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Thread: Can someone walk me through this trig problem?

  1. #1
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    Can someone walk me through this trig problem?

    sin(pi/12)=(1/2)(sqrt(A-sqrt(B))). I get the part about the half-angle identity sin(x/2)= +-Sqrt((1-cosx)/(2))). But I dont even know where to start with this problem. I'm supposed to figure out what A and B are.
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  2. #2
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    Re: Can someone walk me through this trig problem?

    Let's write it out. $\displaystyle \sin\left( \dfrac{\pi}{12} \right) = \dfrac{1}{2}\sqrt{A - \sqrt{B}} = \sin\left( \dfrac{x}{2} \right) = \pm \sqrt{\dfrac{1-\cos x}{2}}$. So, let's equate things that look similar. $\displaystyle \sin\left( \dfrac{\pi}{12} \right) = \sin\left( \dfrac{x}{2} \right)$ means that $\displaystyle \dfrac{x}{2} = \dfrac{\pi}{12}$ or $\displaystyle x = \dfrac{\pi}{6}$. So plug that in.

    $\displaystyle \dfrac{1}{2}\sqrt{A - \sqrt{B}} = \sqrt{\dfrac{1-\cos \left( \tfrac{\pi}{6} \right)}{2}}$

    Multiply both sides by 2:

    $\displaystyle \sqrt{A - \sqrt{B}} = 2\sqrt{\dfrac{1-\cos \left( \tfrac{\pi}{6} \right)}{2}}$

    Square both sides:

    $\displaystyle A - \sqrt{B} = 2\left( 1-\cos \left( \tfrac{\pi}{6} \right) \right) = 2 - 2\cos \left( \tfrac{\pi}{6} \right)$

    Give me a possible value for $\displaystyle A$. Once you choose a value for $\displaystyle A$, subtract it from both sides, then solve for $\displaystyle B$.
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  3. #3
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    Re: Can someone walk me through this trig problem?

    so I can just choose anything for A?
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  4. #4
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    Re: Can someone walk me through this trig problem?

    Quote Originally Posted by rhcprule3 View Post
    so I can just choose anything for A?
    Of course not, you said yourself, you are trying to find out what A and B are. You have the half-angle identity, now apply it!
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  5. #5
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    Re: Can someone walk me through this trig problem?

    Quote Originally Posted by rhcprule3 View Post
    so I can just choose anything for A?
    I disagree with Prove It. You have a single equation with two variables. Given any $\displaystyle A \in \mathbb{R}$, there exists a solution for $\displaystyle B\in \mathbb{R}$. So, based on the problem in your original post, yes, you can choose anything for A.

    Edit: I am mistaken. You can choose any $\displaystyle A\in \mathbb{R}$ such that $\displaystyle A \ge 2 - 2\cos\left( \dfrac{\pi}{6}\right) $ since $\displaystyle \sqrt{B}\ge 0$.
    Last edited by SlipEternal; Oct 7th 2013 at 07:53 PM.
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  6. #6
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    Re: Can someone walk me through this trig problem?

    Can someone walk me through this trig problem?-08-oct-13.png
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  7. #7
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    Re: Can someone walk me through this trig problem?

    Thanks guys, I got her!!
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