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

• May 7th 2009, 04:33 PM
brentwoodbc
[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)) $
= 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.
• May 7th 2009, 04:55 PM
skeeter
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)) $\textcolor{red}{
= cos^2(pi/6)times sin^2(pi/6)}$
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 ...

$\cos^2\left(\frac{\pi}{6}\right) - \sin^2\left(\frac{\pi}{6}\right)$
• May 7th 2009, 04:59 PM
brentwoodbc
thanks.