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Math Help - Compound Angle Identities

  1. #1
    Junior Member Morphayne's Avatar
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    Exclamation Compound Angle Identities

    Problem:

    Determine the exact value of each trigonometric ratio.

    tan\frac{23\pi}{12}<br />

    I know that i have to use either: tan(A+B)=\frac{tan A + tan B}{1-tan Atan B} OR tan(A-B)=\frac{tan A-tan B}{1+tan A tan B}

    I'm stuck after that. Please help me.
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  2. #2
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    First find \tan{\left(\frac{\pi}{12}\right)}

    using the expansion for \tan{\left(\frac{\pi}{3} - \frac{\pi}{4}\right)}

    because 1/3 - 1/4 = 1/12.

    Now find \tan{\left(\frac{23\pi}{12}\right)} by expanding
    \tan{\left(2\pi - \frac{\pi}{12}\right)}
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  3. #3
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    hint ...

    \frac{23\pi}{12} = \frac{8\pi}{12} + \frac{15\pi}{12} = \frac{2\pi}{3} + \frac{5\pi}{4}

    last two values are on the unit circle ...

    \tan\left(\frac{2\pi}{3}\right) = -\sqrt{3}

    \tan\left(\frac{5\pi}{4}\right) = 1
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  4. #4
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    Hello, Morphayne!

    Determine the exact value of: . \tan\frac{23\pi}{12}
    Your game plan is excellent.

    Now we must express \frac{23\pi}{12} as the sum of two "familiar" angles.


    Make a list . . .

    . . \frac{23\pi}{12} \;=\;\begin{array}{ccccccc}\frac{\pi}{12} + \frac{22\pi}{12} &=& \frac{\pi}{12} + \frac{11\pi}{6} & & \text{We don't know }\frac{\pi}{12}\\ \\[-3mm]\frac{2\pi}{12} + \frac{21\pi}{12} &=&\frac{\pi}{6} + \frac{7\pi}{4} && \text{But we know these two!}\\ \\[-3mm] \frac{3\pi}{12} + \frac{20\pi}{12} \\ \vdots \end{array}

    So we have: . \tan\frac{23\pi}{12} \;=\;\tan\left(\frac{\pi}{6} + \frac{7\pi}{4}\right) \;=\;\frac{\tan\frac{\pi}{6} + \tan\frac{7\pi}{4}}{1 - \tan\frac{\pi}{6}\tan\frac{7\pi}{4}}

    . . . . . . . . . . . . . . . = \;\frac{\frac{1}{\sqrt{3}} + (-1)}{1 - \left(\frac{1}{\sqrt{3}}\right)(-1)} \;=\;\frac{\frac{1}{\sqrt{3}} - 1}{1 + \frac{1}{\sqrt{3}}}

    Multiply by \frac{\sqrt{3}}{\sqrt{3}}\!:\;\;\frac{\sqrt{3}\lef  t(\frac{1}{\sqrt{3}} - 1\right)} {\sqrt{3}\left(1 + \frac{1}{\sqrt{3}}\right)} \;=\;\frac{1-\sqrt{3}}{\sqrt{3}+1}

    Multiply by \frac{1-\sqrt{3}}{1-\sqrt{3}}\!:\quad\frac{1-\sqrt{3}}{1 + \sqrt{3}}\cdot\frac{1-\sqrt{3}}{1-\sqrt{3}} \;=\;\frac{1 - 2\sqrt{3} + 3}{1 - 3} \;=\;\frac{4-2\sqrt{3}}{-2}

    . . =\;\frac{-2(\sqrt{3}-2)}{-2} \;=\;\sqrt{3}-2

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  5. #5
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    Quote Originally Posted by Morphayne View Post
    Problem:

    Determine the exact value of each trigonometric ratio.

    tan\frac{23\pi}{12}<br />

    I know that i have to use either: tan(A+B)=\frac{tan A + tan B}{1-tan Atan B} OR tan(A-B)=\frac{tan A-tan B}{1+tan A tan B}

    I'm stuck after that. Please help me.
    Hi Morphayne,

    Since \frac{23\pi}{12}=\frac{24\pi-\pi}{12}=2\pi-\frac{\pi}{12},

    \tan \frac{23\pi}{12}=\tan (-\frac{\pi}{12})=-\tan \frac{\pi}{12}<br />

    And you will probably need to use the half-angle formula for tangent here.
    \tan \frac{\theta}{2}=\frac{\sin \theta}{1+cos\theta}

    See if you can do it from here.
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