# Another Trig Identity problem.

• Nov 28th 2010, 08:25 PM
Diesal
Another Trig Identity problem.
I need to know in steps how to prove this identity, if anyone minds helping. Thanks!

http://img4.imageshack.us/img4/3883/codecogseqn2.gif
• Nov 28th 2010, 08:30 PM
harish21
$\displaystyle \tan(x)=\dfrac{\sin(x)}{\cos(x)}$

$\displaystyle \therefore \cos(x)\sin(x)\tan(x)=\cos(x)\sin(x)\dfrac{\sin(x) }{\cos(x)}=....$
• Nov 28th 2010, 08:41 PM
Diesal
Quote:

Originally Posted by harish21
$\displaystyle \tan(x)=\dfrac{\sin(x)}{\cos(x)}$

$\displaystyle \therefore \cos(x)\sin(x)\tan(x)=\cos(x)\sin(x)\dfrac{\sin(x) }{\cos(x)}=....$

Can't solve?
• Nov 28th 2010, 08:45 PM
harish21
Are you even attempting?

You should be able to recognize that the cos(x) in the numerator and the denominator cancel each other out. So you are left with $\displaystyle \cos(x)\sin(x)\dfrac{\sin(x)}{\cos(x)} = \sin(x)\sin(x)=\sin^2(x)$

You have been told in your previous post that $\displaystyle sin^2(x)+cos^2(x)=1$. based on that, $\displaystyle \sin^2(x)= ??$
• Nov 28th 2010, 08:53 PM
Diesal
I didn't realize they canceled each other out, but when I did the equation myself I canceled them out anyway. When I was left with http://latex.codecogs.com/gif.latex?...space;1-cos^2x I just added the negative to the other side to make a positive, thus making it all equal one. I thought I was doing it wrong though, since I didn't know they canceled. Thanks for clarification.