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Question: Evaluate the following expression: \[{{\left( \dfrac{1+\cos \dfrac{\pi }{8}-i\sin \dfrac{\pi }{8}}...

Evaluate the following expression:
(1+cos⁡π8−isin⁡π81+cos⁡π8+isin⁡π8)8={{\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) 11
(b) −1-1
(c) 22
(d) 12\dfrac{1}{2}

Explanation

Solution

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

Complete step-by-step solution -
We have to find the value of
A=(1+cos⁡π8−isin⁡π81+cos⁡π8+isin⁡π8)8....(i)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⁡2θ=2cos⁡2θ−1\cos 2\theta =2{{\cos }^{2}}\theta -1 when 2θ=π82\theta =\dfrac{\pi }{8}, then θ=π16\theta =\dfrac{\pi }{16}.
Therefore, cos⁡π8=2cos⁡2π16−1\cos \dfrac{\pi }{8}=2{{\cos }^{2}}\dfrac{\pi }{16}-1
Also, sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta
sin⁡π8=2sin⁡π16cos⁡π16\sin \dfrac{\pi }{8}=2\sin \dfrac{\pi }{16}\cos \dfrac{\pi }{16}
Putting the values of cos⁡π8\cos \dfrac{\pi }{8} and sin⁡π8\sin \dfrac{\pi }{8}in equation (i), we get,
A=(1+2cos⁡2π16−1−i(2sin⁡π16cos⁡π16)1+2cos⁡2π16−1+i(2sin⁡π16cos⁡π16))8A={{\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}}
⇒A=(2cos⁡2π16−i(2sin⁡π16cos⁡π16)2cos⁡2π16+i(2sin⁡π16cos⁡π16))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 2cos⁡π162\cos \dfrac{\pi }{16} common from numerator and denominator and cancelling it, we get
A=(cos⁡π16−isin⁡π16cos⁡π16+isin⁡π16)8...(ii)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 eiθ=cos⁡θ+isin⁡θ....(iii){{e}^{i\theta }}=\cos \theta +i\sin \theta ....\left( iii \right)
Therefore, eiπ16=cos⁡π16+isin⁡π16{{e}^{i\dfrac{\pi }{16}}}=\cos \dfrac{\pi }{16}+i\sin \dfrac{\pi }{16}
ei(−π16)=cos⁡(−π16)+isin⁡(−π16){{e}^{i\left( -\dfrac{\pi }{16} \right)}}=\cos \left( -\dfrac{\pi }{16} \right)+i\sin \left( -\dfrac{\pi }{16} \right)
As sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta and cos⁡(−θ)=cos⁡θ\cos \left( -\theta \right)=\cos \theta
Therefore, e−iπ6=cos⁡(π16)−isin⁡(π16){{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=[e−iπ16eiπ16]8A={{\left[ \dfrac{{{e}^{-i\dfrac{\pi }{16}}}}{{{e}^{i\dfrac{\pi }{16}}}} \right]}^{8}}
We know that, aman=am−n\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}
Therefore, A=[e−iπ16−iπ16]8A={{\left[ {{e}^{-\dfrac{i\pi }{16}-\dfrac{i\pi }{16}}} \right]}^{8}}
⇒A=[e−iπ8]8\Rightarrow A={{\left[ {{e}^{-\dfrac{i\pi }{8}}} \right]}^{8}}
As (am)n=am.n{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m.n}}
We get, A=[e−iπ8.8]=e−iπA=\left[ {{e}^{\dfrac{-i\pi }{8}.8}} \right]={{e}^{-i\pi }}
From equation (iii),
A=e−iπ=cos⁡(−π)+isin⁡(−π)A={{e}^{-i\pi }}=\cos \left( -\pi \right)+i\sin \left( -\pi \right)
As, cos⁡(−π)=cos⁡(π)=−1\cos \left( -\pi \right)=\cos \left( \pi \right)=-1 and sin⁡(−π)=−sin⁡π=0\sin \left( -\pi \right)=-\sin \pi =0
We get, A=e−iπ=−1A={{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 π8\dfrac{\pi }{8} into π4\dfrac{\pi }{4} instead of π16.\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 π,π2,π4,etc.\pi ,\dfrac{\pi }{2},\dfrac{\pi }{4},etc.