How do you prove sin (A + B) × sin (A - B) = sin² A - sin² B ?

The above expression can be proved by using algebraic identity as well as trigonometric identity.

Answer:  sin (A + B) × sin (A - B) = sin² A - sin² B

Let us proceed step by step.

Explanation:

We already know the formula of sin (A + B) and sin (A − B) from trigonometric identity.

• sin (A + B) = sin (A + B) = sin (A) cos (B) + cos (A) sin (B)
• sin (A − B) = sin (A) cos (B) − cos (A)  sin (B)

Therefore, sin (A + B) × sin (A − B) can be written as 

sin (A + B) × sin (A − B)  = (sin A cos B + cos A sin B) × (sin A cos B − cos A sin B) ------(1)

From algebraic identity, (a + b) × (a − b) = a² – b²

Using the above identity in equation (1) we get,

sin (A + B) × sin (A − B) = (sin A cos B)² − (cos A sin B)²

= sin² A cos² B − sin² B cos² A  [ from trigonometric identity sin² θ + cos² θ = 1 ]

= sin²A (1− sin² B) - sin² B (1− sin² A)

= sin² A − sin² B − sin² A sin² B + sin² B sin²A

= sin²A − sin²B  [ on simplifying the above expression ]

Hence proved, sin (A + B) × sin (A - B) = sin² A – sin² B