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Math Help - can you please help me with this high school trigonometry problem?

  1. #1
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    Nov 2008
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    can you please help me with this high school trigonometry problem?

    we took down a note, where the example showed us how to find the exact value of

    tan (-5π / 12)

    then she wrote down....

    tan (-π/4) - tan (π/6)
    ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
    1+tan(-π/4)tan(π/6)


    and then from that, she jumped to the final answer, which is...

    (-√3-1) / (-√3-1)

    How did she do that?
    Maybe I copied down her note wrong, cuz I keep getting a different answer...
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  2. #2
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
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    New Orleans
    Posts
    894
    Quote Originally Posted by lanvin View Post
    we took down a note, where the example showed us how to find the exact value of

    tan (-5π / 12)

    then she wrote down....

    tan (-π/4) - tan (π/6)
    ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
    1+tan(-π/4)tan(π/6)


    and then from that, she jumped to the final answer, which is...

    (-√3-1) / (-√3-1)

    How did she do that?
    Maybe I copied down her note wrong, cuz I keep getting a different answer...
    \tan{\frac{-5\pi}{12}} = \tan{225}

    Using a reference angle we see that it is an angle of 75 degrees

    So now sum difference formula for tan

    \frac{\tan{45}+\tan{30}}{1-\tan{45}\tan{30}}

    \frac{(1) + (\frac{\sqrt{3}}{3})}{1 - (1)(\frac{\sqrt{3}}{3})}

    Then multiply by 3 to numerator and denominator

    \frac{3 +\sqrt{3}}{3- \sqrt{3}}

    then rationalize

    \frac{6\sqrt{3} + 12}{6}

    2 + \sqrt{3}

    Now multiply a neg in because it is in the 4th quad

    -2 - \sqrt{3}
    Last edited by 11rdc11; December 8th 2008 at 06:43 PM.
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