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Math Help - prove

  1. #1
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    152

    prove

    \sin 47 + \sin 61 - \sin 11 - \sin 25 = \cos 7.
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  2. #2
    Senior Member
    Joined
    Dec 2007
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    Anchorage, AK
    Posts
    276

    Sum to Product

    First, note that 47-11=61-25=36, so rearrange the left hand side to get (\sin 47^\circ - \sin 11^\circ) + (\sin 61^\circ - \sin 25^\circ). From this, we use the sum-to-product formulas, specifically the formula

    \sin A - \sin B = 2\cos \left ( \frac{A+B}{2} \right ) \sin \left ( \frac{A-B}{2} \right ).

    Thus we get
    \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = (\sin 47^\circ - \sin 11^\circ) + (\sin 61^\circ - \sin 25^\circ)
    \,\,= 2 \cos 29^\circ \sin 18^\circ + 2 \cos 43^\circ \sin 18^\circ
    \,\,= 2(\cos 29^\circ + \cos 43^\circ)\sin 18^\circ

    We use the sum-to-product formulas again, this time using \cos A + \cos B = 2\cos \left ( \frac{A+B}{2} \right ) \cos \left ( \frac{A-B}{2} \right ), so
    \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2(\cos 29^\circ + \cos 43^\circ)\sin 18^\circ
    \,\,= 2(2\cos 36^\circ \cos (-7^\circ))\sin 18^\circ
    \,\,= 4\cos 7^\circ \cos 36^\circ \sin 18^\circ, as \cos(-x)=\cos x.

    Working from the geometry of the pentagon, one can find the sine and cosine for angles that are multiples of 18 degrees, and thus find that:
    \cos 36^\circ = \frac{1 + \sqrt{5}}{4}, \sin 18^\circ = \frac{-1 + \sqrt{5}}{4}, and \cos 36^\circ \sin 18^\circ = \frac{1 + \sqrt{5}}{4}\frac{-1 + \sqrt{5}}{4}=\frac{5-1}{16}=\frac{1}{4}

    So we get:
    \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 4\cos 7^\circ \cos 36^\circ \sin 18^\circ
    \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 4\cos 7^\circ \cdot \frac{1}{4}=\cos 7^\circ

    --Kevin C.
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