Can somebody help me simplify this?

(cos2x+sin^2x)/(cos^2x)

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- Jun 1st 2008, 06:14 PMunmcintoshHomework help
Can somebody help me simplify this?

(cos2x+sin^2x)/(cos^2x) - Jun 1st 2008, 06:25 PMtopsquark
$\displaystyle \frac{cos(2x) + sin^2(x)}{cos^2(x)}$

$\displaystyle = \frac{(2~cos^2(x) - 1) + sin^2(x)}{cos^2(x)}$

Upon some rearranging in the numerator:

$\displaystyle \frac{(cos^2(x) + sin^2(x)) + cos^2(x) - 1}{cos^2(x)}$

$\displaystyle \frac{1 + cos^2(x) - 1}{cos^2(x)}$

$\displaystyle \frac{cos^2(x)}{cos^2(x)}$

$\displaystyle = 1$

-Dan - Jun 1st 2008, 06:37 PMunmcintosh
thanks for the help and everything, but could you or someone else explain how you got from the first step to the second please?

- Jun 1st 2008, 06:46 PMtopsquark
$\displaystyle cos(2x) = 2~cos^2(x) - 1$

We also have that

$\displaystyle cos(2x) = 1 - 2~sin^2(x) = cos^2(x) - sin^2(x)$

as well.

See here.

-Dan - Jun 1st 2008, 07:18 PMunmcintosh
thank you. my test is tomorrow so im basically screwed. Also, could somebody help me verify

-2cos^2xtan(-x)=sin2x

please? - Jun 1st 2008, 09:38 PMChris L T521
$\displaystyle -2\cos^2x\tan(-x)=\sin(2x)$

$\displaystyle \implies -2\cos^2x\frac{\sin(-x)}{\cos(-x)}=\sin(2x)$

$\displaystyle \implies -2\cos^2x\frac{-\sin(x)}{\cos(x)}=\sin(2x)$

$\displaystyle \implies -2\cos(x)(-\sin(x))=\sin(2x)$

$\displaystyle \implies 2\sin(x)\cos(x)=\sin(2x)$

This is the case, for this is a double angle identity.

Hope this helped! :D