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TS EAMCET 5 May 2018 Shift 2 Solved Paper

Section: Mathematics

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Question : 9 of 160

Marks: +1, -0

\left( \frac{1+\cos\frac{\pi}{8}-i\sin\frac{\pi}{8}}{1+\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}} \right)^{12}=

−1
i
−i
2

Solution:

Consider the expression.

\left( \frac{1+\cos\frac{\pi}{8}-i\sin\frac{\pi}{8}}{1+\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}} \right)^{12}

It is solved as,

\left( \frac{1+\cos\frac{\pi}{8}-i\sin\frac{\pi}{8}}{1+\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}} \right)^{12}

= \left[ \frac{2\cos^2\frac{\pi}{16} - \sin\frac{\pi}{16}\cos\frac{\pi}{16} i}{2\cos^2\frac{\pi}{16} + 2\sin\frac{\pi}{16}\cos\frac{\pi}{16} i} \right]^{12}

= \left[ \frac{\cos\frac{\pi}{16}\left(\cos\frac{\pi}{16} - i\sin\frac{\pi}{16}\right)}{\cos\frac{\pi}{16}\left(\cos\frac{\pi}{16} + i\sin\frac{\pi}{16}\right)} \right]^{12}

= \left[ \frac{\cos\frac{\pi}{16} - i\sin\frac{\pi}{16}}{\cos\frac{\pi}{16} + i\sin\frac{\pi}{16}} \right]^{12}

= \left( \frac{e^{-i\pi/16}}{e^{i\pi/16}} \right)^{12}

Solve further,

\left( \frac{1+\cos\frac{\pi}{8}-i\sin\frac{\pi}{8}}{1+\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}} \right)^{12}

= \left( e^{i\pi/8} \right)^{12}

= \cos\frac{3\pi}{2} - i\sin\frac{3\pi}{2}

= 0 - (-i)

= i

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