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.