-2cos(B+C).sin(B-C)=sin2C-sin2B

Question

Prove that

Answer 1

cos(B+C)=cosBcosC-sinBsinC; sin(B-C)=sinBcosC-cosBsinC (trig identities).

So cos(B+C).sin(B-C)=sinBcosBcos^2C-cos^2BsinCcosC-sin^2BsinCcosC+sinBcosBsin^2C=

sinBcosB(cos^2C+sin^2C)-sinCcosC(cos^2B+sin^2B)=sinBcosB-sinCcosC=

(1/2)sin2B-(1/2)sin2C or -(1/2)(sin2C-sin2B). Therefore -2cos(B+C).sin(B-C)=sin2C-sin2B.

Answer 2

Given -2cos(B+C).sin(B-C)=sin2C-sin2B

LHS -2cos(B+C).sin(B-C)

-2(cos B cos c-sin B sin c)(sin b cos c-cos B sin c)

Solve it further we'll have

sin2C-sin2B

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