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Thread: finding the exact value

  1. #1
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    finding the exact value

    Hi guys, trying to find the exact value of $\displaystyle sin\frac{\pi}{8}$
    any help much appreciated
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jvignacio View Post
    Hi guys, trying to find the exact value of $\displaystyle sin\frac{\pi}{8}$
    any help much appreciated
    Note that $\displaystyle \sin\left(\frac{\pi}{8}\right)=\sin\left(\frac{1}{ 2}\left(\frac{\pi}{4}\right)\right)=\pm\sqrt{\frac {1-\cos\left(\frac{\pi}{4}\right)}{2}}=\pm\sqrt{\frac {1-\frac{\sqrt{2}}{2}}{2}}=\pm\frac{\sqrt{2-\sqrt{2}}}{2}$.

    Does this make sense?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Note that $\displaystyle \sin\left(\frac{\pi}{8}\right)=\sin\left(\frac{1}{ 2}\left(\frac{\pi}{4}\right)\right)=\pm\sqrt{\frac {1-\cos\left(\frac{\pi}{4}\right)}{2}}=\pm\sqrt{\frac {1-\frac{\sqrt{2}}{2}}{2}}=\pm\frac{\sqrt{2-\sqrt{2}}}{2}$.

    Does this make sense?
    okay you lost me on the 3rd step..
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jvignacio View Post
    okay you lost me on the 3rd step..
    From the second step, you apply the half angle formula $\displaystyle \sin\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1-\cos\alpha}{2}}$ where in our case, $\displaystyle \alpha=\frac{\pi}{4}$.

    Does this clarify things?
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  5. #5
    Member sinewave85's Avatar
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    Quote Originally Posted by jvignacio View Post
    Hi guys, trying to find the exact value of $\displaystyle sin\frac{\pi}{8}$
    any help much appreciated
    Hi, jvignacio. To find the exact value of sin(pi/8), you need to know two things: The exact value of cos(pi/4) and the half-angle identity for sin(x).

    $\displaystyle \sin\frac{\pi}{8} = \sin\left({\frac{1}{2}*\frac{\pi}{4}}\right)$

    $\displaystyle \sin\frac{\pi}{8} = \pm\sqrt{\frac{1 - \cos\frac{\pi}{4}}{2}}$

    $\displaystyle \sin\frac{\pi}{8} = \pm\sqrt{\frac{1 -\frac{\sqrt{2}}{2}}{2}}$

    $\displaystyle \sin\frac{\pi}{8} = \pm\sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}}$

    $\displaystyle \sin\frac{\pi}{8} = \pm\sqrt{\frac{2 - \sqrt{2}}{4}}$

    $\displaystyle \sin\frac{\pi}{8} = \pm\frac{\sqrt{2 - \sqrt{2}}}{2}$

    and since pi/8 is in Quadrant I, you need the positive value. Hope that helps!
    Last edited by sinewave85; Mar 10th 2009 at 07:43 PM. Reason: Took out a Latex snafu.
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  6. #6
    Member sinewave85's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    From the second step, you apply the half angle formula $\displaystyle \sin\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1-\cos\alpha}{2}}$ where in our case, $\displaystyle \alpha=\frac{\pi}{4}$.

    Does this clarify things?
    You got two posts typed out while I unawares was wading through the jungle of Latex on one. Ah, redundancies.
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  7. #7
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    Quote Originally Posted by jvignacio View Post
    Hi guys, trying to find the exact value of $\displaystyle sin\frac{\pi}{8}$
    any help much appreciated
    A slightly alternatively approach is to note that $\displaystyle \cos (2A) = 1 - 2 \sin^2 (A)$.

    Let $\displaystyle A = \frac{\pi}{8}$: $\displaystyle \cos \frac{\pi}{4} = 1 - 2 \sin^2 \frac{\pi}{8} \Rightarrow \frac{\sqrt{2}}{2} = 1 - 2 \sin^2 \frac{\pi}{8}$.

    Re-arrange and solve for $\displaystyle \sin \frac{\pi}{8}$ (the negative root solution is rejected - why?).
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