# Thread: Sometimes, trig identities irk me.

1. ## Sometimes, trig identities irk me.

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

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

$\sin(2t) - \tan(t) = \tan(t)\cos(2t)$

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

$\frac{\cos^2(x) - \sin^2(x)}{1-\tan^2(x)} = \cos^2(x)$
start with the LHS. write $\tan^2 x$ as $\frac {\sin^2 x}{\cos^2 x}$ and combine the fractions in the denominator. you should see the answer almost immediately.

$\sin(2t) - \tan(t) = \tan(t)\cos(2t)$
start with the LHS. write $\sin 2t$ as $2 \sin t \cos t$ and $\tan t$ as $\frac {\sin t}{\cos t}$. 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

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

4. Originally Posted by Pyruvate
Oh wow, that first one was pretty obvious. Thank you very much. Unfortunately, for the second problem, I'm getting $2\tan(t)\cos(2t)$ as an answer.
that's strange. i got the right answer. check your work, make sure you are using the identities correctly

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

6. 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 ${\color{red}2} \cos^2 t - 1$. that is where the cos(2t) came from. that red 2 is the same 2 that came from the sin(2t)

7. It appears that for an hour, I'd forgotten how to factor. Once again, I thank you for your help.