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Math Help - exact value check

  1. #1
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    exact value check

    find the exact value of cos(\frac{\pi}{8})

    \cos\frac{\pi}{8} = \cos\left({\frac{1}{2}*\frac{\pi}{4}}\right)

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

    \cos\frac{\pi}{8} = \pm\sqrt{\frac{1 -\frac{\sqrt{2}}{2}}{2}}

    \cos\frac{\pi}{8} = \pm\sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}}

    \cos\frac{\pi}{8} = \pm\sqrt{\frac{2 - \sqrt{2}}{4}}

    \cos\frac{\pi}{8} = \pm\frac{\sqrt{2 - \sqrt{2}}}{2}

    therefore: we take the positive one because \frac{\pi}{8} is in the first quadrant and its positive.

    this correct ? much appreciated..
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  2. #2
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    Quote Originally Posted by jvignacio View Post
    find the exact value of cos(\frac{\pi}{8})

    \cos\frac{\pi}{8} = \cos\left({\frac{1}{2}*\frac{\pi}{4}}\right)

    \cos\frac{\pi}{8} = \pm\sqrt{\frac{1 - \sin\frac{\pi}{4}}{2}}
    Could you explain the way you go from the first identity to the second one ?

    \cos (2x) = 1-2\sin^2(x)

    \sin (x) = \pm\sqrt{\frac{1-\cos (2x)}{2}}

    Or \cos (2x) = 2\cos^2(x)-1

    \cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Could you explain the way you go from the first identity to the second one ?

    \cos (2x) = 1-2\sin^2(x)

    \sin (x) = \pm\sqrt{\frac{1-\cos (2x)}{2}}

    Or \cos (2x) = 2\cos^2(x)-1

    \cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}
    well its sin^2x+cos^2x = 1 (fundamental identity)

    then

     <br />
sin^2x = 1-cos^2x<br />
     <br />
sinx = \sqrt{1-cos^2x}<br />

    and in this case x = \frac{\pi}{4}

    i split up the \frac{\pi}{8} to \frac{\pi}{4} to work with 45 degrees
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  4. #4
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    Quote Originally Posted by running-gag View Post
    Could you explain the way you go from the first identity to the second one ?

    \cos (2x) = 1-2\sin^2(x)

    \sin (x) = \pm\sqrt{\frac{1-\cos (2x)}{2}}

    Or \cos (2x) = 2\cos^2(x)-1

    \cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}
    sorry yes..

    cos2x = 1-2sin^2(x)

    2sin^2(x)=1-cos(2x)

    sin^2(x)=\frac{1-cos(2x)}{2}
     <br /> <br />
sin(x) = \pm\sqrt{\frac{1-cos(2x)}{2}}<br />
    Last edited by jvignacio; March 14th 2009 at 06:17 AM.
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  5. #5
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    You should rather take this one

    \cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}

    with x=\frac{\pi}{8}
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  6. #6
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    Quote Originally Posted by running-gag View Post
    You should rather take this one

    \cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}

    with x=\frac{\pi}{8}
    ahh darn how would i know which one to use ?
    Last edited by mr fantastic; March 15th 2009 at 05:14 AM. Reason: m --> r
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  7. #7
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    Which one among what ?

    This one

    \cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}

    is useful because you know the value of \cos \left(\frac{\pi}{4}\right)
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  8. #8
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    Quote Originally Posted by running-gag View Post
    Which one among what ?

    This one

    \cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}

    is useful because you know the value of \cos \left(\frac{\pi}{4}\right)
    yeh sorry i was just getting confused but now i understand. thanks!!
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  9. #9
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    Quote Originally Posted by running-gag View Post
    Which one among what ?

    This one

    \cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}

    is useful because you know the value of \cos \left(\frac{\pi}{4}\right)
    so should i do \cos (x) = \pm\sqrt{\frac{1+\cos (\frac{2\pi}{4})}{2}}

    ?
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  10. #10
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    \cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}

    \cos \left(\frac{\pi}{8}\right) = \pm\sqrt{\frac{1+\cos \left(\frac{\pi}{4}\right)}{2}}
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