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Math Help - cos(tan^(-1) v)

  1. #1
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    cos(tan^(-1) v)

    Show that cos(tan^(-1) v) = 1/(sqrt{1 + v^2})
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by magentarita View Post
    Show that cos(tan^(-1) v) = 1/(sqrt{1 + v^2})
    Let \vartheta=\tan^{-1}v

    We can construct a right triangle representing this angle.

    On the opposite side of the angle, it has a value of \text{opp}=v, and the adjacent side has a value of \text{adj}=1. Now, we can find the value of the hypotenuse:

    (\text{hyp})^2=(\text{opp})^2+(\text{adj})^2\impli  es (\text{hyp})^2=v^2+1^2\implies \text{hyp}=\sqrt{v^2+1}

    Now, what do you think \cos\vartheta equals?

    --Chris
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  3. #3
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    I prefer the way Chris did this but since he got in first (and I just have to put in my oar!) here's another way:

    Use trig identities. [tex]tan^2(\theta)+1= sec^2(\theta)= 1/cos^2(\theta)[/itex] so cos(\theta)= \sqrt{\frac{1}{1+ tan^2(\theta)}= \frac{1}{\sqrt{1+ tan^2(\theta)}

    Replacing \theta by tan^{-1}(v), that gives [tex]cos(tan^{-1}(v))= \frac{1}{\sqrt{1+ tan^2(tan^{1}(v))}= \frac{1}{\sqrt{1+ v^2}}.
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  4. #4
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    cos(theta)

    Quote Originally Posted by Chris L T521 View Post
    Let \vartheta=\tan^{-1}v

    We can construct a right triangle representing this angle.

    On the opposite side of the angle, it has a value of \text{opp}=v, and the adjacent side has a value of \text{adj}=1. Now, we can find the value of the hypotenuse:

    (\text{hyp})^2=(\text{opp})^2+(\text{adj})^2\impli  es (\text{hyp})^2=v^2+1^2\implies \text{hyp}=\sqrt{v^2+1}

    Now, what do you think \cos\vartheta equals?

    --Chris
    cos(theta) = sqrt{v^2 + 1}, right?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by magentarita View Post
    cos(theta) = sqrt{v^2 + 1}, right?
    You almost have it!

    Note that \cos\vartheta=\frac{\text{adj}}{\text{hyp}}

    They adjacent value is \text{adj}=1 and the hypotenuse value is \text{hyp}=\sqrt{1+v^2}

    So \cos\vartheta=\frac{\text{adj}}{\text{hyp}}=\dots

    --Chris
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