Trigonometric sum asked in IMO-Do you speak Mathematics? group by Dan Sitaru

Exercise:

Solve,$\sum\limits_{k=1}^{10}{\sin \left( 1+kx \right)=}0$ , $x\in \left( 0,2\pi  \right)$

Solution: Let $C=\sum\limits_{k=1}^{10}{\sin \left( 1+kx \right)}$ multiply by $2\sin \frac{x}{2}$ both sides to get

$2\sin \frac{x}{2}C=\sum\limits_{k=1}^{10}{2\sin \left( 1+kx \right)\sin }\left( \frac{x}{2} \right)$

Notice that $-2\sin p\sin q=\cos \left( p+q \right)-\cos \left( p-q \right)\Leftrightarrow 2\sin p\sin q=\cos \left( p-q \right)-\cos \left( p+q \right)$

So $2\sin \left( 1+kx \right)\sin \left( \frac{x}{2} \right)=\cos \left( 1+kx-\frac{x}{2} \right)-\cos \left( 1+kx+\frac{x}{2} \right)$

                             $=\cos \left( 1+\frac{\left( 2k-1 \right)x}{2} \right)-\cos \left( 1+\frac{\left( 2k+1 \right)x}{2} \right)$

Hence $2\sin \frac{x}{2}C=\sum\limits_{k=1}^{10}{\cos \left( 1+\frac{\left( 2k-1 \right)x}{2} \right)-\sum\limits_{k=1}^{10}{\cos \left( 1+\frac{\left( 2k+1 \right)x}{2} \right)}}$

\[\Rightarrow 2\sin \frac{x}{2}C=\left\{ \cos \left( 1+\frac{x}{2} \right)+\cos \left( 1+\frac{3x}{2} \right)+\cos \left( 1+\frac{5x}{2} \right)+...+\cos \left( 1+\frac{19x}{2} \right) \right\}\]

                 $-\left\{ \cos \left( 1+\frac{3x}{2} \right)+\cos \left( 1+\frac{5x}{2} \right)+\cos \left( 1+\frac{7x}{2} \right)+....+\cos \left( 1+\frac{21x}{2} \right) \right\}$

$\Rightarrow 2\sin \frac{x}{2}C=\cos \left( 1+\frac{x}{2} \right)-\cos \left( 1+\frac{21x}{2} \right)$

Observe that, $p+q=1+\frac{x}{2}\,\,\,\And \,\,\,p-q=1+\frac{21x}{2}\Leftrightarrow p=1+\frac{11x}{2}\,\,\And \,\,q=-5x$

$\Rightarrow 2\sin \frac{x}{2}C=2\sin \left( 1+\frac{11x}{2} \right)\sin \left( -5x \right)$

Thus $C=\frac{\sin \left( -5x \right)}{\sin \frac{x}{2}}\sin \left( 1+\frac{11x}{2} \right)$

Therefore $\sum\limits_{k=1}^{10}{\sin \left( 1+kx \right)=\frac{-\sin 5x}{\sin \frac{x}{2}}\sin \left( 1+\frac{11x}{2} \right)=0}$

So $-\sin 5x\sin \left( 1+\frac{11x}{2} \right)=0\Leftrightarrow \sin 5x=0\,\,or\,\,\sin \left( 1+\frac{11x}{2} \right)=0$

$\Leftrightarrow \sin 5x=\sin \left( 0 \right)\,\,\,or\,\,\sin \left( 1+\frac{11x}{2} \right)=\sin 0$

So $5x=0+2k\pi \,\,\,or\,\,5x=\pi +2k\pi \,\,::\,\,\,1+\frac{11x}{2}=0+2k\pi \,\,or\,\,1+\frac{11x}{2}=\pi +2k\pi $

$x=\frac{2k\pi }{5}\,\,\,or\,\,x=\frac{\pi +2k\pi }{5}\,\,\,::\,\,\,\,\frac{11x}{2}=-1+2k\pi \,\,or\,\,\frac{11x}{2}=\pi -1+2k\pi $

$x=\left\{ \frac{\pi }{5},\frac{2\pi }{5},\frac{3\pi }{5},\frac{4\pi }{5},\pi ,\frac{6\pi }{5},\frac{7\pi }{5},\frac{8\pi }{5},\frac{9\pi }{5} \right\}\,\,=\frac{\left( k+1 \right)\pi }{5}\,\,,k=0,1,2,...,9$  so 9 roots

Or $x=\left\{ \frac{-2+2\pi }{11} \right.,\frac{-2+4\pi }{11},\frac{-2+6\pi }{11},\frac{-2+8\pi }{11},\frac{-2+10\pi }{11},\frac{-2+12\pi }{11},\frac{-2+14\pi }{11},\frac{-2+16\pi }{11},$

        $\frac{-2+18\pi }{11},\frac{-2+20\pi }{11},\left. \frac{-2+22\pi }{11} \right\}=\frac{-2+2k\pi }{11}\,\,\,,\,\,k=1,2,...,11$ so 11 roots

Hence there are 20 roots in $\left( 0,2\pi  \right)$ for this equation.