What trigonometric identity makes the following true? $\displaystyle \sin(x)*\sin(y)=\frac{1}{2}[\cos(x-y)-\cos(x+y)]$
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Originally Posted by Mike9182 What trigonometric identity makes the following true? $\displaystyle \sin(x)*\sin(y)=\frac{1}{2}[\cos(x-y)-\cos(x+y)]$ $\displaystyle \frac{1}{2}[\cos(x-y)-\cos(x+y)]$ use the sum\difference identity for cosine ... $\displaystyle \frac{1}{2}[\cos{x}\cos{y} + \sin{x}\sin{y} - (\cos{x}\cos{y} - \sin{x}\sin{y})]$
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