Question

Please show that; Sin(x+y)cos(x-y) = sin x cos x + sin y cos y

Answer 1

Apply the rules of sum/difference for trigonometric functions:

$\sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y)$

$\cos(x-y) = \sin(x) \sin(y) + \cos(x) \cos(y)$

Multiply the two expressions:

$(\sin(x) \cos(y) + \cos(x) \sin(y)) (\sin(x) \sin(y) + \cos(x) \cos(y)) =$

$\sin(x) \cos(y)\sin(x) \sin(y) + \sin(x) \cos(y)\cos(x) \cos(y) + \cos(x) \sin(y)\sin(x) \sin(y) + \cos(x) \sin(y)\cos(x) \cos(y) =$

$\sin^2(x) \cos(y)\sin(y) + \sin(x) \cos^2(y)\cos(x)+ \cos(x) \sin^2(y)\sin(x) + \cos^2(x) \sin(y)\cos(y)$

You can factor $\cos(y)\sin(y)$ from the first and last term, and $\sin(x)\cos(x)$ from the two middle terms: you have

$\cos(y)\sin(y)(\sin^2(x)+\cos^2(x)) + \sin(x) \cos(x)(\cos^2(y)+\sin^2(y))$

Since $\sin^2(x)+\cos^2(x)=1$, the identity is proven.