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Math Help - Proving sin 15 deg

  1. #1
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    Proving sin 15 deg

    Hello! Need help on this problem

    prove that sin 15deg = (sqrt(6) - sqrt(2)) / 4

    using

    sinC= sinA cosB + cosA sinB
    sinB= sinA cosC + cosA sinC
    sinA= sinB cosC + cosB sinC

    Hoping for your attention!=)
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  2. #2
    Senior Member
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    Dec 2008
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    what you want to use is the addintion formulae sin(45-30)
    =sin45cos30-sin30cos45
    substitute in your exact values of sin45=1/sqrt2 sin30=0.5 cos45=1/sqrt2 and cos30=sqrt3/2 and then rearrange the fraction you get by multiplying by sqrt2 and that should be it.
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  3. #3
    Junior Member
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    Aug 2008
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    Dubai, UAE
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    Recall that sin (A - B) = sin A cos B - cos A sin B

    sin 15 = sin (45 - 30)

    = sin 45 cos 30 - cos 45 sin 30

     = \frac {\sqrt {2}}{2} \cdot \frac {\sqrt {3}}{2} - \frac {\sqrt {2}}{2} \cdot \frac {1}{2}

     = \frac {\sqrt {6}}{4} - \frac {\sqrt {2}}{4}

     = \frac {\sqrt {6} - \sqrt {2}}{4}. ---> Q.E.D.

    I hope that helps.

    ILoveMaths07.
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