# [SOLVED] Proving Trig Identities

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• May 10th 2009, 03:42 PM
demo1
[SOLVED] Proving Trig Identities
I can't seem to solve this one. Could someone please show me?

1 - tan^2X
__________ = cos^2X - sin^2X
1 + tan^2X

Thanks!
• May 10th 2009, 03:51 PM
TheEmptySet
Quote:

Originally Posted by demo1
I can't seem to solve this one. Could someone please show me?

1 - tan^2X
__________ = cos^2X - sin^2X
1 + tan^2X

Thanks!

Multiply the numerator and denominator by $\displaystyle \cos^2(x)$

$\displaystyle \frac{\cos^2(x)}{\cos^2(x)}\left(\frac{1-\tan^2(x)}{1+\tan^2(x)} \right)=\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)+\sin^2(x)}=\cos^2(x)-\sin^2(x)$
• May 10th 2009, 04:02 PM
demo1
Quote:

Originally Posted by TheEmptySet
Multiply the numerator and denominator by $\displaystyle \cos^2(x)$

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

How do you get from $\displaystyle \frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)+\sin^2(x)}$ to here $\displaystyle \cos^2(x)-\sin^2(x)$? Thanks btw!
• May 10th 2009, 04:15 PM
TheEmptySet
Quote:

Originally Posted by demo1
How do you get from $\displaystyle \frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)+\sin^2(x)}$ to here $\displaystyle \cos^2(x)-\sin^2(x)$? Thanks btw!

The pythagorean identity

$\displaystyle \sin^2(x)+\cos^2(x)=1$
• May 10th 2009, 05:22 PM
demo1
Quote:

Originally Posted by TheEmptySet
The pythagorean identity

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

(Headbang)(Headbang)(Headbang) Ah, I really need to review these. Can't believe I didn't see that. Thanks!