Prove sin(α+β)sin(α-β)=cos^2β-cos^2α
First off these are usually not this hard...just saying...ill be changing your alphas to x's and your betas to y's
we will work with the left side because it is easier.
use identity sum and difference
sin(x+y)sin(x+y)=cos^2y-cos^2x
s
(sinxcosy+cosxsiny)(sinxcosy-cosxsiny)=
Then, foil those and your middle two terms should cancel out leaving you with
sin^2xcos^2y-cos^2xsin^2y=
now using the pythagereom theorem change all the sin^2's to cos^2 giving you
cos^2y-cos^2xcos^2y-[cos^2xcos^2xcos^2y]=
cos^2y-cos^2xcos^2y-cos^2x+cos^2xcos^2y=
cos^2y-cos^2x
To prove the identity sin(α + β)sin(α - β) = cos^2β - cos^2α, we can use the trigonometric identities. Let's break it down step by step:
1. Start with the left side of the equation: sin(α + β)sin(α - β).
2. Expand the left side using the product-to-sum identity: sin(α + β)sin(α - β) = (1/2)[cos((α + β) - (α - β)) - cos((α + β) + (α - β))].
3. Simplify the equation further: (1/2)[cos(2β) - cos(2α)].
4. To further simplify, use the identity cos(2θ) = 1 - 2sin^2θ: (1/2)[1 - 2sin^2β - 1 + 2sin^2α].
5. Combine the terms: (1/2)[2sin^2α - 2sin^2β].
6. Factor out a -1: (1/2)[-2(sin^2β - sin^2α)].
7. Use the identity sin^2θ - sin^2φ = cos(2φ) - cos(2θ): (1/2)[-2(cos2α - cos2β)].
8. Simplify further: (1/2)[-2(cos2α - cos2β)] = -[cos2α - cos2β].
9. Finally, use the identity cos(2θ) = 2cos^2θ - 1: -[cos2α - cos2β] = -(2cos^2α - 1 - 2cos^2β + 1).
10. Combine like terms: -2cos^2α + 2cos^2β.
11. Rearrange the terms: 2cos^2β - 2cos^2α.
12. This expression is equivalent to cos^2β - cos^2α, which is the right side of the equation.
Therefore, we have proven the identity sin(α + β)sin(α - β) = cos^2β - cos^2α.