The amplitude of sin frac{pi}{5} + i(1 - cos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $z = \sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$.Using the trigonometric identities $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	he

Vedclass · Accepted Answer

$\frac{\pi}{10}$

Vedclass · Answer

(B) Let $z = \sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$.Using the trigonometric identities $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ and $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$,we get:$z = 2 \sin \frac{\pi}{10} \cos \frac{\pi}{10} + i(2 \sin^2 \frac{\pi}{10})$$z = 2 \sin \frac{\pi}{10} (\cos \frac{\pi}{10} + i \sin \frac{\pi}{10})$Since $2 \sin \frac{\pi}{10} > 0$,the complex number is in the polar form $r(\cos 	heta + i \sin 	heta)$ where $r = 2 \sin \frac{\pi}{10}$ and $	heta = \frac{\pi}{10}$.Thus,the amplitude of the given complex number is $\frac{\pi}{10}$.

The amplitude of $\sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$ is

Similar Questions

If $z = \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}$,then $arg(z) = $ ............. $^\circ$

The amplitude of $\frac{1 + \sqrt{3}i}{\sqrt{3} + i}$ is

Convert the following into polar form: $\frac{1+3i}{1-2i}$

Convert the following complex number into polar form: $\frac{1+7i}{(2-i)^2}$

The amplitude of $0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module