# Thread: Proof of trig identities

1. ## Proof of trig identities

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

used double angle formulae except seemed to only get up to
cossquaredx cossquared y - sin squared x sin squared y ??

any help is GREATLY APPRECIATED

thankss !!

2. Hello, iiharthero!

Prove: .$\displaystyle \cos(x+y)\cos(x-y) \:=\: \cos^2\!y - \sin^2\!x \:=\:\cos^2\!x - \sin^2\!y$

Used compound-angle formulae except seemed to only get up to:

. . $\displaystyle \cos^2\!x\cos^2\!y - \sin^2\!x\sin^2\!y$ . .
This is correct -- Good work!

$\displaystyle \text{We have: }\;\,\cos^2\!x\underbrace{\cos^2\!y} \quad\; - \;\quad\sin^2\!x\underbrace{\sin^2\!y}$
. . . .$\displaystyle = \;\cos^2\!x\overbrace{(1-\sin^2\!y)} - \sin^2y\overbrace{(1 - \cos^2\!x)}$

. . . .$\displaystyle = \;\cos^2\!x - \cos^2\!x\sin^2\!y - \sin^2\!y + \cos^2\!x\sin^2\!y$

. . . .$\displaystyle =\;\cos^2\!x - \sin^2\!y$

Got it?

3. THANKU !
yupp that was a great helpp, realli realli appreciate it !
thanx for takin da time to help me =DD

4. Originally Posted by iiharthero
THANKU !
yupp that was a great helpp =DD
thank you.