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Math Help - Evaluating inverse trig function

  1. #1
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    Evaluating inverse trig function

    dv/dt = 3/[t((t^2)-1)] t>0, v(2) = 0

    dv = integral 3/[t((t^2)-1)] dt

    v = 3sect^-1 + C

    3sec(-2)^-1 = -C


    How do I evaluate inverse secant?! Is there any simple rule to go by just looking at the unit circle? My teacher says the answer is c = -pi.
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  2. #2
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    Hello, tom ato!

    My answer is: . -{\color{red}2}\pi


    3\sec^{-1}(\text{-}2) \:=\: -C

    How do I evaluate inverse secant?

    We are expected to reason like this . . .

    Let \theta \:=\:\sec^{-1}(\text{-}2)
    . . Then \theta is the angle whose secant is -2.

    Hence, \theta is the angle whose cosine is \text{-}\tfrac{1}{2}

    The "first" such angle is: . \theta \:=\:\frac{2\pi}{3}

    Therefore: . \sec^{-1}(\text{-}2) \;=\;\frac{2\pi}{3}

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  3. #3
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    Thanks. And I just realized that the answer should be negative pi because I'm supposed to take the inverse secant of 2 (I had put -2)
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