Download Trignometric Identities

1.3 Trigonometric identities and integrals.
In this section we look at three trigonometric identities that are very useful for integrating products of sine's
and cosine's. First recall the addition and subtraction identities for cosine.
(1)
cos(A + B) = cos A cos B - sin A sin B
(2)
cos(A - B) = cos A cos B + sin A sin B
If we add (1) and (2) and divide by 2 we get
(3)
cos A cos B =
cos(A + B) + cos(A - B)
2
If we subtract (1) from (2) and divide by 2 we get
(4)
sin A sin B =
cos(A - B) - cos(A + B)
2
The addition and subtraction identities for sine are
(5)
sin(A + B) = sin A cos B + cos A sin B
(6)
sin(A - B) = sin A cos B - cos A sin B
If we add (1) and (2) and divide by 2 we get
(7)
sin A cos B =
sin(A + B) + sin(A - B)
2
The identities (3), (4) and (7) are very useful for integrating products of sine's and cosine's. Here they are
grouped together.
(8)
cos A cos B =
cos(A + B) + cos(A - B)
2
sin A sin B =
cos(A - B) - cos(A + B)
2
sin A cos B =
sin(A + B) + sin(A - B)
2
A special case is when A = B. Since cos 0 = 1 and sin 0 = 1, the identities (8) become
1.3 - 1
cos2 A =
1 + cos 2A
2
sin2A =
1 - cos 2A
2
(9)
sin A cos A =
sin 2A
2
/2

 sin x cos 2x dx.
Example 1. Find
0
1
1
1
1
Solution. Using the third identity in (8) one has sin x cos 2x = 2 sin 3x + 2 sin(- x) = 2 sin 3x - 2 sin x, where
/2
we also used sin(- x) = - sin x. So
/2

 sin x cos 2x dx =
1
2
0
/2
1
- 6 cos 3x |
0
1
/2
1
/2

 sin 3x dx 0
3
1
1
1
2

 sin x dx =
0

1
1
1
1
+ 2 cos x |
= - 6 cos 2 + 6 cos 0 + 2 cos 2 - 2 cos 0 = 6 - 2 = - 3.
0
When we compute the Fourier series of a function f(x) we need to evaluate the integrals 
 f(x) cos nx dx
and 
 f(x) sin nx dx. In the case where f(x) is a quadratic function we have the following two formulas.
(ax2 + bx + c) sin nx
(2ax + b) cos nx 2a sin nx
+
n
n2
n3
(10)
2

 (ax + bx + c) cos nx dx =
(11)
 (ax2 + bx + c) sin nx dx = 
(ax2 + bx + c) cos nx
(2ax + b) sin nx
2a cos nx
+
+
n
n2
n3
To prove (10), integrate by parts letting u = ax2 + bx + c, du = (2ax + b) dx, dv = cos nx dx and v =
This gives
2

 (ax + bx + c) cos nx dx =
(ax2 + bx + c) sin nx
n
1.3 - 2


(2ax + b) sin nx
dx
n
sin nx
.
n
Integrate by parts again letting u = 2ax + b, du = 2a dx, dv =
2

 (ax + bx + c) cos nx dx =
=
sin nx
cos nx
dx and v = giving
n
n2
(ax2 + bx + c) sin nx
(2ax + b) cos nx
+
n
n2


2a cos nx
dx
n2
(ax2 + bx + c) sin nx
(2ax + b) cos nx 2a sin nx
+
n
n2
n3
which proves (10). The proof of (11) is similar.
1.3 - 3