Download Section –V Trigonometric Functions of multiple angles Almost all

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
2
A
1  tan A
1  tan 2 A
1  tan 2
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
SOLVED EXAMPLES
Example 1 :
Prove that cot  - cot 2 = cosec 2
Solution :
cot   cot 2 

cos  cos 2 sin 2 cos   cos 2 sin 


sin  sin 2
sin  sin 2
sin 2   
sin 

 cos ec 2  RHS
sin  sin 2 sin  sin 2
Example 2 :
Prove that 1 + cot 2 cot  = cosec 2 cot 
Solution :
1  cot 2 cot   1 

cos 2 cos 
sin 2 sin 
sin 2 sin   cos 2 cos  cos2    cos 
1


.
sin 2 sin 
sin 2 sin  sin  sin 2
cot  cos ec2  RHS
Example 3 :
tan  + cot  = 2cosec 2
Solution :
LHS  tan   cot  
2
2 sin  cos 
sin  cos  sin 2   cos 2 
1



cos  sin 
sin  cos 
sin  cos 

2
 2 cos ec 2
sin 2
2
Example 4 :
A
A

 sin  cos   1  sin A
2
2

Solution :
LHS  sin 2
Example 5 :
Prove that
Solution :
LHS 
A
A
A
A
 cos 2  2 sin cos  1  sin A
2
2
2
2
cos 2
 tan  45   
1  sin 2
cos 2
cos 2   sin 2   cos   sin   cos   sin  


2
1  sin 2  cos   sin  2
 cos  sin  
cos   sin  1  tan 
tan 45  tan 


 tan 45   
cos   sin  1  tan  1  tan 45 tan 
OR



 

sin   2 
2 sin     cos   
cos 2
2
 
4
 4
  tan     



4
1  sin 2



2 


1  cos  2 
2 cos    
2
4




(Applying sin A  2 sin
1  cos A  2 cos 2
A
A

cos where role of A is being played by  2 and
2
2
2

A
with A as  2 )
2
2
Example 6 :
Prove that  cos  cos     sin   sin    4cos2
Solution :
On squaring
2
LHS
2
 
2
 cos 2   cos 2   2 cos  cos   sin 2   sin 2   2 sin  sin 
 2  2cos  cos   sin  sin  
 2  2 cos     21  cos   
 2.2 cos 2
 4 cos 2
 
2
 
2
(Applying 1  cos A  2 cos 2
 RHS
A
with A as     )
2
Example 7 :
Prove that cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
Solution :
cos 6x  2 cos 2 3x  1

( cos 2 A  2 cos 2 A  1 with A as 3x )

2
 2 4cos3 x  3cos x  1  32cos6 x  48cos4 x  18cos2 x  1
Example 8 :
A

Prove that sin A  1  2sin 2  45  
2

Solution :
A

RHS  1  2 sin 2  45    1  2 sin 2 
2

A

  45  
2

 cos 2  cos90  A  sin A
Example 9 :




Prove that cos2      sin 2      sin 2
4

4

Solution :
On removing squares


1  cos 2   


4

cos 2     
4
2




1  cos  2 
2
  1  sin 2

2
2


1  cos  2 


2
  1  sin 2
sin 2     
2
2
4

On subtracting we get the desired result
Example 10 :
Prove that
Solution :
LHS
1  sin 2  cos 2
 tan 
1  sin 2  cos 2

1  cos 2  sin 2 2sin 2   2sin  cos

1  cos 2  sin 2 2cos 2   2sin  cos

2 sin  sin   cos  
 tan   RHS
2 cos  cos   sin  
Example 11 :
Prove that sin 2 x  2sin 4 x  sin 6 x  4cos 2 x sin 4 x
Solution :
LHS
 sin 2 x  sin 6 x  2sin 4 x
 2 sin 4x cos 2x  2 sin 4x
 2 sin 4 x1  cos 2 x   2 sin 4 x.2 cos 2 x
 4 cos 2 x sin 4 x  RHS
Example 12 :
Prove that 2 cos  
Solution :
1  cos 4  2 cos 2 2
2  2  2 cos 4
 2  2 cos 4  4 cos 2 2
On taking square root  2  2 cos 4  2 cos 2
On adding 2 on both sides we get
2  2  2 cos 4  2  2 cos 2  21  cos 2   2.2 cos 2 
On taking the square root again we get
2  2  2 cos 4  2 cos  as desired
Example 13 :
Prove that tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
Solution :
LHS
 tan A  2 tan 2 A  4 tan 4 A  8cot 8 A
  cot A  tan A  2 tan 2 A  4 tan 4 A  8cot 8 A  cot A
(Subtracting and adding cot A )
Now
 cot A  tan A  

cos A sin A sin 2 A  cos 2 A


sin A cos A
sin A cos A
 cos 2 A
.2  2 cot 2 A
2 sin A cos A
(*)
 LHS  2cot 2 A  2 tan 2 A  4 tan 4 A  8cot 8 A  cot A
 4cot 4 A  4 tan 4 A  8cot 8 A  cot A (using (*) for A replaced by 2 A )
 8cot 8 A  8cot 8 A  cot A
(Again using (*))
cot A  RHS
Example 14 :
Prove that
Solution :
sec 8 A  1
sec 4 A  1
sec8 A  1 tan 8 A

sec 4 A  1 tan 2 A
1
1
1  cos 8 A cos 4 A
 cos 8 A

.
1
cos
8
A
1  cos 4 A
1
cos 4 A

2 sin 2 4 A cos 4 A 2 sin 4 A cos 4 A sin 4 A
,

.
2 sin 2 2 A cos 8 A
cos 8 A
2 sin 2 2 A
 tan 8 A .
2 sin 2 A cos 2 A tan 8 A

2 sin 2 2 A
tan 2 A
A
if tan A  3 / 4
2
Example 15 :
Find the values of tan
Solution :
A
2  2 x where x  tan A
tan A 
A 1  x2
2
1  tan 2
2
2 tan
Now tan A  3 / 4  A is either in first quadrant (or 2n to 2n 
quadrant
(i.e. from 2n   to 2n 
If A is from 2n to 2n 

2
then
If A is from 2n   to 2n 
 tan
2.
2
) OR in the third
3
)
2

A
A 1
A
 tan  0  tan 
is from n to n 
4
2
2 3
2

3
A
3
then is from n  to n 
2
2
4
2
A
A
 0  tan  3
2
2
EXERCISE
1.

x
x
x
Find sin ,cos and tan in each of the following :
2
2
2
4
1
(1).
(2).
tan x   , x in quadrant II
cos x   , in quadrant III
3
3
1
(3).
tan x  , in quadrant II
4
Prove the following identities:
(1)
cosec 2 - cot 2 = tan 
(2)
1 + tan 2 tan  = sec 2 
(3)
2 cosec 2 = sec  cosec 
(4)
cos4  - sin4  = cos 2
(5)
cot  - tan  = 2cot 2
(6)
cot 2 A 
(7)
cot A  tan A
 cos 2 A
cot A  tan A
(8)
1  cot 2 A
 cos ec 2 A
2cot A
(9)
cot 2 A  1
 sec 2 A
cot 2 A  1
(10)
1  sec

 2cos2
sec
2
(11)
2  sec2 
 cos 2
sec2 
(12)
cos ec 2  2
 cos 2
cos ec 2
cot 2 A  1
2cot A
cos 2
 cot  45   
1  sin 2
2
(13)
A
A

 sin  cos   1  sin A
2
2

(15)
sin 8A = 8sin A cos A cos 2A cos 4A
(17)
cos 2 sin 2

 cos3
sec cosec
2
(14)
2
(16)
1 + cos 4A = 2cos 2A(1 – 2sin2 A)
tan 4 

(18)
cos 4x = 1 – 8 sin x cos x
(20)
(sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
(21)
cot  
(22)
cosec  (sec  - 1) – cot (1 – cos ) = tan  - sin 
(23)
tan(45 + A)– tan(45 - A) = 2tan 2A
(24)
tan (45 + A) + tan(45 - A) = 2 sec 2A
(26)
cot 3 A 
(28)
sin 3  sin 3 
 cot 
cos3   cos3
(30)
cos 5A cos 2A – cos 4A cos 3A = -sin 2A sin A
(31)
sin 4 cos  - sin 3 cos 2 = sin  cos 2
(32)
4sin A sin(60 + A)sin(60 - A) = sin 3A
(33)
 2
  2

4cos cos 
   cos 
    cos3
 3
  3

(34)
 2

 2

cos  cos 
    cos 
   0
3
3




(35)
cos2 A + cos2 (60 + A) + cos2 (60 – A) = 3/2
(36)
sin2 A + sin2(120 + A) + sin2(120 - A) = 3/2
(37)
(38)
(39)
(19)
4 tan  1  tan 2 

1  6 tan   tan 
2
4
sin 
 cos ec
1  cos 
cot 3 A  3cot A
3cot 2 A  1
sin   sin   sin    
sin   sin   sin    
 cot

2
cot
(25)
sin 3 A cos3 A

2
sin A cos A
(27)
3cos   cos3
 cot 3 
3sin   sin 3
(29)
cos3   cos3 sin 3   sin 3

3
cos 
sin 

2
(cos A – sin A)(cos 2A – sin 2A) = cos A – sin 3A
sin 2 A  cos 2 A 
1  tan A2  2 tan 2 A
1  tan A
2
(40)
sin 4 A 

4 tan A 1  tan 2 A
1  tan A
2
2

4cos 2 A
1  2sin 2 A
(41)
cot 15  A  tan 15  A 
(42)
cot 15  A   tan 15  A  
(43)
tan  A  30 tan  A  30  
(44)
(2cos A + 1)(2cos A – 1)(2cos 2A – 1) = 2cos 4A + 1
(45)
cos2( - ) + cos2( - ) + cos2( - ) = 1 + 2cos ( - )cos( - )cos( - )
(46)
2cos2 (45 - A) = 1 + sin 2A
(47)
2cos 2x cosec 3x = cosec x – cosec 3x
(49)
2tan 2x =
(50)
(cos x + cos y)2 + (sin x – sin y)2 = 4cos2
x y
2
(51)
(cos x – cos y)2 + (sin x – sin y)2 = 4sin 2
x y
2
(52)
1  sin 2 x


 tan 2   x 
1  sin 2 x
4

(54)
(sin3x  sin x)sin x  (cos3x  cos x)cos x  0
(55)
1
1
sin A  2sin 3 A  sin 5 A sin 3 A

 cot 2 A. (56)

tan 3 A  tan A cot 3 A  cot A
sin 3 A  2sin 5 A  sin 7 A sin 5 A
(57)
sin 6
(58)
3A
A
  A   A
4 cos    sin     sin
cos ec
2
2
6 2 3 2
(59)
cos 5 = 16 cos5  - 20 cos3  + 5 cos 
(60)
cos A cos (60 – A) cos(60 + A) = ¼ cos 3A
(61)
cos8 A cos 5 A  cos12 A cos 9 A
 tan 4 A
sin 8 A cos 5 A  cos12 A sin 9 A
(62)
cos3 A  cos3 120  A   cos3  240  A 
4
cos 2 A  3 sin 2 A
1  2cos 2 A
1  2cos 2 A
(48)
cos A cos 2A cos 4A cos 8A =
sin16 A
16sin A
cos x  sin x cos x  sin x

cos x  sin x cos x  sin x
(53)
cos3 sin 3

 2cot 2
sin 
cos 
A
A 3
 cos 6   sin 2 A  4  cos A.
2
2 4
3
cos 3 A
4
(63)
2sin A cos3 A – 2sin3 A cos A =
1
sin 4A
2
(64)
sin 3A cos3 A + cos 3A sin3 A =
3
sin 4A
4
(65)
tan 3 
cot 3 
1  2sin 2  cos 2 


sin  cos 
1  tan 2  1  cot 2 
(66)
2cos  cos cos ( + ) = cos2  + cos2  - sin2 ( + ).
sin 2 3 cos 2 3

 8cos 2
sin 2 
cos 2 
(67)
 3

1  sin 4 A  cot   2 A  cos 4 A  0
 4

(69)
2cos  - cos 3 - cos 5 = 16cos3  sin2 
(70
cos3 ( + ) sin3 ( - ) + cos3 ( + ) sin3 ( - ) + cos3 ( + ) sin3 ( - )
(68)
= 3cos ( + ) cos ( + ) cos( + ) sin( - ) sin( - ) sin ( - )
2.
(71)
cos2 A + cos2 B – 2cos A cos B cos(A + B) = sin2 (A + B).
(72)
tan 3 A
cot 3 A

 sec A cos ec A  sin 2 A
1  tan 2 A 1  cot 2 A
(73)
tan  tan    60  tan  tan    60  tan    60  tan    60   3
(74)
2sin4 - sin 10  +sin2 = 16sin  cos  cos 2 sin 2 3 .
Prove the following numerical equalities (based on last three sections)
(1)
cos 80 + cos 40 – cos 20 = 0
(2)
cos 130 cos 40 + sin 130 sin 40 = 0
(3)
cos 5 – sin 25 = sin 35
(4)
cos 70 cos 10 + sin 70 sin 10 = ½
(5)
sin 65 + cos 65 = 2 cos 20
(6)
sin 78 – sin 18 + cos 132 = 0
(7)
cot 15 + cot 75 + cot 135 – cosec 30 = 1
(8)
(9)
sin 10 sin 30 sin 50 sin 70 = 1/16
1
sin 20 sin 40 sin 80 =
3
8
(10)
tan 40 + cot 40 = 2sec 10
2
(11)
tan 70 + tan 20 = 2 cosec 40
(12)
(13)
tan 6 tan 42 tan 66 tan 78 = 1
(14)
cos 33  cos 57
1
(16)

sin 21  cos 21
2
cos 20  2sin 2 55  1  2 sin 65
2
(15)
(17)
2
 1  cot 60  1  cos30

 
 1  cot 60  1  cos30
cosec 10 - 3 sec10 = 4
1
cos ec10  2sin 70  1
2
3.
(18)
sin 70  8cos 20 cos 40 cos80  2 cos 2 10
(19)
cos 2 73  cos 2 47  cos 73 cos 47 
(20)
tan10  tan 70  tan 50  3
(21)
tan 20 + tan 40 +
3
4
3 tan 20tan 40   3
If sin A + sin B + sin C = cos A + cos B + cos C = 0, prove that
sin2 A + sin2 B + sin2 C = cos2 A + cos2 B + cos2 C = 3/2
ANSWERS
1.
(1).
2 5
5
,
,2
5
5
(2).
6
3
,
, 2
3
3
(3).
8  2 15
8  2 15
,
, 4  15
4
4