Question & Answer
QUESTION

Simplify: $\dfrac{{{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}}{{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}}$

ANSWER Verified Verified
Hint: To simplify the above expression, first multiply and divide by $\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)$ inside the bracket to change base into real value and then change into Euler form of complex number to further simplify.

Euler form:

${{e}^{i\theta }}=\cos \theta +i\sin \theta $

Complete step-by-step answer:


We know that $\dfrac{{{a}^{n}}}{{{b}^{n}}}={{\left( \dfrac{a}{b} \right)}^{n}}$.

Using this, $\dfrac{{{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}}{{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}}={{\left( \dfrac{\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}} \right)}^{8}}$

Now, multiplying and dividing by $\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)$ inside the bracket, we will get,

\[\Rightarrow \dfrac{{{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}}{{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}}={{\left( \dfrac{\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}}\times \dfrac{\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}} \right)}^{8}}\]

Using $\left( a-b \right)\times \left( a+b \right)={{a}^{2}}-{{b}^{2}}$ in the denominator,

$\begin{align}

  & ={{\left( \dfrac{{{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{2}}}{{{\left( \sin \dfrac{\pi }{8} \right)}^{2}}-{{\left( i\cos \dfrac{\pi }{8} \right)}^{2}}} \right)}^{8}} \\

 & ={{\left( \dfrac{{{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{2}}}{{{\sin }^{2}}\dfrac{\pi }{8}-\left( -{{\cos }^{2}}\dfrac{\pi }{8} \right)} \right)}^{8}} \\

 & ={{\left( \dfrac{{{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{2}}}{{{\sin }^{2}}\dfrac{\pi }{8}+{{\cos }^{2}}\dfrac{\pi }{8}} \right)}^{8}} \\

\end{align}$

Using ${{\sin }^{2}}\dfrac{\pi }{8}+{{\cos }^{2}}\dfrac{\pi }{8}=1.\ \ \left( Identity\ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \right)$

$\Rightarrow \dfrac{{{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}}{{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}}={{\left( {{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{2}} \right)}^{8}}$

Now, using ${{\left( {{a}^{n}} \right)}^{m}}={{a}^{n\times m}}$

$\Rightarrow \dfrac{{{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}}{{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}}={{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{16}}..............\left( 1 \right)$

We know, $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta \ and\ \sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $.

 $\begin{align}

  & \Rightarrow \sin \dfrac{\pi }{8}=\cos \left( \dfrac{\pi }{2}-\dfrac{\pi }{8} \right)=\cos \dfrac{3\pi }{8} \\

 & and\ \cos \dfrac{\pi }{8}=\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{8} \right)=\sin \dfrac{3\pi }{8} \\

\end{align}$

Using these values of $\sin \dfrac{\pi }{8}$ and $\cos \dfrac{\pi }{8}$ in equation, we get,
$\Rightarrow \dfrac{{{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}}{{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}}={{\left[ \cos \dfrac{3\pi }{8}+i\sin \dfrac{3\pi }{8} \right]}^{16}}$

Using Euler form of $\left( \cos \dfrac{3\pi }{8}+i\sin \dfrac{3\pi }{8} \right)$, we will get,

$\begin{align}

  & \Rightarrow \dfrac{{{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}}{{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}}={{\left( {{e}^{i\left( \dfrac{3\pi }{8} \right)}} \right)}^{16}} \\

 & ={{e}^{i\left( \dfrac{3\pi }{8}\times 16 \right)}} \\

 & ={{e}^{i\left( \dfrac{3\pi }{2} \right)}} \\

\end{align}$

We know,

$\begin{align}

  & {{e}^{i\theta }}=\cos \theta +i\sin \theta \\

 & {{e}^{i\dfrac{3\pi }{2}}}=\cos \theta \left( \dfrac{3\pi }{2} \right)+i\sin \left( \dfrac{3\pi }{2}

\right) \\

 & =0-i \\

 & =-i \\

\end{align}$

Hence, $\dfrac{{{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}}{{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}}=-i$


Note: We can also solve this question without using Euler form and by using the identity,
$''{{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos \left( n\theta \right)+i\sin \left( n\theta \right)''$.


This can be proved using the Euler form of complex numbers.

$\begin{align}

  & {{e}^{i\theta }}=\cos \theta +i\sin \left( \theta \right) \\

 & {{\left( {{e}^{i\theta }} \right)}^{n}}={{e}^{in\theta }}\cos \left( n\theta \right)+i\sin \left(

n\theta \right) \\

\end{align}$