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Math Help - Inverse Trig Ratios

  1. #1
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    Inverse Trig Ratios

    How would I evaluate cos(2 arctan(2))?
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  2. #2
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    Re: Inverse Trig Ratios

    Use \cos(2\theta)=1-2\sin^2(\theta)\,.

    Then can you find \sin(\arctan(2))\,?
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  3. #3
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    Re: Inverse Trig Ratios

    Do you mind working it out for me?
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  4. #4
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    Re: Inverse Trig Ratios

    Hello, Manni!

    \text{Evaluate: }\,\cos(2 \arctan 2 )

    Let \theta \,=\, \arctan 2 \quad\Rightarrow\quad \tan\theta \,=\,2

    Then: . \tan\theta \:=\:\tfrac{2}{1} \:=\:\tfrac{opp}{adj}

    That is, \theta is in a right triangle with opp = 2,\;adj = 1.
    . . Hence: . hyp \,=\,\sqrt{5} \quad\Rightarrow\quad \cos\theta \,=\,\tfrac{adj}{hyp} \,=\,\tfrac{1}{\sqrt{5}}

    We have: . \cos 2\theta \;=\;2\cos^2\!\theta - 1 \;=\;2\!\left(\tfrac{1}{\sqrt{5}}\right)^2 - 1 \;=\;\tfrac{2}{5} - 1 \;=\;\text{-}\tfrac{3}{5}

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  5. #5
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    Re: Inverse Trig Ratios

    Thank you! This clarified everything!
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