# Sometimes, trig identities irk me.

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• Jun 2nd 2009, 03:56 PM
Pyruvate
Sometimes, trig identities irk me.
Here are two that I cannot seem to prove. Any and all help is appreciated.

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

$\displaystyle \sin(2t) - \tan(t) = \tan(t)\cos(2t)$
• Jun 2nd 2009, 04:05 PM
Jhevon
Quote:

Originally Posted by Pyruvate
Here are two that I cannot seem to prove. Any and all help is appreciated.

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

start with the LHS. write $\displaystyle \tan^2 x$ as $\displaystyle \frac {\sin^2 x}{\cos^2 x}$ and combine the fractions in the denominator. you should see the answer almost immediately.

Quote:

$\displaystyle \sin(2t) - \tan(t) = \tan(t)\cos(2t)$
start with the LHS. write $\displaystyle \sin 2t$ as $\displaystyle 2 \sin t \cos t$ and $\displaystyle \tan t$ as $\displaystyle \frac {\sin t}{\cos t}$. simplify. you should see what to do from there

as you are seeing, it is very important to know your standard trig identities for these problems
• Jun 2nd 2009, 04:22 PM
Pyruvate
Oh wow, that first one was pretty obvious. Thank you very much. Unfortunately, for the second problem, I'm getting $\displaystyle 2\tan(t)\cos(2t)$ as an answer.
• Jun 2nd 2009, 04:34 PM
Jhevon
Quote:

Originally Posted by Pyruvate
Oh wow, that first one was pretty obvious. Thank you very much. Unfortunately, for the second problem, I'm getting $\displaystyle 2\tan(t)\cos(2t)$ as an answer.

that's strange. i got the right answer. check your work, make sure you are using the identities correctly
• Jun 2nd 2009, 04:35 PM
Pyruvate
Well, the 2 originated in the problem from the double-angle identity. How did you get rid of it?
• Jun 2nd 2009, 04:49 PM
Jhevon
Quote:

Originally Posted by Pyruvate
Well, the 2 originated in the problem from the double-angle identity. How did you get rid of it?

i didn't get rid of it. i ended up using it in an identity. after simplifying, i could pull out a $\displaystyle {\color{red}2} \cos^2 t - 1$. that is where the cos(2t) came from. that red 2 is the same 2 that came from the sin(2t)
• Jun 2nd 2009, 05:01 PM
Pyruvate
It appears that for an hour, I'd forgotten how to factor. Once again, I thank you for your help.