Evaluate the following expression:
\[{{\left( \dfrac{1+\cos \dfrac{\pi }{8}-i\sin \dfrac{\pi }{8}}{1+\cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8}} \right)}^{8}}=\]
(a) \[1\]
(b) \[-1\]
(c) \[2\]
(d) \[\dfrac{1}{2}\]

- Hint: First of all, eliminate 1 from numerator and denominator by using half angle formulas that are \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] and \[\sin 2\theta =2\sin \theta \cos \theta \] and then use \[{{e}^{i\theta }}=\cos \theta +i\sin \theta \].

Complete step-by-step solution -
We have to find the value of
\[A={{\left( \dfrac{1+\cos \dfrac{\pi }{8}-i\sin \dfrac{\pi }{8}}{1+\cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8}} \right)}^{8}}....\left( i \right)\]
We know that, \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] when \[2\theta =\dfrac{\pi }{8}\], then \[\theta =\dfrac{\pi }{16}\].
Therefore, \[\cos \dfrac{\pi }{8}=2{{\cos }^{2}}\dfrac{\pi }{16}-1\]
Also, \[\sin 2\theta =2\sin \theta \cos \theta \]
\[\sin \dfrac{\pi }{8}=2\sin \dfrac{\pi }{16}\cos \dfrac{\pi }{16}\]
Putting the values of \[\cos \dfrac{\pi }{8}\] and \[\sin \dfrac{\pi }{8}\]in equation (i), we get,
\[A={{\left( \dfrac{1+2{{\cos }^{2}}\dfrac{\pi }{16}-1-i\left( 2\sin \dfrac{\pi }{16}\cos \dfrac{\pi }{16} \right)}{1+2{{\cos }^{2}}\dfrac{\pi }{16}-1+i\left( 2\sin \dfrac{\pi }{16}\cos \dfrac{\pi }{16} \right)} \right)}^{8}}\]
\[\Rightarrow A={{\left( \dfrac{2{{\cos }^{2}}\dfrac{\pi }{16}-i\left( 2\sin \dfrac{\pi }{16}\cos \dfrac{\pi }{16} \right)}{2{{\cos }^{2}}\dfrac{\pi }{16}+i\left( 2\sin \dfrac{\pi }{16}\cos \dfrac{\pi }{16} \right)} \right)}^{8}}\]
Taking \[2\cos \dfrac{\pi }{16}\] common from numerator and denominator and cancelling it, we get
\[A={{\left( \dfrac{\cos \dfrac{\pi }{16}-i\sin \dfrac{\pi }{16}}{\cos \dfrac{\pi }{16}+i\sin \dfrac{\pi }{16}} \right)}^{8}}...\left( ii \right)\]
We know that \[{{e}^{i\theta }}=\cos \theta +i\sin \theta ....\left( iii \right)\]
Therefore, \[{{e}^{i\dfrac{\pi }{16}}}=\cos \dfrac{\pi }{16}+i\sin \dfrac{\pi }{16}\]
\[{{e}^{i\left( -\dfrac{\pi }{16} \right)}}=\cos \left( -\dfrac{\pi }{16} \right)+i\sin \left( -\dfrac{\pi }{16} \right)\]
As \[\sin \left( -\theta \right)=-\sin \theta \] and \[\cos \left( -\theta \right)=\cos \theta \]
Therefore, \[{{e}^{-i\dfrac{\pi }{6}}}=\cos \left( \dfrac{\pi }{16} \right)-i\sin \left( \dfrac{\pi }{16} \right)\]
Putting these values in equation (ii), we get
\[A={{\left[ \dfrac{{{e}^{-i\dfrac{\pi }{16}}}}{{{e}^{i\dfrac{\pi }{16}}}} \right]}^{8}}\]
We know that, \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\]
Therefore, \[A={{\left[ {{e}^{-\dfrac{i\pi }{16}-\dfrac{i\pi }{16}}} \right]}^{8}}\]
\[\Rightarrow A={{\left[ {{e}^{-\dfrac{i\pi }{8}}} \right]}^{8}}\]
As \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m.n}}\]
We get, \[A=\left[ {{e}^{\dfrac{-i\pi }{8}.8}} \right]={{e}^{-i\pi }}\]
From equation (iii),
\[A={{e}^{-i\pi }}=\cos \left( -\pi \right)+i\sin \left( -\pi \right)\]
As, \[\cos \left( -\pi \right)=\cos \left( \pi \right)=-1\] and \[\sin \left( -\pi \right)=-\sin \pi =0\]
We get, \[A={{e}^{-i\pi }}=-1\]
Therefore, option (b) is the correct answer.

Note: In this question, students must take the utmost care of angles and their transformation. Students often make this mistake of converting \[\dfrac{\pi }{8}\] into \[\dfrac{\pi }{4}\] instead of \[\dfrac{\pi }{16}.\] So this must be taken care of. Also always try to reduce the angles into sine and cosine of familiar angles like \[\pi ,\dfrac{\pi }{2},\dfrac{\pi }{4},etc.\]