Question

If $A=\sin \frac{2\pi }{7}+\sin \frac{4\pi }{7}+\sin \frac{8\pi }{7}$ and $B=\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{8\pi }{7}$, then $A^2+B^2$ is equal to

• A. $2$
• B. $3$
• C. $\sqrt{2}$
• D. $1$

Answer 1

The correct option is A $2$

Let

z=$e^{\frac{2i\pi }{7}}=e^{2ia}$

where a=$\frac{\pi }{7}$

then $A^2+B^2$ is the real part of the sum

$3+2(z+z^2+z^3)=3+2z\frac{1-z^3}{1-z}$

which is

$=3+2cos\left(4a\right)\frac{sin\left(3a\right)}{sin\left(a\right)}$

$=3+\frac{1}{sin\left(a\right)}\left(sin\right(7a)-sin(a\left)\right)$

=3−1=2.