Another Trig Identity problem.

**Diesal** · November 28th 2010, 08:25 PM

I need to know in steps how to prove this identity, if anyone minds helping. Thanks!

**harish21** · November 28th 2010, 08:30 PM

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

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

**Diesal** · November 28th 2010, 08:41 PM

Originally Posted by harish21

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

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

Can't solve?

**harish21** · November 28th 2010, 08:45 PM

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 $\cos(x)\sin(x)\dfrac{\sin(x)}{\cos(x)} = \sin(x)\sin(x)=\sin^2(x)$

You have been told in your previous post that $sin^2(x)+cos^2(x)=1$ . based on that, $\sin^2(x)= ??$

**Diesal** · November 28th 2010, 08:53 PM

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 $sin^2x = 1-cos^2x$ 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.

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