Question 138819
Note: I'm only manipulating the left side. I'm not touching the right side. I'm only showing it for comparison.




*[Tex \LARGE \sin\left(a+B\right)\sin\left(a-B\right)=\sin^2\left(a\right)-\sin^2\left(B\right)] Start with the given equation



*[Tex \LARGE \left(\sin\left(a\right)\cos\left(B\right)+\cos\left(a\right)\sin\left(B\right)\right)\left(\sin\left(a\right)\cos\left(B\right)-\cos\left(a\right)\sin\left(B\right)\right)=\sin^2\left(a\right)-\sin^2\left(B\right)] Expand by using the Sum-Difference Formulas



*[Tex \LARGE \sin^2\left(a\right)\cos^2\left(B\right)-\sin\left(a\right)\cos\left(a\right)\sin\left(B\right)\cos\left(B\right)+\sin\left(a\right)\cos\left(a\right)\sin\left(B\right)\cos\left(B\right)-\cos^2\left(a\right)\sin^2\left(B\right)=\sin^2\left(a\right)-\sin^2\left(B\right)] Foil



*[Tex \LARGE \sin^2\left(a\right)\cos^2\left(B\right)-\cos^2\left(a\right)\sin^2\left(B\right)=\sin^2\left(a\right)-\sin^2\left(B\right)] Combine like terms



*[Tex \LARGE \sin^2\left(a\right)\left(1-\sin^2\left(B\right)\right)-\left(1-\sin^2\left(a\right)\right)\sin^2\left(B\right)=\sin^2\left(a\right)-\sin^2\left(B\right)]  Replace *[Tex \LARGE \cos^2\left(B\right)] with *[Tex \LARGE 1-\sin^2\left(B\right)]. Replace *[Tex \LARGE \cos^2\left(a\right)] with *[Tex \LARGE 1-\sin^2\left(a\right)]





*[Tex \LARGE  \sin^2\left(a\right)-\sin^2\left(a\right)\sin^2\left(B\right)-\sin^2\left(B\right)+\sin^2\left(a\right)\sin^2\left(B\right)=\sin^2\left(a\right)-\sin^2\left(B\right)] Distribute



*[Tex \LARGE  \sin^2\left(a\right)-\sin^2\left(B\right)=\sin^2\left(a\right)-\sin^2\left(B\right)] Combine like terms





So we've just proven that 


*[Tex \LARGE \sin\left(a+B\right)\sin\left(a-B\right)=\sin^2\left(a\right)-\sin^2\left(B\right)] 


is an identity.