Question

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

• A. $\frac{2\mathrm{\pi }}{5}$
• B. $\frac{\mathrm{\pi }}{15}$
• C. $\frac{\mathrm{\pi }}{10}$
• D. $\frac{\mathrm{\pi }}{5}$

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{10}$

Step1: Given expression

The given expression is as follows

$\sin \frac{\pi }{5}+i\left(1-\cos \frac{\pi }{5}\right)$

Step2: Formula used

It is understood that the amplitude $\left(z\right)={\tan }^{-1}\left(\frac{y}{x}\right)$

Step3: Obtaining the amplitude

On comparing the expression $\sin \frac{\pi }{5}+i\left(1-\cos \frac{\pi }{5}\right)$ with $z=x+iy$,

we get $x=\sin \frac{\pi }{5}$ and $y=\left(1-\cos \frac{\mathrm{\pi }}{5}\right)$

Substitute the known values in the formula $\left(z\right)={\tan }^{-1}\left(\frac{y}{x}\right)$,

$\begin{array}{rcl}amplitude& =& {\tan }^{-1}\left(\frac{1-\cos \frac{\pi }{5}}{\sin \frac{\pi }{5}}\right)\left[\mathrm{from}\mathrm{f}\text{ormula:}\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\&1-\cos \theta =2{\sin }^2\frac{\theta }{2}\right]\\ & =& {\tan }^{-1}\left(\frac{2{\sin }^2{\displaystyle \frac{\mathrm{\pi }}{10}}}{2\sin \frac{\pi }{10}\cos \frac{\pi }{10}}\right)\\ & =& {\tan }^{-1}\left(\tan \frac{\mathrm{\pi }}{10}\right)\\ & =& \frac{\mathrm{\pi }}{10}\end{array}$

So, the amplitude of $\sin \frac{\pi }{5}+i\left(1-\cos \frac{\pi }{5}\right)$ is $\frac{\pi }{10}$.

Hence, option C is the correct answer.