Question

If $x=\sin \frac{2\pi }{7}+\sin \frac{4\pi }{7}+\sin \frac{8\pi }{7}andy=\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{8\pi }{7}then{x}^2+y^2=$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

Let $\alpha =\cos \left(\frac{2\pi }{7}\right)+i\sin \left(\frac{2\pi }{7}\right)$

Then:${\alpha }^7=\cos \left(2\pi \right)+i\sin \left(2\pi \right)=1+i0$

So:$0={\alpha }^7-1=(\alpha -1)({\alpha }^6+{\alpha }^5+{\alpha }^4+{\alpha }^3+{\alpha }^2+\alpha +1)$

Consider ${({\alpha }^4+{\alpha }^2+\alpha )}^2$

$={\alpha }^2+{\alpha }^4+{\alpha }^8+2({\alpha }^3+{\alpha }^6+{\alpha }^5)$

$=\alpha +{\alpha }^2+{\alpha }^4+2(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5+{\alpha }^6-\alpha -{\alpha }^2-{\alpha }^4)$

$=\alpha +{\alpha }^2+{\alpha }^4+2(-1-\alpha -{\alpha }^2-{\alpha }^4)$

$=-(1+{\alpha }^2+{\alpha }^4)-2$

So,$\alpha +{\alpha }^2+{\alpha }^4$ is a root of:

$t^2+t+2=0$ which we can solve by completing the square:

$\Rightarrow t^2+2\times \frac{1}{2}t+\frac{1}{4}-\frac{1}{4}+2=0$

$\Rightarrow {(t+\frac{1}{2})}^2-\frac{1-8}{4}=0$

$\Rightarrow {(t+\frac{1}{2})}^2-\frac{-7}{4}=0$

$\Rightarrow {(t+\frac{1}{2})}^2-{\left(\frac{\sqrt{7}i}{2}\right)}^2=0$

$\Rightarrow (t+\frac{1}{2}-\frac{\sqrt{7}i}{2})(t+\frac{1}{2}+\frac{\sqrt{7}i}{2})=0$

So,$\alpha +{\alpha }^2+{\alpha }^4=t=-\frac{1}{2}\pm \frac{\sqrt{7}i}{2}$

$\Rightarrow (\cos \left(\frac{2\pi }{7}\right)+\cos \left(\frac{4\pi }{7}\right)+\cos \left(\frac{8\pi }{7}\right))+i(\sin \left(\frac{2\pi }{7}\right)+\sin \left(\frac{4\pi }{7}\right)+\sin \left(\frac{8\pi }{7}\right))=-\frac{1}{2}\pm \frac{\sqrt{7}i}{2}$

Equating the real and imaginary parts,we get

$\cos \left(\frac{2\pi }{7}\right)+\cos \left(\frac{4\pi }{7}\right)+\cos \left(\frac{8\pi }{7}\right)=-\frac{1}{2}$

and $\sin \left(\frac{2\pi }{7}\right)+\sin \left(\frac{4\pi }{7}\right)+\sin \left(\frac{8\pi }{7}\right)=\pm \frac{\sqrt{7}}{2}$

$\therefore x=\sin \left(\frac{2\pi }{7}\right)+\sin \left(\frac{4\pi }{7}\right)+\sin \left(\frac{8\pi }{7}\right)=\pm \frac{\sqrt{7}}{2}$

and $y=\cos \left(\frac{2\pi }{7}\right)+\cos \left(\frac{4\pi }{7}\right)+\cos \left(\frac{8\pi }{7}\right)=-\frac{1}{2}$

$\therefore x^2+y^2={(\pm \frac{\sqrt{7}}{2})}^2+{(-\frac{1}{2})}^2$

$\Rightarrow x^2+y^2=\frac{7}{4}+\frac{1}{4}=\frac{8}{4}=2$

Hence $x^2+y^2=2$