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

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.