Question #d6d64

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Answer 1 · 2017-12-06T15:18:49

see below

to prove

#sin(A+B)sin(A-B)=sin^2A-sin^2B#

#LHSrarrsin(A+B)sin(A-B)#

#=(sinAcosB+sinBcosA)(sinAcosB-sinBcosA)#

by difference of squares

#=sin^2Acos^2B-sin^2Bcos^2A--(1)#

using

#sin^2x+cos^2x=1#

#(1)rarrsin^2A (1-sin^2B)-sin^2B(1-sin^2A)#

#sin^2A-cancel(sin^2Asin^2B)-sin^2B+cancel(sin^2Bsin^2A)#

#=sin^2A-sin^2B#

#=RHS#