Q. The trigonometric form of $\mathcal{z =}\left( 1 - i\cot 8 \right)^{3}\ ($where$\ i = \sqrt{- 1})$ is
A) $cosec^{3}\ 8.e^{i\left( 24 - \frac{3\pi}{2} \right)}$
B) $cosec^{3}\ 8.e^{- i\left( 2u - \frac{3\pi}{2} \right)}$
C) $cosec^{3}\ 8.e^{i\left( 36 - \frac{\pi}{2} \right)}$
D) $cosec^{2}\ 8.e^{- i24 + \frac{\pi}{2}}$
Explanation
(a)
$\mathcal{z} = (1 - i{\cot{8)}}^{3}$
$= cosec^{3}8(\sin{8 - i{\cos{8)}}^{3}}$
$= cosec^{3}8\left( \cos{\left( \frac{\pi}{2} - 8 \right) - i\sin\left( \frac{\pi}{2} - 8 \right)} \right)^{3}$
$= cosec^{3}8\left( \cos{\left( \frac{3\pi}{2} - 24 \right) - i\sin\left( \frac{3\pi}{2} - 24 \right)} \right)$
$= cosec^{3}8.e^{- i\left( \frac{3\pi}{2}\ - 24 \right)}$
$= cosec^{3}\ 8.e^{i\left( 24 - \ \frac{3\pi}{2} \right)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.