Question

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

Answer 1

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

Answer 2

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

Answer 3

Well once it's broken down where do I go from there?

Answer 4

you should get what is on the right side so proving the identity

Answer 5

its a bit long winded but that is the way to do it

Answer 6

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.

Answer 7

Ehhh I can maybe show you a few steps -_- let's see here....

Answer 8

\[\large\rm \color{orangered}{\sin(x + y) = \sin x \cos y + \sin y \cos x}\]\[\large\rm \color{royalblue}{\cos (x - y) = \cos x \cos y + \sin x \sin y}\]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? :)

Answer 9

Yes I understand that!

Answer 10

So hmm.. I guess we have to FOIL from here.

Answer 11

\[\rm =\color{orangered}{\sin x \cos y}\color{royalblue}{\cos x \cos y}+\color{orangered}{\sin x \cos y}\color{royalblue}{\sin x \sin y}\\+\color{orangered}{\sin y \cos x}\color{royalblue}{\cos x \cos y}+\color{orangered}{\sin y \cos x}\color{royalblue}{\sin x \sin y}\]Which becomes this I suppose.