Question

Using eular's formula prove that $\frac{cos\pi }{11}+\frac{cos3\pi }{11}+\frac{co5\pi }{11}+\frac{cos7\pi }{11}+\frac{cos9\pi }{11}=\frac{1}{2}sinx$

Answer 1

$cos\frac{\pi }{11}+cos\frac{3\pi }{11}+cos\frac{5\pi }{11}+cos\frac{7\pi }{11}+cos\frac{9\pi }{11}$
Put $x=\frac{\pi }{11}\to cosx+cos3x+cos5x+cos7x+cos9x$
As per Eulor's formula: $cosnx=\frac{e^{nx}+e^{-nx}}{2}$
$\to \frac{1}{2}\{e^{ix}+e^{-ix}+e^{3ix}+e^{-3ix}+e^{5x}+e^{-5ix}+e^{7ix}+e^{-7ix}+e^{9ix}+e^{-9ix}\}\to \frac{1}{2}\{e^{ix}+e^{i3x}+e^{i5x}+e^{i7x}+e^{i9x}+e^{-x}+e^{-i5x}+e^{-i5x}+e^{-i7x}+e^{-9ix}\}$
$\to \frac{1}{2}\left\{e^{ix}\right(1+e^{2ix})+(e^{2ix}{)}^3+(e^{2ix}{)}^4+e^{-x}(1+e^{-2ix})+(e^{-2ix}{)}^2+\left(e^{-2ix}{)}^3\right\}$
It is know that $1+x+x^2+x^3....x^{n-1}=\frac{x^n-1}{x-1}$
$\to \frac{1}{2}\{e^{ix}\left(\frac{(e^{2ix}{)}^5-1}{e^{2ix}-1}\right)+e^{-ix}\left(\frac{(e^{-2ix}{)}^5-1}{e^{-2ix}-1}\right)\}\to \frac{1}{2}\{\frac{1}{e^{-ix}}\frac{e^{10ix}-1}{(e^{2ix}-1)}+\frac{1}{e^{ix}}.\frac{e^{-10ix}-1}{(e^{-2x}-1)}\}$
$\to \frac{1}{2}\{\frac{e^{10ix}-1}{e^{ix}-e^{-ix}}+\frac{e^{-10ix}-1}{e^{-ix}-e^{ix}}\}=\frac{1}{2}\left\{\frac{e^{10ix}-v-e^{-10ix}+x}{e^{ix}-e^{-ix}}\right\}$
$\to \frac{1}{2}\left\{\frac{\frac{(e^{i10x}-e^{-i10x})}{2}}{\frac{(e^{ix}-e^{-ix})}{2}}\right\}$
$\to \frac{1}{2}\frac{sin10x}{sinx}=\frac{1}{2}\frac{sin10\frac{\pi }{11}}{sin\frac{\pi }{11}}$(By eucleirs formula for sin x )
$=\frac{1}{2}(sincesin10x=sinx)$-proved