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Abbie23

  • 4 years ago

Use a trigonometric substitution involving tangent to write √(961 + x^2) as a function of theta in the first quadrant without a radical So confused pleaseeee help

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  1. DoomDude
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=%E2%88%9A%28961+%2B+x%5E2%29

  2. anonymous
    • 4 years ago
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    would it help to know that \(\sqrt{961}=31\)?

  3. anonymous
    • 4 years ago
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    put \(x=32\tan(\theta)\) so \(x^2=961\tan^2(\theta)\) and you get \[\sqrt{961+x^2}=\sqrt{961+961\tan^2(\theta)}=\sqrt{961(1+\tan^2(\theta)}=31\sqrt{1+\tan^2(\theta)}\]

  4. anonymous
    • 4 years ago
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    sorry i meant \[x=31\tan(\theta)\]

  5. anonymous
    • 4 years ago
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    and now a miracle occurs, namely that \[1+\tan^2(\theta)=\sec^2(\theta)\] so you can get rid of the radical

  6. anonymous
    • 4 years ago
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    when you see \[\sqrt{a^2+x^2}\] use \[x=a\tan(\theta)\] and when you see \[\sqrt{a^2-x^2}\] use \[x=a\sin(\theta)\]

  7. anonymous
    • 4 years ago
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    we can do another if you like

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