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# Show that one value of the expression ${{\left( \dfrac{1+\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{1+\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}} \right)}^{\dfrac{8}{3}}}$ is -1.  Verified
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Hint: Use Euler’s identity ${{e}^{i\theta }}=\cos \theta +i\sin \theta$and ${{i}^{2}}=-1$. Try converting the numerator and denominator in $\cos \theta +i\sin \theta$ form. Use the trigonometric identities $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ and $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$.

Let $L=\dfrac{1+\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{1+\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}}$
Using $\sin x=\cos \left( \dfrac{\pi }{2}-x \right)$ and $\cos x=\sin \left( \dfrac{\pi }{2}-x \right)$ we get
$L=\dfrac{1+\cos \dfrac{3\pi }{8}+i\sin \dfrac{3\pi }{8}}{1+\cos \dfrac{3\pi }{8}-i\sin \dfrac{3\pi }{8}}$
Using $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ and $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ we get
$L=\dfrac{2{{\cos }^{2}}\dfrac{3\pi }{16}+2i\sin \dfrac{3\pi }{16}\cos \dfrac{3\pi }{16}}{2{{\cos }^{2}}\dfrac{3\pi }{16}-2i\sin \dfrac{3\pi }{16}\cos \dfrac{3\pi }{16}}$
Taking $2\cos \dfrac{3\pi }{16}$ common from numerator and denominator, we get
$L=\dfrac{2\cos \dfrac{3\pi }{16}\left( \cos \dfrac{3\pi }{16}+i\sin \dfrac{3\pi }{16} \right)}{2\cos \dfrac{3\pi }{16}\left( \cos \dfrac{3\pi }{16}-i\sin \dfrac{3\pi }{16} \right)}$
Simplifying we get
\begin{align} & L=\dfrac{{{e}^{i\dfrac{3\pi }{16}}}}{{{e}^{-i\dfrac{3\pi }{16}}}} \\ & ={{e}^{i\dfrac{3\pi }{16}+i\dfrac{3\pi }{16}}} \\ & ={{e}^{i\dfrac{6\pi }{16}}} \\ & ={{e}^{\dfrac{3\pi }{8}}} \\ \end{align}
Hence ${{\left( \dfrac{1+\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{1+\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}} \right)}^{\dfrac{8}{3}}}={{L}^{\dfrac{8}{3}}}={{e}^{i\dfrac{3\pi }{8}\times \dfrac{8}{3}}}={{e}^{i\pi }}=\cos \pi +i\sin \pi =-1$
Hence, we have ${{\left( \dfrac{1+\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{1+\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}} \right)}^{\dfrac{8}{3}}}=-1$.

Note:
 Euler’s identity ${{e}^{i\theta }}=\cos \theta +i\sin \theta$

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