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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sin \theta=12/13, \cos \theta>0 Find \cos(2\theta). \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[cos(2\theta) = sin^2\theta  cos^2\theta\] \[cos\theta = +\sqrt{1sin^2\theta}\] \[sin\theta = 12/3\] Plug the bottom 2 into the top 1 and solve.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where did 12/3 come from? I'm confused

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, should have been 12/13. It's given to you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you mean by plug the bottom two into the top 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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 \(cos 2\theta\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm still trying to figure it out! Lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay so how do I verify that this is an identity \[\cos(4u)= 2\cos^2(2u)1\]
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