Question

Verify sin(x+y)cos(x-y)=sin(x)cos(x)+sin(y)cos(y) using trigonometric functions and identities. I keep getting the same answer as I start out with...

Answer 1

use sum and difference identities
\[\sin(a+b) = \sin(a)\cos(b)+\sin(b)\cos(a)\]
\[\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)\]

Answer 2

\[\sin(x+y)\cos(x-y)\]\[[\sin(x)\cos(y)+\cos(x)\sin(y)]*[\cos(x)\cos(y)+\sin(x)\sin(y)]\]\[[\sin(x)\cos(x)\cos^2(y)+\sin^2(x)\cos(y)\sin(y)+\cos^2(x)\cos(y)\sin(y)+\sin^2(y)\cos(x)\sin(x)]\]\[[(\cos^2y+\sin^2y)(sinXcosX+sinXcosX)+(\cos^2x+\sin^2x)(sinYcosY+sinYcosY)]\]\[\cos^2x+\sin^2x=1 and \cos^2y+\sin^2y=1\]\[[2sinXcosX+2sinYcosY]\]I don't know but it is not equal to that. Please check

Answer 3

That's what I got... I appreciate your time and efforts on my behalf! ;D

Answer 4

lol you're welcome

Answer 5

quick correction
when you factored out the (sin^2 + cos^2) parts , you should be left with
\[(\sin^{2} x + \cos^{2}x)\sin y \cos y + (\sin^{2} y + \cos^{2} y) \sin x \cos x\]
\[= \sin y \cos y + \sin x \cos x\]

Answer 6

Thank you SOO much! I spent about an hour trying to figure out what I did wrong. I can finally go to bed!

Answer 7

yw