Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Let $S={\sum }_{k=1}^n{\cos }^2\left(\frac{k\pi }{2n+1}\right)$. Using the identity ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$, we have:$$S=\sum _{k=1}^n\frac{1+\cos \left(\frac{2k\pi }{2n+1}\right)}{2}=\frac{n}{2}+\frac{1}{2}\sum _{k=1}^n\cos \left(\frac{2k\pi }{2n+1}\right)$$ Let $N=2n+1$. We know that for any integer $N>1$, ${\sum }_{k=0}^{N-1}\cos \left(\frac{2k\pi }{N}\right)=0$. This implies $1+{\sum }_{k=1}^{N-1}\cos \left(\frac{2k\pi }{N}\right)=0$. Since $\cos \left(\frac{2k\pi }{N}\right)=\cos \left(\frac{2(N-k)\pi }{N}\right)$, the sum ${\sum }_{k=1}^{N-1}\cos \left(\frac{2k\pi }{N}\right)$ is symmetric. Thus, $2{\sum }_{k=1}^n\cos \left(\frac{2k\pi }{N}\right)=-1$, which means ${\sum }_{k=1}^n\cos \left(\frac{2k\pi }{N}\right)=-\frac{1}{2}$. Substituting this back into the expression for $S$: $$S=\frac{n}{2}+\frac{1}{2}\left(-\frac{1}{2}\right)=\frac{n}{2}-\frac{1}{4}=\frac{2n-1}{4}$$