1. ## Homework help

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

2. Originally Posted by unmcintosh
Can somebody help me simplify this?
(cos2x+sin^2x)/(cos^2x)
$\frac{cos(2x) + sin^2(x)}{cos^2(x)}$

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

Upon some rearranging in the numerator:
$\frac{(cos^2(x) + sin^2(x)) + cos^2(x) - 1}{cos^2(x)}$

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

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

$= 1$

-Dan

3. thanks for the help and everything, but could you or someone else explain how you got from the first step to the second please?

4. 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?
$cos(2x) = 2~cos^2(x) - 1$

We also have that
$cos(2x) = 1 - 2~sin^2(x) = cos^2(x) - sin^2(x)$
as well.

See here.

-Dan

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

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

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

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

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

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

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

Hope this helped!