Verify: sin (x+y)cos(x-y)=sin(x) cos(x)+cos(y)sin(y)

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Verify: sin (x+y)cos(x-y)=sin(x) cos(x)+cos(y)sin(y)

Mathematics
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use the compound angle formula to expand the left side then simplify. I think you will be able to make use of the identity sin^2 x + cos^2 x = 1
sin(x + y) = sin x cos y + sin y cos x cos (x - y) = cos x cos y + sin x sin y
Well once it's broken down where do I go from there?

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you should get what is on the right side so proving the identity
its a bit long winded but that is the way to do it
Can u show me all the way to the answer? My brain is ready to fall outta my ears from trying to figure this out.
Ehhh I can maybe show you a few steps -_- let's see here....
\[\large\rm \color{orangered}{\sin(x + y) = \sin x \cos y + \sin y \cos x}\]\[\large\rm \color{royalblue}{\cos (x - y) = \cos x \cos y + \sin x \sin y}\]We'll apply these identities to our problem:\[\large\rm \color{orangered}{\sin(x+y)}\color{royalblue}{\cos(x-y)}\]Which will give us:\[\large\rm \color{orangered}{\left[\sin x \cos y + \sin y \cos x\right]}\color{royalblue}{\left[\cos x \cos y + \sin x \sin y\right]}\]Ok with that first step? :)
Yes I understand that!
So hmm.. I guess we have to FOIL from here.
\[\rm =\color{orangered}{\sin x \cos y}\color{royalblue}{\cos x \cos y}+\color{orangered}{\sin x \cos y}\color{royalblue}{\sin x \sin y}\\+\color{orangered}{\sin y \cos x}\color{royalblue}{\cos x \cos y}+\color{orangered}{\sin y \cos x}\color{royalblue}{\sin x \sin y}\]Which becomes this I suppose.

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