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 !!
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 !!
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?