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) #

Related questions
See all questions in Proving Identities
Impact of this question
15442 views around the world
You can reuse this answer
Creative Commons License