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Thread: sum-product/product-sum

  1. #1
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    sum-product/product-sum

    solve the following using sum-product/product-sum identities

    $\displaystyle cos20+cos100+cos140=0$

    $\displaystyle sin19cos11+sin71sin11=\frac{1}{2}$

    $\displaystyle cos20cos40cos60cos180=\frac{1}{16}$

    $\displaystyle cos20cos40cos80=\frac{1}{8}$
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  2. #2
    Member srirahulan's Avatar
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    Re: sum-product/product-sum

    In this case,

    Q1,
    $\displaystyle cos20+cos100+cos140=0$
    L.H.S
    =$\displaystyle 2cos60cos40+cos140$

    =$\displaystyle cos40+cos140$

    =$\displaystyle 2cos90cos50$

    =$\displaystyle 0$ (Since cos90=0)

    =R.H.S

    Q2,
    $\displaystyle sin19cos11+sin71sin11=\frac{1}{2}$
    L.H.S
    =$\displaystyle \frac{2sin19cos11+2sin71sin11}{2}$

    =$\displaystyle \frac{sin30+sin8+cos60-cos82}{2}$

    =$\displaystyle \frac{\frac{1}{2}+\frac{1}{2}+sin(90-82)-cos82}{2}$

    =$\displaystyle \frac{1+cos82-cos82}{2}$

    =$\displaystyle \frac{1}{2}$

    =R.H.S

    Q3,
    I think you mention cos180 or cos 80 I can't get any answer.

    Q4,
    $\displaystyle cos20cos40cos80=\frac{1}{8}$
    L.H.S
    =$\displaystyle \frac{2cos20cos40cos80}{2}$

    =$\displaystyle \frac{(cos60+cos20)cos80}{2}$

    =$\displaystyle \frac{(\frac{1+2cos20}{2})cos80}{2}$

    =$\displaystyle \frac{\frac{cos80+cos100+cos60}{2}}{2}$

    =$\displaystyle \frac{\frac{cos(180-100)+cos100+cos60}{2}}{2}$

    =$\displaystyle \frac{\frac{-cos100+cos100+\frac{1}{2}}{2}}{2}$

    =$\displaystyle \frac{1}{8}$

    =R.H.S
    Last edited by srirahulan; Aug 2nd 2012 at 09:07 AM.
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  3. #3
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    Re: sum-product/product-sum

    $\displaystyle \frac{2cos20cos40cos80}{2}$

    Where did the 2 in the numerator and denominator come from???
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  4. #4
    Member srirahulan's Avatar
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    Re: sum-product/product-sum

    you want to multiply by 2 to numerator and denominator,This is a way to get your answer quickly.
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