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Math Help - trigo (1)

  1. #1
    Senior Member
    Joined
    Jan 2009
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    381

    trigo (1)

    Show for all values of a and b

     <br />
\sin^2 a + \sin^2 (a+b) + \sin^2 (a+2b) = \frac{3}{2}-\frac{1}{2} \cos (2a+2b) (2 {\color{red}\cos b}+1)<br />

    This is what i did :

     <br />
\frac{1}{2} [1 - \cos (2a)] + \frac{1}{2} [1 - \cos (2a+2b)] + \frac{1}{2} [1 - \cos(2a+4b)]<br />

    =\frac{1}{2}[3-cos(2a+2b)-(cos2a+cos(2a+4b))]

    =\frac{1}{2}[3-cos(2a+2b)-2cos(2a+2b)cos2b]

    =\frac{1}{2}[3-cos(2a+2b)(1+2cos2b)]

     <br />
= \frac{3}{2} - \frac{1}{2} \cos (2a+2b) (2{\color{red}\cos (2b)} +1)<br />

    Why the difference in red ? Where did i go wrong ? Thanks for pointing that out .
    Last edited by mr fantastic; August 13th 2009 at 04:23 AM. Reason: Fixed latex
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  2. #2
    Senior Member
    Joined
    Jul 2009
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    Hi thereddevils

    I think you're right and the question is wrong
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