$x=e^\frac{2i\pi}{11}$, show that $i\tan\frac{3\pi}{11}=\frac{x^3-1}{x^3+1}$.
I don't know how the solution jump to this. Please help. Thank you.
$x=e^\frac{2i\pi}{11}$, show that $i\tan\frac{3\pi}{11}=\frac{x^3-1}{x^3+1}$.
I don't know how the solution jump to this. Please help. Thank you.
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?
$\frac{x^3-1}{x^3+1} = \frac{x^{1.5}-x^{-1.5}}{x^{1.5}+x^{-1.5}}$ 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 $i$
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 $n=3,m=\frac1{11}$