Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

For option (A): Consider the sum $S_n={\sum }_{k=1}^n\cos \left(\frac{k\pi }{n+1}\right)$. This is a sum of cosines in an arithmetic progression. Using the formula ${\sum }_{k=1}^n\cos (a+kd)=\frac{\cos (a+(n+1)d/2)\sin (nd/2)}{\sin (d/2)}$, with $a=\frac{\pi }{n+1}$ and $d=\frac{\pi }{n+1}$.$$S_n=\frac{\cos \left(\frac{\pi }{n+1}+\frac{(n-1)\pi }{2(n+1)}\right)\sin \left(\frac{n\pi }{2(n+1)}\right)}{\sin \left(\frac{\pi }{2(n+1)}\right)}=\frac{\cos \left(\frac{2\pi +(n-1)\pi }{2(n+1)}\right)\sin \left(\frac{n\pi }{2(n+1)}\right)}{\sin \left(\frac{\pi }{2(n+1)}\right)}=\frac{\cos \left(\frac{(n+1)\pi }{2(n+1)}\right)\sin \left(\frac{n\pi }{2(n+1)}\right)}{\sin \left(\frac{\pi }{2(n+1)}\right)}$$ $$S_n=\frac{\cos (\pi /2)\sin \left(\frac{n\pi }{2(n+1)}\right)}{\sin \left(\frac{\pi }{2(n+1)}\right)}=\frac{0\cdot \sin \left(\frac{n\pi }{2(n+1)}\right)}{\sin \left(\frac{\pi }{2(n+1)}\right)}=0$$

Alternatively, using complex numbers, $S_n=\text{Re}\left({\sum }_{k=1}^n{e}^{ik\pi /(n+1)}\right)$. This is a geometric series with ratio $r=e^{i\pi /(n+1)}$. The sum is $r\frac{1-r^n}{1-r}=e^{i\pi /(n+1)}\frac{1-e^{in\pi /(n+1)}}{1-e^{i\pi /(n+1)}}=e^{i\pi /(n+1)}\frac{1-e^{i(\pi -\pi /(n+1))}}{1-e^{i\pi /(n+1)}}=e^{i\pi /(n+1)}\frac{1-(-1)e^{-i\pi /(n+1)}}{1-e^{i\pi /(n+1)}}=\frac{1+e^{i\pi /(n+1)}}{1-e^{i\pi /(n+1)}}$. This simplifies to $i\cot (\pi /(2(n+1)))$, whose real part is 0. So (A) is correct.

For option (B): ${\sum }_{k=1}^n{\sin }^2\left(\frac{k\pi }{n+1}\right)={\sum }_{k=1}^n\frac{1-\cos \left(\frac{2k\pi }{n+1}\right)}{2}=\frac{n}{2}-\frac{1}{2}{\sum }_{k=1}^n\cos \left(\frac{2k\pi }{n+1}\right)$. The sum ${\sum }_{k=1}^n\cos \left(\frac{2k\pi }{n+1}\right)$ is a sum of cosines with $a=d=\frac{2\pi }{n+1}$. Using the formula, this sum equals $\frac{\cos (\pi )\sin (n\pi /(n+1))}{\sin (\pi /(n+1))}=\frac{-1\cdot \sin (\pi -\pi /(n+1))}{\sin (\pi /(n+1))}=-1$. So the total sum is $\frac{n}{2}-\frac{1}{2}(-1)=\frac{n+1}{2}$. Thus, (B) is incorrect.

For option (C): This sum does not have a simple general closed form. For $n=2$, the sum is $\tan (\pi /6)+\tan (2\pi /6)=1/\sqrt{3}+\sqrt{3}=4/\sqrt{3}$. The option suggests $\cot (\pi /6)=\sqrt{3}$. These are not equal. Thus, (C) is incorrect.

For option (D): ${\sum }_{k=1}^n\cos \left(\frac{k\pi }{n+1}\right)\sin \left(\frac{k\pi }{n+1}\right)=\frac{1}{2}{\sum }_{k=1}^n\sin \left(\frac{2k\pi }{n+1}\right)$. This is a sum of sines in an arithmetic progression with $a=d=\frac{2\pi }{n+1}$. Using the formula ${\sum }_{k=1}^n\sin (a+kd)=\frac{\sin (a+(n+1)d/2)\sin (nd/2)}{\sin (d/2)}$, this sum equals $\frac{\sin (\pi )\sin (n\pi /(n+1))}{\sin (\pi /(n+1))}=0$ since $\sin (\pi )=0$. So (D) is correct.