Question

$Howdoyouprove:sin(A+B)\cdot sin(A-B)=sin2A-sin2B?$

Answer 1

$Thestandardformulaforsin(A+B)is:sin(A+B)=sin\left(A\right)cos\left(B\right)+cos\left(A\right)sin\left(B\right)Nowsin(-B)=-sin\left(B\right)andcos(-B)=cos\left(B\right),sosin(A-B)=sin\left(A\right)cos\left(B\right)-cos\left(A\right)sin\left(B\right)So:sin(A+B)\cdot sin(A-B)=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)=(sinAcosB{)}^2-(cosAsinB{)}^2...$

$usingtheidentity$

$(p+q)(p-q)=p^2-q^2$

$=sin2Acos2B-sin2Bcos2A$

$=sin2A(1-sin2B)-sin2B(1-sin2A)$

...using $si{n}^2\theta +co{s}^2\theta =1$

$=sin2A-sin2B-sin2Asin2B+sin2Bsin2A$

$=sin2A-sin2B$