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Math Help - Polar form

  1. #1
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    Polar form

    Let z=2exp(-Pi/6) and w=sqrt(2)exp(Pi/4)

    Find zw in polar form and hence express tan(Pi/12) in surd form

    So I manage to find zw in polar form, which is 2sqrt(2)exp(Pi/12), but I don't know how to express tan(Pi/12) in surd form. Please help me thanks.
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  2. #2
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    Quote Originally Posted by Rine198 View Post
    Let z=2exp(-Pi/6) and w=sqrt(2)exp(Pi/4)
    Find zw in polar form and hence express tan(Pi/12) in surd form
    Do you know a formula for \tan(\frac{\theta}{2})}~?
    Note that \frac{\pi}{12}=\frac{1}{2}\frac{\pi}{6}.
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  3. #3
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    do u mean the t-formula?
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  4. #4
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    Hello, Rine198!

    \text{Express }\tan\frac{\pi}{12}\text{ in surd form.}

    I'm not sure about the "hence".
    We can get the answer with some old=fashioned trig.

    \tan\frac{\pi}{12} \;=\;\tan\left(\frac{\pi}{4}-\frac{\pi}{6}\right) \;=\;\dfrac{\tan\frac{\pi}{4} - \tan\frac{\pi}{6}}{1 + \tan\frac{\pi}{4}\tan\frac{\pi}{5}} \;=\;\dfrac{1-\frac{1}{\sqrt{3}}} {1 + \frac{1}{\sqrt{3}}}

    \text{Multiply by }\frac{\sqrt{3}}{\sqrt{3}}\!: \;\;\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}

    \text{Rationalize: }\;\dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}\cdot \dfrac{\sqrt{3} - 1}{\sqrt{3}+1} \;=\;\dfrac{3 - 2\sqrt{3} + 1}{3-1}

    . . . . . . . . =\;\dfrac{4 - 2\sqrt{3}}{2} \;=\;2 - \sqrt{3}

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