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Math Help - [SOLVED] exact answer using trigonometric difference identity

  1. #1
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    [SOLVED] exact answer using trigonometric difference identity

    I'm afraid I'm lost attempting to solve what is probably a really simple problem. The idea is to get an exact answer using the trigonometric difference identity for tangent.

    \tan(\frac{-\pi}{12})

    which is the same as
    \tan(\frac{\pi}{3} - \frac{\pi}{4})

    the formula is

    \frac{tanA - tanB}{1+tanAtanB}
    thus
    \frac{\tan(\frac{pi}{3}) - \tan(\frac{pi}{4})}{1+\tan(\frac{pi}{3})tan(\frac{  pi}{4})}
    ergo
    \frac{\sqrt3 - 1}{1+\sqrt3}

    and this is where I'm stuck. The book has the answer as -2 + \sqrt3, so either I'm completely wrong, or I just don't know how to simplify properly. Any assistance would be greatly appreciated.
    Last edited by satis; October 21st 2009 at 03:24 AM. Reason: corrected all my math tags
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  2. #2
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    Quote Originally Posted by satis View Post
    I'm afraid I'm lost attempting to solve what is probably a really simple problem. The idea is to get an exact answer using the trigonometric difference identity for tangent.

    \tan(\frac{-\pi}{12})

    which is the same as
    \color {red} \tan(\frac{\pi}{3} - \frac{\pi}{4}) \quad \\edit: \color {blue} \tan(\frac{\pi}{4}-\frac{\pi}{3})
    the formula is

    \frac{tanA - tanB}{1+tanAtanB}
    thus
    \frac{\tan(\frac{\pi}{3}) - \tan(\frac{\pi}{4})}{1+\tan(\frac{\pi}{3})tan(\fra  c{\pi}{4})}
    ergo
    \frac{\sqrt3 - 1}{1+\sqrt3} \quad \\edit \color {blue}\frac{1- \sqrt3}{1+\sqrt3} \quad

    and this is where I'm stuck. The book has the answer as -2 + \sqrt3, so either I'm completely wrong, or I just don't know how to simplify properly. Any assistance would be greatly appreciated.
    rationalize \color {blue}\frac{1- \sqrt3}{1+\sqrt3} \quad
    hint : \frac{1- \sqrt3}{1+\sqrt3}  \times \frac{1- \sqrt3}{1-\sqrt3}
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  3. #3
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    Quote Originally Posted by satis View Post
    I'm afraid I'm lost attempting to solve what is probably a really simple problem. The idea is to get an exact answer using the trigonometric difference identity for tangent.

    \tan(\frac{-\pi}{12})

    which is the same as
    \tan(\frac{\pi}{3} - \frac{\pi}{4})

    the formula is

    \frac{tanA - tanB}{1+tanAtanB}
    thus
    \frac{\tan(\frac{pi}{3}) - \tan(\frac{pi}{4})}{1+\tan(\frac{pi}{3})tan(\frac{  pi}{4})}
    ergo
    \frac{\sqrt3 - 1}{1+\sqrt3}

    and this is where I'm stuck. The book has the answer as -2 + \sqrt3, so either I'm completely wrong, or I just don't know how to simplify properly. Any assistance would be greatly appreciated.
    Multiply the numerator and denominator of your answer by the conjugate surd 1 - \sqrt{3} and simplify to what the book has.
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  4. #4
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    Thank you both for your assistance. I realize (now that it's been pointed out) I made a mistake right from the beginning. And rationalizing still poses me some difficulties, apparently, even though that's really more algebra than trig. However, I was able to work it out thanks to the feedback. Once again, thanks... I'm feeling much less lost now.
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