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Math Help - [SOLVED] Trigonometric identity

  1. #1
    Newbie fanofandrew's Avatar
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    Red face [SOLVED] Trigonometric identity

    cosA + cos(A + B) + cos(A + 2B) + cos(A + 3B) +...+cos{A +(n-1)B} =

    sin nB/2
    ---------- . cos{A + (n-1)B/2}
    sin B/2

    Can any of you guys prove this for me?
    Yours truly,
    fanofandrew
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by fanofandrew View Post
    cosA + cos(A + B) + cos(A + 2B) + cos(A + 3B) +...+cos{A +(n-1)B} =

    sin nB/2
    ---------- . cos{A + (n-1)B/2}
    sin B/2

    Can any of you guys prove this for me?
    Yours truly,
    fanofandrew
    \cos A + \cos (A + B) + \cos (A + 2B) + \cos (A + 3B) + ... + \cos (A +[n-1]B)



    = \cos A + \{\cos A \cos B - \sin A \sin B\} + \{ \cos A \cos (2B) - \sin A \sin (2B)\}


     + \{\cos A \cos (3B) - \sin A \sin (3B)\} + ... + \{\cos A \cos ([n-1]B) - \sin A \sin ([n-1]B)\}




    = \cos A + \cos A \{\cos B + \cos (2B) + \cos (3B) ... + \cos ([n-1]B) \}


     - \sin A \{ \sin B + \sin (2B) + \sin (3B) + ... + \sin ([n-1]B) \} .




    Now use the two useful results given and proved by topsquark in this thread ( aha! I knew they'd be useful in the future )
    Last edited by mr fantastic; March 7th 2008 at 01:43 PM.
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  3. #3
    Newbie fanofandrew's Avatar
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    Thank you!

    Thank you for the solution.
    Since then I have seen another solution.
    Let S = cosA + cos(A + B) + cos(A + 2B) + cos(A + 3B) + ... + cos(A + (n - 1)B).

    2S sin(B/2) = sin(A + B/2) - sin(A - B/2) + sin(A + 3B/2) - sin(A + B/2) + sin(A + 5B/2) - sin(A + 3B/2) + ... + sin(A + (n - 1/2)B) - sin(A + (n - 3/2)B)
    = -sin(A - B/2) + sin(A + (n - 1/2)B)
    = 2cos(A + (n - 1)B/2)sin(nB/2).

    sin(nB/2)
    So S = ----------- . cos(A + (n - 1)B/2).
    sin(B/2)
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