Finding an exact value using half angle.

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May 2nd 2010, 01:35 PM
JennyFlowers

Finding an exact value using half angle.

Hello, all.

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

Here's what I've done.

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

So I've got:

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

I evaluate this to:

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

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

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

Thanks for the help!
May 2nd 2010, 02:19 PM
pickslides

I get

$\cos \frac{17\pi}{12} = \cos\left(\frac{1}{2} \frac{17\pi}{6}\right) \pm\sqrt{ \frac{1+\cos\frac{17\pi}{6}}{2} }$

$=\pm\sqrt{ \frac{1+\cos(\frac{12\pi}{6}+\frac{5\pi}{6})}{2} }<br />$

$=\pm\sqrt{ \frac{1+\cos\frac{5\pi}{6}}{2} }<br />$

$=\pm\sqrt{ \frac{1-\cos\frac{\pi}{6}}{2} }<br />$

$=\pm\sqrt{ \frac{1-\frac{\sqrt{3}}{2}}{2} }<br />$

simplify.
May 2nd 2010, 03:59 PM
JennyFlowers

Since the reference angle is negative, shouldn't we use the ( - ) sign in front of the equation and then place the reference angle inside the equation with a ( + ) sign?

Obviously, my thinking is wrong, but I was hoping someone could explain. :)

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