I need to find gamma but I'm stuck...

Question

anonymous · Answer

you need to find what:

anonymous · Answer

$$\cos ^{2}(\frac{ \pi }{ 3 })+\cos ^{2}(\frac{ \pi }{ 3 })+\cos ^{2}(\gamma)=1$$
$$\frac{ 1 }{ 4 }+\frac{ 1 }{ 4 }+\cos ^2(\gamma)=1$$
$$\cos^2(\gamma)=\frac{ 1 }{ 2 }$$

anonymous · Answer

oh

luigi0210 · Answer

Have you tried converting the $cos^2(\gamma)$?

anonymous · Answer

I don't know which trig identity to use

luigi0210 · Answer

$\Large \frac{1}{2}(cos(2u)+1)$

anonymous · Answer

How did you get that?

luigi0210 · Answer

http://www.purplemath.com/modules/idents.htm Under "Half-Angles"

anonymous · Answer

my mama did

anonymous · Answer

Ok now I have $$\cos (2\gamma)=1$$ How would I be able to get gamma by itself if the 2 next to it?

anonymous · Answer

$$\cos^2(\gamma)=\frac{ 1 }{ 2 }$$
Take the square root:
$$\cos(\gamma)=\pm\sqrt\frac{ 1 }{ 2 }$$
There are two angles in the unit circle that have a cosine of $\dfrac{1}{\sqrt2}$ and $-\dfrac{1}{\sqrt2}$. What are they?

anonymous · Answer

Or, if you're sticking to the identity Luigi posted,
$$\cos2\gamma=1$$
Only one angle has a cosine of 1. What is it?

anonymous · Answer

angle 0

anonymous · Answer

Right, but the angle in this case is $2\gamma=0$, which means $\gamma=0$. Note that $2\gamma=2\pi$ is also a solution, as are $2\gamma=-2\pi,\pm4\pi,...$. This means the general solution would be $\gamma=0,\pm\pi,\pm2\pi...$. This is usually written as $\gamma=n\pi$, where $n$ is an integer.

anonymous · Answer

I was thrown off because my tutor told me I couldn't use square root and the way the identity was written threw me off too. Thx!