# Trigo question. Help needed. Need to pass up on tomorrow.

• Jul 15th 2008, 06:53 AM
simonsong
Trigo question. Help needed. Need to pass up on tomorrow.
Given that $\displaystyle \cos x = -\frac{2}{3}$ and $\displaystyle \sin y = -\frac{1}{\sqrt 6}$, that $\displaystyle 0^\circ\le x \le360^\circ$ and that $\displaystyle x$ and $\displaystyle y$ are in the same quadrant, find, without calculator, the values of $\displaystyle \cos\frac{x}{2}$.

Man. This question is killing me. I only get to know $\displaystyle x$ and $\displaystyle y$ are in quadrant 3.
• Jul 15th 2008, 07:08 AM
flyingsquirrel
Hi
Quote:

Originally Posted by simonsong
Given that $\displaystyle \cos x = -\frac{2}{3}$ and $\displaystyle \sin y = -\frac{1}{\sqrt 6}$, that $\displaystyle 0^\circ\le x \le360^\circ$ and that $\displaystyle x$ and $\displaystyle y$ are in the same quadrant, find, without calculator, the values of $\displaystyle \cos\frac{x}{2}$.

Man. This question is killing me. I only get to know $\displaystyle x$ and $\displaystyle y$ are in quadrant 3.

$\displaystyle \cos x=\cos \left(\frac{x}{2}+\frac{x}{2}\right)=2\cos^2\frac{ x}{2}-1$ hence $\displaystyle \cos \frac{x}{2}=\ldots$

This should give you two values for $\displaystyle \cos \frac{x}{2}$. As you also know in which quadrant is $\displaystyle \frac{x}{2}$, you can choose the right solution among the two you've found.
• Jul 15th 2008, 07:14 AM
simonsong
Omg. Never thought of that. Haha. Really thanks pal. (Rofl)