prove sin(α+β)sin(α-β)=cos^2β-cos^2α
To prove the identity sin(α+β)sin(α-β) = cos^2β - cos^2α, we need to start with the left-hand side (LHS) and manipulate it until we get the right-hand side (RHS). Here's how:
Starting with LHS:
sin(α+β)sin(α-β)
We can use the product-to-sum identity for sine to rewrite this expression:
(sin(α)cos(β) + cos(α)sin(β))(sin(α)cos(β) - cos(α)sin(β))
Now, let's expand and simplify the expression:
(sin(α)cos(β))^2 - (cos(α)sin(β))^2
Using the trigonometric identity sin^2(θ) + cos^2(θ) = 1, we get:
(sin^2(α)cos^2(β) - cos^2(α)sin^2(β))
Rearranging the terms:
(sin^2(α)cos^2(β)) - (cos^2(α)sin^2(β))
Using the identity sin^2(θ) = 1 - cos^2(θ), we can further simplify:
(cos^2(β) - cos^2(α))(1 - sin^2(β))
Finally, applying the trigonometric identity 1 - sin^2(θ) = cos^2(θ), we obtain:
(cos^2(β) - cos^2(α))(cos^2(β))
This is the same as the RHS, which completes the proof:
cos^2(β) - cos^2(α) = cos^2(β) - cos^2(α)
Thus, we have proved the identity sin(α+β)sin(α-β) = cos^2β - cos^2α.