How do you prove sin(A+B)×sin(A-B)=sin2A-sin2B?
Prove the given equation
We know that sin(A+B)=sin(A)cos(B)+cos(A)sin(B)
And, sin(A-B)=sin(A)cos(B)-cos(A)sin(B)
∴sin(A+B)×sin(A-B)=sin(A)cos(B)+cos(A)sin(B)sin(A)cos(B)-cos(A)sin(B)
=sinAcosB2-cosAsinB2 ∵a+ba-b=a2-b2
=sin2Acos2B-cos2Asin2B
=sin2A1-sin2B-1-sin2Asin2B ∵sin2θ+cos2θ=1⇒1-sin2θ=cos2θ
=sin2A-sin2A×sin2B-sin2B+sin2A×sin2B
⇒sin(A+B)×sin(A-B)=sin2A-sin2B
Hence proved
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