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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?

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  1. anonymous
    • 5 years ago
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    \[\sin \theta=12/13, \cos \theta>0 Find \cos(2\theta). \]

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

  3. anonymous
    • 5 years ago
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    Where did 12/3 come from? I'm confused

  4. anonymous
    • 5 years ago
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    Sorry, should have been 12/13. It's given to you

  5. anonymous
    • 5 years ago
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    What do you mean by plug the bottom two into the top 1?

  6. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Does that make sense?

  8. anonymous
    • 5 years ago
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    I'm still trying to figure it out! Lol

  9. anonymous
    • 5 years ago
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    Okay so how do I verify that this is an identity \[\cos(4u)= 2\cos^2(2u)-1\]

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