Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Let $S={\sum }_{k=1}^{10}\sin \left(\frac{k\pi }{11}\right)\sin \left(\frac{(k+1)\pi }{11}\right)$. Using the identity $2\sin A\sin B=\cos (A-B)-\cos (A+B)$, we have:$$2S=\sum _{k=1}^{10}\left[\cos \left(\frac{(k+1)\pi }{11}-\frac{k\pi }{11}\right)-\cos \left(\frac{(k+1)\pi }{11}+\frac{k\pi }{11}\right)\right]$$ $$2S=\sum _{k=1}^{10}\left[\cos \left(\frac{\pi }{11}\right)-\cos \left(\frac{(2k+1)\pi }{11}\right)\right]$$ $$2S=10\cos \left(\frac{\pi }{11}\right)-\sum _{k=1}^{10}\cos \left(\frac{(2k+1)\pi }{11}\right)$$

The sum${\sum }_{k=1}^{10}\cos \left(\frac{(2k+1)\pi }{11}\right)$ is an arithmetic progression of cosines with first term $a=\frac{3\pi }{11}$, common difference $d=\frac{2\pi }{11}$, and $n=10$ terms. Using the formula ${\sum }_{r=1}^n\cos (a+(r-1)d)=\frac{\cos (a+(n-1)d/2)\sin (nd/2)}{\sin (d/2)}$:$$\frac{\cos \left(\frac{3\pi }{11}+\frac{9\cdot (2\pi /11)}{2}\right)\sin \left(\frac{10\cdot (2\pi /11)}{2}\right)}{\sin \left(\frac{2\pi /11}{2}\right)}=\frac{\cos \left(\frac{12\pi }{11}\right)\sin \left(\frac{10\pi }{11}\right)}{\sin \left(\frac{\pi }{11}\right)}$$ Since$\cos \left(\frac{12\pi }{11}\right)=\cos \left(\pi +\frac{\pi }{11}\right)=-\cos \left(\frac{\pi }{11}\right)$ and $\sin \left(\frac{10\pi }{11}\right)=\sin \left(\pi -\frac{\pi }{11}\right)=\sin \left(\frac{\pi }{11}\right)$, the sum simplifies to $\frac{-\cos (\pi /11)\sin (\pi /11)}{\sin (\pi /11)}=-\cos (\pi /11)$. Substituting this back into the expression for $2S$:$$2S=10\cos \left(\frac{\pi }{11}\right)-\left(-\cos \left(\frac{\pi }{11}\right)\right)=11\cos \left(\frac{\pi }{11}\right)$$ Thus,$S=\frac{11}{2}\cos \left(\frac{\pi }{11}\right)$.