 2sin A)
(24)
sin 3 - sin  - sin 5 = sin 3(1 – 2cos 2)
(25)
cos  + cos 2 + cos 5 = cos 2(1 + 2cos 3)
(26)
sin  - sin 2 + sin 3 = 4sin /2 cos  cos 3/2
(27)
sin 3 + sin 7 + sin 10 = 4sin 5 cos 7/2 cos 3/2
(28)
sin A + 2sin 3A + sin 5A = 4sin 3A cos2 A
(29)
sin 2  sin 5  sin 
 tan 2
cos 2  cos5  cos 
(31)
cos7  cos3  cos5  cos
 cot 2
sin 7  sin 3  sin 5  sin 
(32)
sin( +  + ) + sin( -  - ) + sin( +  - ) + sin( -  + ) = 4sin  cos  cos 
(33)
cos( +  - ) – cos( +  - ) + cos( +  - ) – cos( +  + ) = 4sin  cos  sin 
(34)
sin 2 + sin 2 + sin 2 - sin 2( +  + ) = 4sin ( + )sin( + )sin( + )
(35)
cos  + cos  + cos  + cos( +  + ) = 4cos
(36)
cos( + )sin( - ) + cos( + )sin( - ) + cos( + )sin( - ) + cos( + )sin( - ) = 0
(37)
sin  cos( + ) – sin  cos( + ) = cos  sin( - )
(38)
cos  cos( + ) – cos  cos( + ) = sin  sin( - )
(39)
sin( - ) + sin( - ) + sin( - ) + 4sin
(40)
sin (sin 3 + sin 5 + sin 7 + sin 9) = sin 6 sin 4
(41)
sin   sin 3  sin 5  sin 7
 tan 4
cos  cos3  cos5  cos7
(42)
sin x  sin3x  sin5x  sin 7 x  4cos x cos 2 x sin 4 x
(43)
2 cos
(44)
tan 60    tan 60    
(45)
cos 20 cos100  cos100 cos140  cos140 cos 200  
(46)
sin

2

13
cos
.sin
(30)
sin   sin 2  sin 4  sin 5
 tan 3
cos  cos 2  cos 4  cos5
 
2
 
2
sin
cos
 
2
 
2
sin
cos
 
 
2
2
0
9
3
5
 cos
 cos
0
13
13
13
2 cos 2  1
2 cos 2  1
3
4
7
3
11
 sin
sin
 sin 2 sin 5
2
2
2
If cos( A  B).sin C  D  cos( A  B) sin( C  D) prove that tan A tan B tan C  tan D  0
9.
Show that sin x sin y sin(x – y) + sin y sin z sin(y – z) + sin z sin x sin(z – x) + sin(x – y) sin(y – z)
sin(z – x) = 0 for all x, y, z.
Section –V Trigonometric Functions of multiple angles
Almost all trigonometric formula have been generated by sin( A  B ) formula
Now we will observe that most useful formulas being generated by these formula’s.
The formulas express Trigonometric functions of the angle n A in terms of Trigonometric
Functions of angles A where n is an integer ( in general n can be positive rational ) we start deriving
these formula’s.We will put frequently used formulas in a block.
sin 2 A  sin( A  B)  sin A cos A  cos A sin A  2 sin Acos A
Thus sin 2 A  2 sin A cos A
We can similarly derive,
sin 4 A  2 sin 2 A cos 2 A, sin 8 A  2 sin 4 A cos 4 A, sin A  2 sin
A
A
cos
2
2
A
A
A
 2 sin n1 cos n1
n
2
2
2
Again cos 2 A  cos ( A  A)  cos A cos A  sin A sin A  cos 2 A  sin 2 A
sin
But cos 2 A  sin 2 A  cos 2 A  (1  cos 2 A)  2 cos 2 A  1
 1 2 sin 2 A
Thus cos 2 A  cos 2 A  sin 2 A  2 cos 2 A  1  1  2 sin 2 A
We can similarly derive cos 4 A  cos 2 2 A  sin 2 2 A  2 cos 2 2 A  1  1  2 sin 2 2 A
A
A
A
A
cos A  cos 2  sin 2  2 cos 2  1 1  2 sin 2
2
2
2
2
tan A  tan A
2 tan A

Again tan 2 A  tan( A  A) 
.
1  tan A tan A 1  tan 2 A
A
2 tan
2 tan A
2 tan 2 A
2
Thus tan 2 A 
Also tan 4 A 
or tan A 
2
1  tan A
1  tan 2 2 A
2 A
1  tan
2
2
2 tan A
1  tan A
we can easily get sin 2 A 
, cos 2 A 
2
1  tan A
1  tan 2 A
Or cos 2 A  sin 2 A  (1  sin 2 A)  sin 2 A
3A formulas :-
sin 3 A  sin(2 A  A)
 sin A cos 2 A  cos A sin 2 A = sin A (1  2 sin 2 A)  cos A.2 sin A cos A
= sin A (1  2 sin 2 A)  2 sin A(1  sin 2 A) = 3sin A  4 sin 3 A
cos3 A  cos( A  2 A)  cos A cos 2 A  sin Asin 2 A
=
cos A (2 cos 2 A  1)  2 sin 2 A cos A = cos A(2 cos 2 A  1)  2 (1  cos 2 A) cos A
= 4cos3 A  3cos A
tan 3 A  tan( A  2 A) 
tan A  tan 2 A
1  tan A tan 2 A
2 tan A
3
1  tan 2 A = 3tan A  tan A
=
2 tan A
1  3tan 2 A
1  tan A.
1  tan 2 A
sin 3 A  3sin A  4 sin 3 A
tan A 
Thus
cos 3 A  4 cos 3 A  3 cos A
tan 3 A 
3 tan A  tan 3 A
1  3 tan 2 A
(i)
(ii)
The sine, cosine, tangents of angle of n A when n  4 can also be found.
We will observe that if n is a positive integer.
cos nA will be a polynomial of degree n in cos A for all n .
sin nA will be a polynomial of degree n in sin A if n is odd .
(iii)
tan nA 
n
C1 tan A  n C3 tan 3 A  n C5 tan 5 A.....
1  n C2 tan 2 A  n C4 tan 4 A.....
The proof of (i), (ii), (iii) involve three branches of algebra which are complex numbers, binomial
theorem and polynomial theory. We will limit our solves upto n  5 . The general proof is beyond the
scope of this book.


2
For example : cos 4 A  2 cos 2 2 A  1  2 2 cos 2 A  1  1  8 cos 4 A  8 cos 2 A  1
or cos 4 A  Real part of cos 4 A  i sin 4 A
4
 Real part of cos A  i sin A
 Real part of
cos A 
4
4
C1 cos A i sin A  4C2 cos A i sin A  4C3 cos A i sin A  4C4 i sin A
3
2
 cos 4 A  6 cos 2 A sin 2 A  sin 4 A

 
2
1
3

2
 cos 4 A  6 cos 2 A 1  cos 2 A  1  cos 2 A
 8 cos 4 A  8 cos 2 A  1
5
sin 5 A  Imaginary part of cos 5 A  i sin 5 A  I.P of cos A  i sin A
This time writing only Imaginary part we get
i sin 5 A  5C1 cos A i sin A  5C3 cos A i sin A  5C5 i sin A
4
1
2
3
5
 sin 5 A  5 sin Acos A  10 sin 3 A cos 2 A  sin A
4

5

2

 

 5sin A 1  sin 2 A  10sin3 A 1  sin2 A  sin5 A
 16 sin 5 A  20 sin 3 A  5 sin A
(*)
The above discussion may lead to something very basic in mathematics that sin x , or
cos x are solutions of polynomial equations with integer coefficients.
For example If A 18 then 5A  90 from (*)
 1  16 x 5  20 x 3  5 x
where x  sin A
5
3
or 16 x  20 x  5 x  1  0
 sin 18 is a solution of a polynomial equation with integer coefficients. Actually numbers
which are solutions of polynomial equations with integer coefficients are called algebraic
numbers. The number  is not algebraic but cos 20 is algebraic since
4

cos 60  4 cos3 20  3 cos 20 (by formula cos 3 A  4 cos3 A  3 cos A )
 1 / 2  4 x 3  3x
where x  cos 20
3
 8x  6 x  1  0
 cos 20 is algebraic
We will later see that sin 18 is a solution of a polynomial equation with much lesser degree.
SQUARE, CUBE FROM POWER REMOVAL : From 2 A formula’s we can get rid of power 2,3,4
easily. For example :
1  cos 2 A
1  cos 2 A
, sin 2 A 
2
2
cos 3 A  3 cos A
3 sin A  sin 3 A
cos3 A 
, sin 3 A 
4
4
cos 2 A 
1  2 cos 2 A  cos 2 2 A
4
1  cos 4 A
1  2 cos 2 A 
3 1
1
2
  cos 2 A  cos 4 A

8 2
8
4
2
 1  cos 2 A 
cos 4 A  cos 2 A  

2




2

(**)
3 1
1
Similarly sin 4 A   cos 2 A  cos 4 A etc. From these fourth power formula we easily get
8 2
8
cos 4

8
 cos 4
3
5
7 3
 cos 4
 cos 4
 by using (**) several times.
8
8
8
2