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

**simonsong** · Jul 15th 2008, 06:53 AM

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

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

**flyingsquirrel** · Jul 15th 2008, 07:08 AM

Hi

Originally Posted by simonsong

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

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

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

This should give you two values for $\cos \frac{x}{2}$ . As you also know in which quadrant is $\frac{x}{2}$ , you can choose the right solution among the two you've found.

**simonsong** · Jul 15th 2008, 07:14 AM

Omg. Never thought of that. Haha. Really thanks pal.

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