Question

For any interger $k$, let ${\alpha }_k=\cos \left(\frac{k\pi }{7}\right)+i\sin \left(\frac{k\pi }{7}\right)$ where, $i=\sqrt{-1}$. Then value of the expression $\frac{\sum _{k=1}^{12}|{\alpha }_{k+1}-{\alpha }_k|}{\sum _{k=1}^3|{\alpha }_{4k-1}-{\alpha }_{4k-2}|}=$ ______

• A. $3$
• B. $4$
• C. $2$
• D. $1$

Answer 1

Answer: (B)

$4$

Let, ${\alpha }_k=\cos \frac{k\pi }{7}+$ $i\sin \frac{k\pi }{7}$.

Now, $|{\alpha }_{k+1}-{\alpha }_k|=$ $|\{\cos \frac{(k+1)\pi }{7}-$ $\cos \frac{k\pi }{7}\}$ $+i\{\sin \frac{(k+1)\pi }{7}$$-\sin \frac{k\pi }{7}\}|$

$=[\{\cos \frac{(k+1)\pi }{7}-$ ${\cos \frac{k\pi }{7}\}}^2$ $+\{\sin \frac{(k+1)\pi }{7}$${{-\sin \frac{k\pi }{7}\}}^2]}^{\frac{1}{2}}$

$=\sqrt{2-2(\cos \frac{(k+1)\pi }{7}\cos \frac{k\pi }{7}+\sin \frac{(k+1)\pi }{7}\sin \frac{k\pi }{7})}$

$=\sqrt{2-2\left(\cos \frac{\pi }{7}\right)}$ [Since $\cos (A-B)=\cos A\cos B-\sin A\sin B$]

$\therefore \sum _{k=1}^{12}$$|{\alpha }_{k+1}-{\alpha }_k|$ $=12$ $\sqrt{2-2\left(\cos \frac{\pi }{7}\right)}$......(1).

And

Now, $|{\alpha }_{4k+1}-{\alpha }_{4k-2}|=$ $|\{\cos \frac{(4k+1)\pi }{7}-$ $\cos \frac{(4k-2)\pi }{7}\}$ $+i\{\sin \frac{(4k+1)\pi }{7}$$-\sin \frac{(4k-2)\pi }{7}\}|$

$=[\{\cos \frac{(4k+1)\pi }{7}-$ ${\cos \frac{(4k-2)\pi }{7}\}}^2$ $+\{\sin \frac{(4k+1)\pi }{7}$${{-\sin \frac{(4k-2)\pi }{7}\}}^2]}^{\frac{1}{2}}$

$=\sqrt{2-2(\cos \frac{(4k+1)\pi }{7}\cos \frac{(4k-2)\pi }{7}+\sin \frac{(4k+1)\pi }{7}\sin \frac{(4k-2)\pi }{7})}$

$=\sqrt{2-2\left(\cos \frac{\pi }{7}\right)}$ [Since $\cos (A-B)=\cos A\cos B-\sin A\sin B$]

$\therefore \sum _{k=1}^3$$|{\alpha }_{4k+1}-{\alpha }_{4k-2}|$ $=3$ $\sqrt{2-2\left(\cos \frac{\pi }{7}\right)}$........(2).

$\therefore \frac{\sum _{k=1}^{12}|{\alpha }_{k+1}-{\alpha }_k|}{\sum _{k=1}^3|{\alpha }_{4k+1}-{\alpha }_{4k-2}|}$ $=4$ [Using (1) and (2)].