Half Angle Formula problem

• Nov 30th 2009, 07:06 PM
DaGr8Gatzby
Half Angle Formula problem
Given:

$(-1) \over (6)$ = $\cos x$

with 90 < x < 180.

Find:

$\cos (x)$ = $( -1 ) \over (2)$ $X$

So I actually did this:

$\cos^2 (X)$ $=$ $(1) + (-1/6) \over 2$

$(6) - (1) \over 12$

And simplified to:

$1 \over 2$ $\sqrt 15$

But this is not an answer choice. I feel like I have missed a step. Thank you.
• Nov 30th 2009, 07:15 PM
mr fantastic
Quote:

Originally Posted by DaGr8Gatzby
Given:

$(-1) \over (10)$ = $\cos x$

with 90 < x < 180.

Find:

$\cos (x)$ = $( -1 ) \over (2)$ $X$

So I actually did this:

$\cos^2 (X)$ $=$ $(1) + (-1/6) \over 2$

$(6) - (1) \over 12$

And simplified to:

$1 \over 2$ $\sqrt 15$

But this is not an answer choice. I feel like I have missed a step. Thank you.

This makes no sense.
• Nov 30th 2009, 07:23 PM
DaGr8Gatzby
Edited: -1/6
• Nov 30th 2009, 07:26 PM
mr fantastic
Quote:

Originally Posted by DaGr8Gatzby
Edited: -1/6

It still makes no sense. What's X and what's meant to be done with X?
• Nov 30th 2009, 07:48 PM
DaGr8Gatzby
Maybe a screenshot would help:

http://img694.imageshack.us/img694/9...91130at824.png

Still stumped on this one.
• Dec 1st 2009, 02:18 AM
mr fantastic
Quote:

Originally Posted by DaGr8Gatzby
Maybe a screenshot would help:

http://img694.imageshack.us/img694/9...91130at824.png

Still stumped on this one.

You are given $\cos (x) = - \frac{1}{6}$ where $90^0 < x < 180^0$ and asked to find $\cos\left( \frac{1}{2} x\right)$.

From the double angle formula: $\cos (x) = 2 \cos^2 \left( \frac{1}{2} x\right) - 1$.

Substitute $\cos (x) = - \frac{1}{6}$ and simplify: $\cos^2\left( \frac{1}{2} x\right) = \frac{5}{12}$.

$90^0 < x < 180^0 \Rightarrow 45^0 < \frac{1}{2} x < 90^0$ and so the positive root is required: $\cos \left( \frac{1}{2} x \right) = \frac{\sqrt{5}}{\sqrt{12}} = \frac{\sqrt{5}}{2 \sqrt{3}}$.

Now rationalise the denominator. The correct option becomes obvious.