# Prove that sin(a+b)*sin(a-b)=sin^2a-sin^2b.

### 3 Answers | Add Yours

We have to prove that sin(a + b)*sin(a - b) = (sin a)^2 - (sin b)^2

sin(a + b)*sin(a - b)

=> (1/2)*[ cos (a + b - a + b) - cos (a + b + a - b)]

=> (1/2)*[ cos (b + b) - cos (a + a)]

=> (1/2)[ cos 2b - cos 2a]

cos 2x = 1 - 2*(sin x)^2

=> (1/2)[1 - 2(sin b)^2 - 1 + 2(sin a)^2]

=> (1/2)[- 2(sin b)^2 + 2(sin a)^2]

=> -(sin b)^2 + (sin a)^2

=> (sin a)^2 - (sin b)^2

**This proves : sin(a + b)*sin(a - b) = (sin a)^2 - (sin b)^2**

L:H:S ≡ sin(A+B).sin(A-B)

= 2sin(A+B).sin(A-B)/2

**⇒ use 2sinA.sinB = cos(A-B) - cos(A+B)**

= [cos (A+B-A+B) - cos(A+B+A-B)]/2

= (cos2B - cos2A)/2

**⇒ use cos 2A = 1 - 2sin²A**

* =* [1 - 2sin²B - (1 - 2sin²A )]/2

= (1 - 2sin²B - 1 + 2sin²A)/2

= (2sin²A - 2sin²B)/2

= sin²A - sin²B

Therefore L:H:S ≡ R:H:S

To solve this problem, we'll have to use the following identities:

sin (a+b) = sin a*cos b + sin b*cos a (1)

sin (a-b) = sin a*cos b - sin b*cos a (2)

We'll multiply (1) by (2):

(sin a*cos b + sin b*cos a)(sin a*cos b - sin b*cos a) = (sin a*cos b)^2 - (sin b*cos a)^2

We'll replace (cos a)^2 by 1 - (sin a)^2 and (cos b)^2 by 1 - (sin b)^2

(sin a*cos b + sin b*cos a)(sin a*cos b - sin b*cos a) = (sin a)^2*[1 - (sin b)^2] - (sin b)^2*[1 - (sin a)^2]

We'll remove the brackets:

(sin a*cos b + sin b*cos a)(sin a*cos b - sin b*cos a) = (sin a)^2 - (sin a*sin b)^2 - (sin b)^2 + (sin a*sin b)^2

We'll eliminate like terms:

(sin a*cos b + sin b*cos a)(sin a*cos b - sin b*cos a) = (sin a)^2 - (sin b)^2

**We notice that managing LHS, we'll get the expression of RHS, therefore the identity sin (a+b)*sin (a-b) = (sin a)^2 - (sin b)^2 is verified.**