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Math Help - Identities

  1. #1
    Junior Member
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    Unhappy Identities

    Prove that Cos2A+Cos2B-cos2C=1-4sinAsinBsinC

    Given that A+B+C=pi
    Last edited by matsci0000; July 14th 2009 at 08:19 AM. Reason: inadequate information
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  2. #2
    Junior Member
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    solution - take a look

    (cos2A+cos2B)+cos2C
    =2cos(A+B)cos(A-B)-cos2C
    =-2cosC cos(A-B)-2cos^2C+1 [ cos(A+B)=cos(pie-C)= -cosC]
    =1-2cosC[cos(A-B)-cosC]
    =1-2cosC[cos(A+B) -cos(A-B)]
    =1-2cosC [2sinA sinB]


    How to get the term sinC from 2cosC?
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    That identity is false.

    It's actually \cos 2a+\cos 2b+\cos 2c=4\sin (a)\sin (b)\sin (c)+1, provided a+b+c=\frac\pi2.

    The solution is quite simple:

    \begin{aligned}<br />
   \cos 2a+\cos 2b+\cos 2c&=2\sin (c)\cos (a-b)-2\sin ^{2}c+1 \\ <br />
 & =2\sin (c)\big(\cos (a-b)-\cos (a+b)\big)+1 \\ <br />
 & =4\sin (a)\sin (b)\sin (c)+1.<br />
\end{aligned}

    In order to have \cos a+\cos b+\cos c=4\sin\frac a2\sin\frac b2\sin\frac c2+1, we require a+b+c=\pi.
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