Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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 \(cos 2\theta\).

Answer 7

Does that make sense?

Answer 8

I'm still trying to figure it out! Lol

Answer 9

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