How do you prove sin(alpha+beta)sin(alpha-beta)=sin^2alpha-sin^2beta?

1 Answer
Will
Aug 19, 2016

= sin^2(alpha) -sin^2(beta)

Explanation:

sin(alpha+beta)sin(alpha-beta) =
sin(alpha)cos(beta) + cos(alpha)sin(beta)*sin(alpha)cos(beta) - cos(alpha)sin(beta)

now multiply
(sin(alpha)cos(beta))^2 + cancel(cos(alpha)sin(beta)sin(alpha)cos(beta)) - cancel(cos(alpha)sin(beta)sin(alpha)cos(beta))-(cos(alpha)sin(beta))^2

then replace the cosine
sin^2(alpha)(1-sin^2(beta)) -sin^2(beta)(1-sin^2(alpha))
= sin^2(alpha) cancel(-sin^2(alpha)sin^2(beta) )-sin^2(beta) + cancel(sin^2(alpha)sin^2(beta))
= sin^2(alpha) -sin^2(beta)

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