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Math Help - What is the exact value of sin 12 degrees?

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    What is the exact value of sin 12 degrees?

    What is the exact value of sin 12 degrees?

    Only hint I was given was to use compound angle formula. Had this on the test but I left it blank, now I just want to know the solution.
    Last edited by mr fantastic; November 27th 2009 at 10:17 PM. Reason: Added post title to main body of post
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    Super Member bigwave's Avatar
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    Quote Originally Posted by ISeriouslyNeedHelp View Post
    What is the exact value of sin 12 degrees?

    Only hint I was given was to use compound angle formula. Had this on the test but I left it blank, now I just want to know the solution.
    12 degrees is \frac{\pi}{15}

    so one way to do this is to use

    sin(U-V) - sinUcosV-cosUsinV

    divide up \frac{\pi}{15} into U and V

    for example \frac{3\pi}{5} - \frac{\pi}{3}  = \frac{\pi}{15}

    then use the difference formula...

    can you take it from there??
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  3. #3
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    Quote Originally Posted by bigwave View Post
    12 degrees is \frac{\pi}{15}

    so one way to do this is to use

    sin(U-V) - sinUcosV-cosUsinV

    divide up \frac{\pi}{15} into U and V

    for example \frac{3\pi}{5} - \frac{\pi}{3} = \frac{\pi}{15}

    then use the difference formula...

    can you take it from there??
    Now of course there's the new problem of finding \sin \frac{3 \pi}{5} and \cos \frac{3 \pi}{5} ....

    I suspect there's more to the original question than was posted.
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    There is nothing more to this question. It is simply to find the exact value of sin 12 degrees. Only thing I could think of was sin(60/5), but these odd numbers really mess with my head.
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  5. #5
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    Quote Originally Posted by ISeriouslyNeedHelp View Post
    What is the exact value of sin 12 degrees?

    Only hint I was given was to use compound angle formula. Had this on the test but I left it blank, now I just want to know the solution.
    To do this you need:

    \sin(12^{\circ})=\sin(30^{\circ}-18^{\circ})

    We know the values of the trig functions for 30^{\circ} and so if we also know the trig function values for 18^{\circ} we are home and dry.

    We can get the values we need for the \sin and \cos of 18^{\circ } from the geometry of a regular pentagon (which we could have guessed as we are dealing with integer fractions of 72^{\circ}=360^{\circ}/5), where we find:

     \sin(18^{\circ})=\frac{1}{2\Phi}

    where \Phi=\frac{1+\sqrt{5}}{2} is the Golden number.

    CB
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