Can somebody help me simplify this?
(cos2x+sin^2x)/(cos^2x)
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Can somebody help me simplify this?
(cos2x+sin^2x)/(cos^2x)
$\displaystyle \frac{cos(2x) + sin^2(x)}{cos^2(x)}$Quote:
Originally Posted by unmcintosh
$\displaystyle = \frac{(2~cos^2(x) - 1) + sin^2(x)}{cos^2(x)}$
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
thanks for the help and everything, but could you or someone else explain how you got from the first step to the second please?
$\displaystyle cos(2x) = 2~cos^2(x) - 1$Quote:
Originally Posted by unmcintosh
We also have that
$\displaystyle cos(2x) = 1 - 2~sin^2(x) = cos^2(x) - sin^2(x)$
as well.
See here.
-Dan
thank you. my test is tomorrow so im basically screwed. Also, could somebody help me verify
-2cos^2xtan(-x)=sin2x
please?
$\displaystyle -2\cos^2x\tan(-x)=\sin(2x)$Quote:
Originally Posted by unmcintosh
$\displaystyle \implies -2\cos^2x\frac{\sin(-x)}{\cos(-x)}=\sin(2x)$
$\displaystyle \implies -2\cos^2x\frac{-\sin(x)}{\cos(x)}=\sin(2x)$
$\displaystyle \implies -2\cos(x)(-\sin(x))=\sin(2x)$
$\displaystyle \implies 2\sin(x)\cos(x)=\sin(2x)$
This is the case, for this is a double angle identity.
Hope this helped! :D
