Question

If $x=CiS\theta $ , then find the value of the expression $\left( {{x}^{6}}+\dfrac{1}{{{x}^{6}}} \right)$.

Answer 1

Hint: In this question $CiS\theta $ mean $\cos \theta +i\sin \theta $. This question will be solved by using the Euler’s formula for complex numbers ($\cos \theta +i\sin \theta ={{e}^{i\theta }}$). Also you can use the negative angle of sine and cosine trigonometric function formula ($\cos (-x)=\cos x$ and $\sin (-x)=-\sin x$).

Complete step-by-step answer:
Euler's formula, named after Leonhard Euler, is a mathematical formula in complex analysis that establishes the fundamental relationship between the trigonometric functions and the complex exponential function
It was around 1740, and mathematicians were interested in imaginary numbers. So Leonhard Euler was enjoying himself one day, playing with imaginary numbers and he took the well-known Taylor series and put it into it. The square of the imaginary number is -1, he simplified it and grouped all the i terms at the end. Finally he found the two groups of the Taylor series for the trigonometric function cosine and sine and after simplifying he got the formula. So, it is now called Euler’s Formula.
It is given that $x=CiS\theta =\cos \theta +i\sin \theta $
We know that, the Euler’s formula $\cos \theta +i\sin \theta ={{e}^{i\theta }}$
$x={{e}^{i\theta }}$
Let us consider the given expression $\left( {{x}^{6}}+\dfrac{1}{{{x}^{6}}} \right)$, put the value of the $x={{e}^{i\theta }}$ and we have
\[\left( {{x}^{6}}+\dfrac{1}{{{x}^{6}}} \right)={{\left( {{e}^{i\theta }} \right)}^{6}}+\dfrac{1}{{{\left( {{e}^{i\theta }} \right)}^{6}}}\]
By using the indices formula, we get
\[\left( {{x}^{6}}+\dfrac{1}{{{x}^{6}}} \right)={{e}^{i6\theta }}+\dfrac{1}{{{e}^{i6\theta }}}\]
We know that, $\dfrac{1}{x}={{x}^{-1}}$
\[\left( {{x}^{6}}+\dfrac{1}{{{x}^{6}}} \right)={{e}^{i6\theta }}+{{e}^{-i6\theta }}\]
Once again, applying the Euler’s formula, we get
\[\left( {{x}^{6}}+\dfrac{1}{{{x}^{6}}} \right)=\cos (6\theta )+i\sin \left( 6\theta \right)+\cos (-6\theta )+i\sin \left( -6\theta \right)\]
We know that, $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x$
\[\left( {{x}^{6}}+\dfrac{1}{{{x}^{6}}} \right)=\cos (6\theta )+i\sin \left( 6\theta \right)+\cos (6\theta )-i\sin \left( 6\theta \right)\]
Cancelling the term $i\sin (6\theta )$, we get
\[\left( {{x}^{6}}+\dfrac{1}{{{x}^{6}}} \right)=\cos (6\theta )+\cos (6\theta )\]
Finally, we get
\[\left( {{x}^{6}}+\dfrac{1}{{{x}^{6}}} \right)=2\cos (6\theta )\]
Hence the value of the given expression \[\left( {{x}^{6}}+\dfrac{1}{{{x}^{6}}} \right)\] is \[2\cos (6\theta )\]

Note: The possibility for the mistake is that you might get confused with the difference between Euler’s formula and Exponential form of the complex number. The Euler’s formula is ${{e}^{i\theta }}$ and the exponential form of the complex number is $z=r{{e}^{i\theta }}$.