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Math Help - Trigonometry - Compound Angles

  1. #1
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    Trigonometry - Compound Angles

    Hi guys i've attempted this question 3 times now yet i keep getting the wrong answer so i turn to you my peers for help!.
    Question:
    Given that cot(30-\theta)=3, find the value of cot(\theta)
    My Workings Out:
    Using tan(30-\theta)=\frac{tan(30)-tan(\theta)}{1+tan(30).tan(\theta)}
    cot(30-\theta)=\frac{1+tan(30).tan(\theta)}{tan(30)-tan(\theta)}
    3=\frac{1+tan(30).tan(\theta)}{tan(30)-tan(\theta)}
    now i did tan30 which gives \frac{\sqrt3}{3} i get
    3=\frac{1+\frac{\sqrt3}{3}tan(\theta)}{\frac{\sqrt  3}{3}-tan(\theta)}
    3(\frac{\sqrt3}{3}-tan(\theta))=1+\frac{\sqrt3}{3}tan(\theta)
    \sqrt3-3tan(\theta)=1+\frac{\sqrt3}{3}tan\theta
    re-arrange gives me
    -1+\sqrt3=\frac{9+\sqrt3}{3}tan(\theta)
    tan\theta=\frac{-1+\sqrt3}{\frac{9+\sqrt3}{3}}=\frac{-29+9\sqrt3}{26}
    so
    cot\theta=\frac{26}{-26+9\sqrt3}
    Answer In back is
    \frac{1+3\sqrt3}{3-\sqrt3}=\frac{6+5\sqrt3}{3}
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  2. #2
    Member sinewave85's Avatar
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    Quote Originally Posted by Kevlar View Post
    Hi guys i've attempted this question 3 times now yet i keep getting the wrong answer so i turn to you my peers for help!.
    Question:
    Given that cot(30-\theta)=3, find the value of cot(\theta)
    My Workings Out:
    Using tan(30-\theta)=\frac{tan(30)-tan(\theta)}{1+tan(30).tan(\theta)}
    cot(30-\theta)=\frac{1+tan(30).tan(\theta)}{tan(30)-tan(\theta)}
    3=\frac{1+tan(30).tan(\theta)}{tan(30)-tan(\theta)}
    now i did tan30 which gives \frac{\sqrt3}{3} i get
    3=\frac{1+\frac{\sqrt3}{3}tan(\theta)}{\frac{\sqrt  3}{3}-tan(\theta)}
    3(\frac{\sqrt3}{3}-tan(\theta))=1+\frac{\sqrt3}{3}tan(\theta)
    \sqrt3-3tan(\theta)=1+\frac{\sqrt3}{3}tan\theta
    re-arrange gives me
    -1+\sqrt3=\frac{9+\sqrt3}{3}tan(\theta)
    tan\theta=\frac{-1+\sqrt3}{\frac{9+\sqrt3}{3}}=\frac{-29+9\sqrt3}{26}
    so
    cot\theta=\frac{26}{-26+9\sqrt3}
    Answer In back is
    \frac{1+3\sqrt3}{3-\sqrt3}=\frac{6+5\sqrt3}{3}
    That was a lot of work! The first thing I saw was that this is wrong:

    tan\theta=\frac{-1+\sqrt3}{\frac{9+\sqrt3}{3}}\not=\frac{-29+9\sqrt3}{26}

    tan\theta=\frac{-1+\sqrt3}{\frac{9+\sqrt3}{3}}=\frac{-3+3\sqrt3}{9 + \sqrt{3}}

    Nothing else jumped out at me, and unfortunately I do not have time to go over the whole thing more carefully. I hope that helped, though, and that someone will have a little more time than I do.
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Kevlar View Post
    Hi guys i've attempted this question 3 times now yet i keep getting the wrong answer so i turn to you my peers for help!.
    Question:
    Given that cot(30-\theta)=3, find the value of cot(\theta)
    My Workings Out:
    Using tan(30-\theta)=\frac{tan(30)-tan(\theta)}{1+tan(30).tan(\theta)}
    cot(30-\theta)=\frac{1+tan(30).tan(\theta)}{tan(30)-tan(\theta)}
    3=\frac{1+tan(30).tan(\theta)}{tan(30)-tan(\theta)}
    now i did tan30 which gives \frac{\sqrt3}{3} i get
    3=\frac{1+\frac{\sqrt3}{3}tan(\theta)}{\frac{\sqrt  3}{3}-tan(\theta)}
    3(\frac{\sqrt3}{3}-tan(\theta))=1+\frac{\sqrt3}{3}tan(\theta)
    {\color{red}\sqrt3-3tan(\theta)=1+\frac{\sqrt3}{3}tan\theta}
    re-arrange gives me
    -1+\sqrt3=\frac{9+\sqrt3}{3}tan(\theta)
    tan\theta=\frac{-1+\sqrt3}{\frac{9+\sqrt3}{3}}=\frac{-29+9\sqrt3}{26}
    so
    cot\theta=\frac{26}{-26+9\sqrt3}
    Answer In back is
    \frac{1+3\sqrt3}{3-\sqrt3}=\frac{6+5\sqrt3}{3}
    Hi Kevlar,

    I'm going to pick it up at the red line above and finish a different way.

    {\color{red}\sqrt3-3\tan \theta=1+\frac{\sqrt3}{3}\tan\theta}

    Multiply by 3

    3\sqrt{3}-3=\sqrt{3}\tan \theta+9 \tan \theta

    3(\sqrt{3}-1)=\tan \theta(\sqrt{3}+9)

    \tan \theta=\frac{3(\sqrt{3}-1)}{\sqrt{3}+9}

    \cot \theta=\frac{\sqrt{3}+9}{3(\sqrt{3}-1)}\cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{3+10\sqrt{3}+9  }{3(3-1)}=\frac{12+10\sqrt{3}}{6}=\frac{6+5\sqrt{3}}{3}
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