Verify that sin(A + B) + sin(A - B) = 2sinA.sinB

To verify this, we will use the trigonometric identity

Answer: Hence, verified that sin(A + B) + sin(A - B) = 2sinA.sinB

Let's see the solution.

Explanation:

Let (A + B) = θ and (A - B) = ϕ

LHS = sin (A + B) + sin (A - B) = sin (θ) + sin (ϕ)

On multiplying and dividing the equation by 2, we get

⇒ 2 (sin (θ) / 2 + sin (ϕ) / 2)

By adding and subtracting θ and ϕ, we get

⇒ θ + ϕ = (A + B) + (A - B) = 2A

⇒ θ - ϕ = (A + B) - (A - B) = 2B

sin X + sin Y = 2 [sin (X + Y) / 2 + sin (X - Y) / 2]

So, sin (A + B) + sin (A - B) = 2 [sin (2A) / 2 + sin (2B) / 2]

Thus, sin (A + B) + sin(A - B) = 2[sin A + sin B]