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Thread: trig equations

  1. #1
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    trig equations

    sin2t=(square root of 2)cost
    [0,360)
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  2. #2
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    Quote Originally Posted by Thelema View Post
    sin2t=(square root of 2)cost
    [0,360)
    Is this your equation,
    $\displaystyle sin(2t) = \sqrt{2}~cos(t)$

    or is it
    $\displaystyle sin^2(t) = \sqrt{2}~cos(t)$

    For the first:
    $\displaystyle 2~sin(t)~cos(t) = \sqrt{2}~cos(t)$

    $\displaystyle 2~sin(t) = \sqrt{2}$

    $\displaystyle sin(t) = \frac{\sqrt{2}}{2}$

    so t = 45, 135.

    For the second:
    $\displaystyle sin^2(t) = \sqrt{2}~cos(t)$

    $\displaystyle 1 - cos^2(t) = \sqrt{2}~cos(t)$

    $\displaystyle cos^2(t) + \sqrt{2}~cos(t) - 1 = 0$

    Using the quadratic formula:
    $\displaystyle cos(t) = \frac{-\sqrt{2} \pm \sqrt{2 + 4\sqrt{2}}}{2}$

    The only valid solution uses the "+" sign, so
    $\displaystyle cos(t) = \frac{-\sqrt{2} + \sqrt{2 + 4\sqrt{2}}}{2}$

    which, numerically, gives t = 58.826, 301.174.

    -Dan
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