Q111 / 160 · AP EAMCET 2015 · Mathematics

$\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^8$ is equal to

  • A
    $1$
  • Correct answer
    B
    $-1$
  • C
    $2$
  • D
    $\frac{1}{2}$

Solution

$\begin{aligned} & \text { }\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^8 \\ & =\left(\frac{2 \cos ^2 \frac{\pi}{16}-2 i \sin \frac{\pi}{16} \cos \frac{\pi}{16}}{2 \cos ^2 \frac{\pi}{16}+2 i \sin \frac{\pi}{16} \cos \frac{\pi}{16}}\right)^8 \\ & =\left(\frac{2 \cos \frac{\pi}{16}\left(\cos \frac{\pi}{16}-i \sin \frac{\pi}{16}\right)}{2 \cos \frac{\pi}{16}\left(\cos \frac{\pi}{16}+i \sin \frac{\pi}{16}\right)}\right)^8 \\ & =\left(\frac{\cos \frac{\pi}{16}-i \sin \frac{\pi}{16}}{\cos \frac{\pi}{16}+i \sin \frac{\pi}{16}}\right)^8 \\ & =\left(\frac{e^{-\frac{\pi}{16} i}}{e^{\frac{\pi}{16} i}}\right)^8 \\ & \end{aligned}$ $\begin{array}{lr} =\left(e^{-\frac{2 \pi}{16} i}\right)^8 & \text { [Using eular's theorem] } \\ =\left(e^{-8 \times \frac{2 \pi}{16} i}\right) & \text { [Using De-Moivre's theorem] } \end{array}$ $\begin{aligned} & =e^{-\pi i} \\ & =\cos \pi-i \sin \pi \\ & =-1\end{aligned}$

Hence, the correct answer is option: 'B'.

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