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Thread: Trigonometric Sum

  1. May 22nd 2009, 01:38 PM #1
    magentarita
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    Trigonometric Sum

    I would like you to show me how to get the amplitude and phase shift of the sum of the trigonometric sum below.

    8sin(50pi*x+pi/4)+7sin(50pi*x-pi/3)
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  2. May 22nd 2009, 10:21 PM #2
    Grandad
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    Trigonometry

    Hello magentarita
    Quote Originally Posted by magentarita View Post
    I would like you to show me how to get the amplitude and phase shift of the sum of the trigonometric sum below.

    8sin(50pi*x+pi/4)+7sin(50pi*x-pi/3)

    8\sin(50\pi x +\pi/4) +7\sin(50\pi x -\pi/3)

    = 8\sin(50\pi x)\cos(\pi/4) +8\cos(50\pi x)\sin(\pi/4) + 7\sin(50\pi x)\cos(\pi/3) -7\cos(50\pi x)\sin(\pi/3)

    = \Big(\frac{8}{\sqrt2}+\frac{7}{2}\Big)\sin(50\pi x) + \Big(\frac{8}{\sqrt2}-\frac{7\sqrt3}{2}\Big)\cos(50\pi x)

    = \frac{1}{2\sqrt2}\Big((16+7\sqrt2)\sin(50\pi x)-(7\sqrt6-16)\cos(50\pi x)\Big)

    = r\sin(50\pi[x-\alpha]), where

    r^2 = \tfrac18\Big((16+7\sqrt2)^2+(7\sqrt6-16)^2\Big)

    and \tan(50\pi\alpha) = \frac{(7\sqrt6-16)}{(16+7\sqrt2)}

    Can you tidy up and simplify now?

    Grandad
    Last edited by Grandad; May 22nd 2009 at 10:47 PM. Reason: Corrected solution
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  3. May 25th 2009, 03:30 AM #3
    magentarita
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    yes...

    Quote Originally Posted by Grandad View Post
    Hello magentarita 8\sin(50\pi x +\pi/4) +7\sin(50\pi x -\pi/3)

    = 8\sin(50\pi x)\cos(\pi/4) +8\cos(50\pi x)\sin(\pi/4) + 7\sin(50\pi x)\cos(\pi/3) -7\cos(50\pi x)\sin(\pi/3)

    = \Big(\frac{8}{\sqrt2}+\frac{7}{2}\Big)\sin(50\pi x) + \Big(\frac{8}{\sqrt2}-\frac{7\sqrt3}{2}\Big)\cos(50\pi x)

    = \frac{1}{2\sqrt2}\Big((16+7\sqrt2)\sin(50\pi x)-(7\sqrt6-16)\cos(50\pi x)\Big)

    = r\sin(50\pi[x-\alpha]), where

    r^2 = \tfrac18\Big((16+7\sqrt2)^2+(7\sqrt6-16)^2\Big)

    and \tan(50\pi\alpha) = \frac{(7\sqrt6-16)}{(16+7\sqrt2)}

    Can you tidy up and simplify now?

    Grandad
    Yes, I can. This question is not easy.
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