# Homework help

• Jun 1st 2008, 06:14 PM
unmcintosh
Homework help
Can somebody help me simplify this?
(cos2x+sin^2x)/(cos^2x)
• Jun 1st 2008, 06:25 PM
topsquark
Quote:

Originally Posted by unmcintosh
Can somebody help me simplify this?
(cos2x+sin^2x)/(cos^2x)

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

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

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 PM
unmcintosh
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 PM
topsquark
Quote:

Originally Posted by unmcintosh
thanks for the help and everything, but could you or someone else explain how you got from the first step to the second please?

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

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 PM
unmcintosh
thank you. my test is tomorrow so im basically screwed. Also, could somebody help me verify
-2cos^2xtan(-x)=sin2x
• Jun 1st 2008, 09:38 PM
Chris L T521
Quote:

Originally Posted by unmcintosh
thank you. my test is tomorrow so im basically screwed. Also, could somebody help me verify
-2cos^2xtan(-x)=sin2x

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

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

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

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

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

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

Hope this helped! :D