# $x=e^\frac{2i\pi}{11}$, show that $i\tan\frac{3\pi}{11}=\frac{x^3-1}{x^3+1}$.

$x=e^\frac{2i\pi}{11}$, show that $i\tan\frac{3\pi}{11}=\frac{x^3-1}{x^3+1}$.

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I suppose you know that $e^{ix} = \cos x + i \sin x$. If you put in $-x$ for $x$ you get $e^{-ix} = \cos x - i\sin x$.

Adding or subtracting these, you get sine and cosine in terms of $e^{ix}$ and $e^{-ix}$:

\begin{align} \sin x & = {e^{ix} - e^{-ix}\over 2i}\\ \cos x & = {e^{ix} + e^{-ix} \over 2 } \end{align}

Once you have sine and cosine in this form, you just divide them to get tangent: $$\tan x = {\sin x \over\cos x} = {1\over i}\cdot{e^{ix} - e^{-ix} \over e^{ix} + e^{-ix}}$$

Can you take it from there?

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$$\frac{x^3-1}{x^3+1} = \frac{x^{1.5}-x^{-1.5}}{x^{1.5}+x^{-1.5}}$$ Now, $x^{1.5} = e^{\frac{3\pi}{11}}$

Also, substituting the well known fact that $e^{iy} = \cos{y}+i\sin{y}$, you can verify that $$\tan{y} = -i \frac{e^{iy}-e^{-iy}}{e^{iy}-e^{-iy}}$$ Just substitute $y = x^{1.5}$ and multiply LHS and RHS by $i$

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If $x=e^{2m\pi i}, x^n=e^{2nm\pi i}$

Applying Componendo and dividendo on $$\frac{x^n}1=\frac{e^{mn\pi i}}{e^{-mn\pi i}},$$

$$\frac{x^n-1}{x^n+1}=\frac{e^{mn\pi i}-e^{-mn\pi i}}{e^{mn\pi i}+e^{-mn\pi i}}$$ $$=\frac{2i\sin mn\pi}{2\cos mn\pi}=i\tan mn\pi$$

Here $n=3,m=\frac1{11}$

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