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Thread: series

  1. #1
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    series

    If $\displaystyle f(r)=\cos 2r\theta$ , simplify $\displaystyle f(r+1)-f(r)$ . Use your result to find the sum of the first n terms of the series : $\displaystyle 3\sin \theta+\sin 5\theta+\sin 7\theta+...$

    For the first part , i got

    $\displaystyle f(r+1)-f(r)=\cos (2r+2)\theta-\cos 2r\theta$ .. is this correct .

    But from here , i couldn't relate to the second problem .
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    If $\displaystyle f(r)=\cos 2r\theta$ , simplify $\displaystyle f(r+1)-f(r)$ . Use your result to find the sum of the first n terms of the series : $\displaystyle \sin {\color{red}3} \theta+\sin 5\theta+\sin 7\theta+...$ Mr F says: Correction in red.

    For the first part , i got

    $\displaystyle f(r+1)-f(r)=\cos (2r+2)\theta-\cos 2r\theta$ .. is this correct .

    But from here , i couldn't relate to the second problem .
    When the first part is answered completely the second part should become clearer:

    $\displaystyle f(r + 1) = \cos(2[r+1] \theta) = \cos(2r \theta + 2 \theta) = \cos (2 r \theta) \cos (2 \theta) - \sin (2 r \theta) \sin (2 \theta)$.

    Therefore:

    $\displaystyle f(r + 1) - f(r) = \cos (2 r \theta) \cos (2 \theta) - \sin (2 r \theta) \sin (2 \theta) - \cos (2 r \theta) $

    $\displaystyle = \cos (2 r \theta) [\cos (2 \theta) - 1] - \sin (2 r \theta) \sin (2 \theta)$

    $\displaystyle = - 2 \cos (2 r \theta) \sin^2 (\theta ) - \sin (2 r \theta) \sin (2 \theta)$

    $\displaystyle = - 2 \sin (\theta) [ \cos (2 r \theta) \sin (\theta ) + \sin (2 r \theta) \cos (\theta)] = -2 \sin (\theta) \sin (2 r \theta + \theta)$

    $\displaystyle = -2 \sin (\theta) \sin ([2 r + 1]\theta) $.

    By the way, the result in the second part of the question has a typo which I have corrected.
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