Hello, all.

I need to find the value of $\displaystyle cos(\frac{17\pi}{12})$ using the half angle formula for cos.

Here's what I've done.

I found that the reference angle for $\displaystyle cos(\frac{17\pi}{12})$ is $\displaystyle cos(\frac{\pi}{12})$ in quadrant II, making the sign for the formula negative.

So I've got:

$\displaystyle -\sqrt{ \frac{1+cos(\frac{\pi}{6})}{2} }$

I evaluate this to:

$\displaystyle -\frac{ \sqrt{ 2+\sqrt{3} } }{2}$

This is incorrect, but I'm not sure how to arrive at the correct answer, which is:

$\displaystyle \frac{\sqrt{2-\sqrt{3}}}{2}$

Thanks for the help!