[SOLVED] Double angle problem 2 (quick squaring question)

it says to solve for cos pi/3 using a double angle identity.
I dont understand whats wrong.
I do
cos(pi/3)
cos(2(pi/6)) $<br /> = cos^2(pi/6)times sin^2(pi/6)$ This is where im stuck, what do you do when you have a squared sin/ cos, I know sin of pi/6 = 1/2 and cos = root 3 over 2 so dont you just square those so
3/4 times 1/4?
that gives the wrong answer though its supposed to be 1/2

thanks.

Quote:

Originally Posted by brentwoodbc

it says to solve for cos pi/3 using a double angle identity.
I dont understand whats wrong.
I do
cos(pi/3)
cos(2(pi/6)) $\textcolor{red}{<br /> = cos^2(pi/6)times sin^2(pi/6)}$ This is where im stuck, what do you do when you have a squared sin/ cos, I know sin of pi/6 = 1/2 and cos = root 3 over 2 so dont you just square those so
3/4 times 1/4?
that gives the wrong answer though its supposed to be 1/2

thanks.

correction ...

$\cos^2\left(\frac{\pi}{6}\right) - \sin^2\left(\frac{\pi}{6}\right)$

thanks.