# need help on prooving trig identities!

• Dec 12th 2007, 04:39 PM
zarlock99
need help on prooving trig identities!
Ok so heres the question,
Prove:
1-cos2x-1/2cos^2x=sec^2x
I got this far but i dont know what do after that

1-(2cos^2x-1)-1/2cos^2x

Can ne one help me???
• Dec 12th 2007, 05:31 PM
zarlock99
?????
anyone?????
• Dec 12th 2007, 06:06 PM
brueneboy
do you really mean to the power of 2x ? maybe im not high enough in math yet then
• Dec 12th 2007, 06:16 PM
Krizalid
Quote:

Originally Posted by zarlock99
Prove:
1-cos2x-1/2cos^2x=sec^2x

Please, use parenthesis. May be that's why the question has not been quickly answered.
• Dec 12th 2007, 06:20 PM
zarlock99
Um
Quote:

Originally Posted by Krizalid

Please, use parenthesis. May be that's why the question has not been quickly answered.

Thats how it shows it in the question, there is no parenthesis. How the hell do u think that question is incomprehensible
• Dec 12th 2007, 06:54 PM
brueneboy
algebraically you can add parenthesis where you need to, also it is hard to tell what is to what power and what is being divided by what, take a picture of it, that might help some of us
• Dec 12th 2007, 07:00 PM
zarlock99
That better?
• Dec 12th 2007, 07:12 PM
brueneboy
is the first 1 above, below, or just to the left of the rest of the problem?
• Dec 12th 2007, 07:14 PM
zarlock99
um
Im trying to prove that the Left hand side equals the right hand side, thus all of that has to equal sec^2x
• Dec 13th 2007, 04:12 AM
Krizalid
Quote:

Originally Posted by zarlock99
$1 - \frac{{\cos 2x - 1}}
{{2\cos ^2 x}} = \sec ^2 x$

Okay, this makes more sense.

Since $\cos 2x = 2\cos ^2 x - 1,$ the LHS becomes

$1 - \frac{{\cos ^2 x - 1}}
{{\cos ^2 x}} = 1 - 1 + \frac{1}
{{\cos ^2 x}} = \sec ^2 x.$

End of the history.