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Thread: Finding an exact expression for tan(pi/12)

  1. #1
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    Finding an exact expression for tan(pi/12)

    The question:
    Let $\displaystyle z = -2 + 2i$ and $\displaystyle w = -1 - \sqrt{3}i$.
    i) Write z and w in polar form and thus write zw in polar form.
    ii) Hence find an exact expression for $\displaystyle tan(\frac{\pi}{12})$

    My attempt:
    i) This is simple,$\displaystyle z = \sqrt{8}e^{\frac{3\pi}{4}i}; w = 2e^{\frac{-2\pi}{3}i} ; zw = 2\sqrt{8}e^{\frac{\pi}{12}i}$
    ii) I'm not sure how to go about this. I notice that the argument of the previous answer matches that of this question. However, I do not know how to attempt it.

    Any help would be great!
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  2. #2
    Pim
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    If you visualize this, you get a triangle, one point on the origin. The vector $\displaystyle zw$ makes an angle of $\displaystyle \frac{1}{12} \pi$ with the x-axis. Therefore, if you get an $\displaystyle a+bi$ form of $\displaystyle 2\sqrt{8}e^{\frac{1}{12 \pi}}$ you have $\displaystyle tan(\frac{1}{12 \pi}) = \frac{b}{a}$
    You can get the $\displaystyle a+bi$ form by just multiplying $\displaystyle z$ and $\displaystyle w$ and working out the brackets.

    I hope this is clear, if not, please say so.
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  3. #3
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    Quote Originally Posted by Glitch View Post
    The question:
    Let $\displaystyle z = -2 + 2i$ and $\displaystyle w = -1 - \sqrt{3}i$.
    i) Write z and w in polar form and thus write zw in polar form.
    ii) Hence find an exact expression for $\displaystyle tan(\frac{\pi}{12})$

    My attempt:
    i) This is simple,$\displaystyle z = \sqrt{8}e^{\frac{3\pi}{4}i}; w = 2e^{\frac{-2\pi}{3}i} ; zw = 2\sqrt{8}e^{\frac{\pi}{12}i}$
    ii) I'm not sure how to go about this. I notice that the argument of the previous answer matches that of this question. However, I do not know how to attempt it.

    Any help would be great!
    $\displaystyle \displaystyle\ tan\left(\frac{\pi}{12}\right)=\frac{sin\left(\fra c{\pi}{12}\right)}{cos\left(\frac{\pi}{12}\right)}$

    $\displaystyle e^{(i\theta)}=cos(\theta)+isin(\theta)$

    so if you multiply out

    $\displaystyle zw=(-2+2i)(-1-\sqrt{3}i)$ you will be able to continue by comparing terms
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  4. #4
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    Aha, that's interesting. Thanks!
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  5. #5
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    Quote Originally Posted by Glitch View Post
    The question:
    Let $\displaystyle z = -2 + 2i$ and $\displaystyle w = -1 - \sqrt{3}i$.
    i) Write z and w in polar form and thus write zw in polar form.
    ii) Hence find an exact expression for $\displaystyle tan(\frac{\pi}{12})$

    My attempt:
    i) This is simple,$\displaystyle z = \sqrt{8}e^{\frac{3\pi}{4}i}; w = 2e^{\frac{-2\pi}{3}i} ; zw = 2\sqrt{8}e^{\frac{\pi}{12}i}$
    ii) I'm not sure how to go about this. I notice that the argument of the previous answer matches that of this question. However, I do not know how to attempt it.

    Any help would be great!
    You don't need to go into the Complex number system here.

    Just use the angle sum identity for tangent...

    $\displaystyle \tan{(\alpha \pm \beta)} = \frac{\tan{\alpha} \pm \tan{\beta}}{1 \mp \tan{\alpha}\tan{\beta}}$.


    Here $\displaystyle \tan{\frac{\pi}{12}} = \tan{\left(\frac{\pi}{3} - \frac{\pi}{4}\right)}$

    $\displaystyle = \frac{\tan{\frac{\pi}{3}} - \tan{\frac{\pi}{4}}}{1 + \tan{\frac{\pi}{3}}\tan{\frac{\pi}{4}}}$
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  6. #6
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    However, the question really is specifically asking you to go the route of the polar form of the complex numbers.
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