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Math Help - Sum and Difference Formulas for Tangents

  1. #1
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    Sum and Difference Formulas for Tangents

    I've been having a lot of trouble understanding how to use the sum and difference formulas for tangents. I can never quite simplify the question completely.

    Sample:

    Find the exact value of tan(19pi/12)

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  2. #2
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    Quote Originally Posted by magentarita View Post
    I've been having a lot of trouble understanding how to use the sum and difference formulas for tangents. I can never quite simplify the question completely.

    Sample:

    Find the exact value of tan(19pi/12)
    \tan \left(\alpha+\beta\right) = \frac{\tan \alpha+\tan \beta}{1-\tan\alpha\tan\beta}

    \frac{19\pi}{12} = \frac{4\pi}{3} + \frac{\pi}{4}

    \therefore \tan \left( \frac{19 \pi}{12} \right) = \tan \left(\frac{4\pi}{3} + \frac{\pi}{4}\right)

    =\frac{\tan <br />
\left(\frac{4\pi}{3}\right)<br />
+\tan \left(\frac{\pi}{4}\right)}{1-\tan<br />
\left(\frac{4\pi}{3}\right)<br />
\tan \left(\frac{\pi}{4}\right)} = \frac{\sqrt{3}+1}{1-\sqrt{3}}

    To simplify \frac{\sqrt{3}+1}{1-\sqrt{3}} , multiply numerator and the denominator by the conjugate of the denominator which is 1+\sqrt{3}.
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  3. #3
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    Smile Fabulous Reply!!

    Quote Originally Posted by Air View Post
    \tan \left(\alpha+\beta\right) = \frac{\tan \alpha+\tan \beta}{1-\tan\alpha\tan\beta}

    \frac{19\pi}{12} = \frac{4\pi}{3} + \frac{\pi}{4}

    \therefore \tan \left( \frac{19 \pi}{12} \right) = \tan \left(\frac{4\pi}{3} + \frac{\pi}{4}\right)

    =\frac{\tan <br />
\left(\frac{4\pi}{3}\right)<br />
+\tan \left(\frac{\pi}{4}\right)}{1-\tan<br />
\left(\frac{4\pi}{3}\right)<br />
\tan \left(\frac{\pi}{4}\right)} = \frac{\sqrt{3}+1}{1-\sqrt{3}}

    To simplify \frac{\sqrt{3}+1}{1-\sqrt{3}} , multiply numerator and the denominator by the conjugate of the denominator which is 1+\sqrt{3}.
    Air, your replies are simply amazing!!
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