sin(x+y)*sin(x-y) = sin2(x) - sin2(y)
Let's work on the left side and make it look like the right side. I see some sines in common, but we have sine squared terms on the right. Maybe we'll use the Pythagorean Identity at some point!
Using sum and difference formulas:
(sinxcosy + sinycosx)(sinxcosy - sinycosx) = sin2(x) - sin2(y)
FOIL and cancel middle terms:
sin2(x)cos2(y) + sinycosx*sinxcosy - sinxcosy*sinycosx - sin2(y)cos2(x)
Replace cosine terms using Pythagorean Identity:
sin2(x) + cos2(x) = 1 → cos2(x) = 1 - sin2(x)
Substitute:
sin2(x)(1 - sin2(y)) - sin2(y)(1 - sin2(x))
Distribute and cancel like terms:
sin2(x) - sin2(x)sin2(y) - sin2(y) + sin2(y)sin2(x) = sin2(x) - sin2(y)
