Question

# 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.

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

[2] Demovire’s identity ${{\left( \cos x+i\sin x \right)}^{n}}=\cos nx+i\sin nx$

[3] We can use identity $\dfrac{1+\cos x+i\sin x}{1+\cos x-i\sin x}=\cos x+i\sin x$ directly to solve the question. The derivation of the above formula will follow the same steps as done in the above solution.

[4] Complex numbers help in finding the values of $\cos nx$ and $\sin nx$ conveniently and easily.

[5] Every complex number can be written in the form of $r{{e}^{ix}}$ where r is called the modulus of the complex number and x is called the argument of the complex number.

[6] Complex numbers play a role of backbone in calculus and are used in many places.

[7] $\left| z \right|=r$ is the equation of circle centred at (0,0) and radius r in the argand plane.

[8] Complex numbers are used in Calculus e.g. Cauchy residue theorem (Theorem for complex integrals) is used to evaluate some difficult real integrals.

[9] Multiplication by $i$ is equivalent to a rotation of $90{}^\circ$ in the argand plane.

[10] A complex number can be thought of as a vector in the Argand plane. This is why complex numbers follow a similar algebra as vector algebra.

[11] Complex numbers are used in AC circuit analysis etc in physics.