Chapter 9.3, Problem 46E

(a)

To determine

To show : $\cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}$ using Euler’s formula.

Explanation of Solution

Given information: The identity $\cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}$ is given.

Concept used :

The Euler’s formula is given by $e^{i\theta }=\cos \theta +i\sin \theta$ .

Calculation:

Solve the right side of the given equation using the Euler’s formula.

  $\begin{array}{c}\frac{e^{i\theta }+e^{-i\theta }}{2}=\frac{\left(\cos \theta +i\sin \theta \right)+\left(\cos \left(-\theta \right)+i\sin \left(-\theta \right)\right)}{2}\\ =\frac{\left(\cos \theta +i\sin \theta \right)+\left(\cos \theta -i\sin \theta \right)}{2}\\ =\frac{\cos \theta +i\sin \theta +\cos \theta -i\sin \theta }{2}\\ =\frac{2\cos \theta }{2}\\ =\cos \theta \end{array}$

This implies that, $\cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}$ .

As a result, the given identity has been proved.

(b)

To determine

To show : $\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}$ using Euler’s formula.

Explanation of Solution

Given information: The identity $\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}$ is given.

Concept used :

The Euler’s formula is given by $e^{i\theta }=\cos \theta +i\sin \theta$ .

Calculation:

Solve the right side of the given equation using the Euler’s formula.

  $\begin{array}{c}\frac{e^{i\theta }-e^{-i\theta }}{2i}=\frac{\left(\cos \theta +i\sin \theta \right)-\left(\cos \left(-\theta \right)+i\sin \left(-\theta \right)\right)}{2i}\\ =\frac{\left(\cos \theta +i\sin \theta \right)+\left(\cos \theta -i\sin \theta \right)}{2i}\\ =\frac{\cos \theta +i\sin \theta -\cos \theta +i\sin \theta }{2i}\\ =\frac{2i\sin \theta }{2i}\\ =\sin \theta \end{array}$

This implies that, $\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}$ .

As a result, the given identity has been proved.