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 $\sqrt{A^2+B^2}$ is equal to

• A. Rational number
• B. Irrational number
• C. A surd
• D. a fraction

Answer 1

The correct option is A Rational number

$A=\sin \frac{2\pi }{7}-\sin \frac{4\pi }{7}+\sin \frac{8\pi }{7}$

$B=\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{8\pi }{7}$

$e^{i\left(2\pi \right)}=1$

$\Rightarrow u^7=1$

$u=e^{i\left(\frac{2\pi }{7}\right)}=\cos \left(\frac{2\pi }{7}\right)+i\sin \left(\frac{2\pi }{7}\right)$

$\therefore u^6+u^5+u^4+u^3+u^2+u+1=0$

Let $m=u+u^2+u^4$

$m^2=u^2+u^4+u^8+2(u^3+u^6+u^5)$

$=u+u^2+u^4+u^3+u^5+u^6+(u^3+u^6+u^5)$

$=-1+(-1-u-u^2-u^4)$

$=-2-m$ As ($m=u+u^2+u^4$)

$m^2+m+2=0$

$m=\frac{-1\pm \sqrt{1-8}}{2}$

$m=\frac{-1}{2}\pm i\frac{\sqrt{7}}{2}$

$\sin \frac{2\pi }{7}+\sin \frac{4\pi }{7}+\sin \frac{8\pi }{7}>0$

$\therefore A=\frac{\sqrt{7}}{2},B=\frac{-1}{2}$

$\sqrt{A^2+B^2}=\sqrt{\frac{7}{4}+\frac{1}{4}}=\sqrt{2}$