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Math Help - Inverse Trig Functions and Advance Trig

  1. #1
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    Inverse Trig Functions and Advance Trig

    hello thx in advance for answering this question


    You are given 5 cos wt = A cos (wt - pi / 6) + 5 cos (wt + Z)

    for all t, where w is a fixed constant

    Find, exactly, all possible values for A > 0 and Z


    Im really having trouble with this question.
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  2. #2
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    5\cos(\omega t) = A\cos\left(\omega t - \frac{\pi}{6}\right) + 5\cos(\omega t + Z)


    5\cos(\omega t) = A\left[\cos(\omega t)\cos\left(\frac{\pi}{6}\right)+\sin(\omega t)\sin\left(\frac{\pi}{6}\right)\right] + 5[\cos(\omega t)\cos(Z) - sin(\omega t)\sin(Z)]

    5\cos(\omega t) = A\left[\cos(\omega t)\left(\frac{\sqrt{3}}{2}\right)+\sin(\omega t)\left(\frac{1}{2}\right)\right] + 5[\cos(\omega t)\cos(Z) - sin(\omega t)\sin(Z)]

    5\cos(\omega t) = \left[\frac{A\sqrt{3}}{2} + 5\cos(z)\right]\cos(\omega t)+ \left[\frac{A}{2} - 5\sin(Z)\right]\sin(\omega t)

    equating coefficients on the left and right side ...

    (1) \frac{A}{2} - 5\sin(Z) = 0

    (2) \frac{A\sqrt{3}}{2} + 5\cos(z) = 5

    from equation (1) ...

    A = 10\sin(Z)

    substitute into equation (2) ...

    5\sqrt{3}\sin(Z) + 5\cos(Z) = 5

    \sqrt{3}\sin(Z) + \cos(Z) = 1

    square both sides ...

    3\sin^2(Z) + 2\sqrt{3}\sin(Z)\cos(Z) + \cos^2(Z) = 1

    2\sin^2(Z) + 2\sqrt{3}\sin(Z)\cos(Z) = 0

    2\sin(Z)[\sin(Z) + \sqrt{3} \cos(Z)] = 0

    o.k. ... all the "grunt" work is done. Take the last equation, solve for Z, then go back and solve for A. Buena suerte.
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