## anonymous 5 years ago I need help solving: If sin(theta)=12/13, cos >0 then what would cos(2theta) be?

1. anonymous

$\sin \theta=12/13, \cos \theta>0 Find \cos(2\theta).$

2. anonymous

$cos(2\theta) = sin^2\theta - cos^2\theta$ $cos\theta = +\sqrt{1-sin^2\theta}$ $sin\theta = 12/3$ Plug the bottom 2 into the top 1 and solve.

3. anonymous

Where did 12/3 come from? I'm confused

4. anonymous

Sorry, should have been 12/13. It's given to you

5. anonymous

What do you mean by plug the bottom two into the top 1?

6. anonymous

You have an equation for $$sin\theta$$ plug that into the equation for $$cos \theta$$. Then solve for $$cos \theta$$ and plug both of those into the first equation to find $$cos 2\theta$$.

7. anonymous

Does that make sense?

8. anonymous

I'm still trying to figure it out! Lol

9. anonymous

Okay so how do I verify that this is an identity $\cos(4u)= 2\cos^2(2u)-1$