Q2P - Mathematical Methods in Physical Sciences

Question

Solve the equations ${\mathit{e}}^{\mathbf{i}\mathbf{\theta }}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{+}\mathit{i}\mathbf{ }\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{,}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{\theta }}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{-}\mathit{i}\mathbf{ }\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$, for cos θ and sin θ and so obtain equations (11.3).

Step-by-Step Solution

Answer

The solution for cos θ and sin θ.

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

Step 1: Given Information.

The given equations are $e^{i\theta }=cos\theta +i sin\theta ,e^{-i\theta }=cos\theta -i sin\theta$.

Step 2: Definition of Power series.

A power series is an infinite series which looks like 

$\mathbf{\sum }_{\mathit{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{a}}_{\mathit{n}}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{c}\mathbf{)}}^{\mathit{n}}\mathbf{=}{\mathit{a}}_{\mathbf{0}}\mathbf{+}{\mathit{a}}_{\mathbf{1}}\mathbf{(}\mathit{x}\mathbf{-}\mathit{c}\mathbf{)}\mathbf{+}{\mathit{a}}_{\mathbf{2}}{\mathbf{(}\mathit{x}\mathbf{-}\mathit{c}\mathbf{)}}^{\mathbf{2}}\mathbf{+}\mathbf{·}\mathbf{·}\mathbf{·}$
where represents the coefficient of the nth term and c is a constant.

Step 3: Add the given functions.

Write the two equations.

$\begin{array}{l} e^{i\theta }=\cos \theta +i \sin \theta ,\\ e^{-i\theta }=\cos \theta -i \sin \theta \end{array}$

 

Add the functions.

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

 

Hence the $\cos \left(\theta \right)$ is, $\frac{e^{\left(\theta i\right)}+e^{\left(-\theta i\right)}}{2}$.

Step 4: Subtract the given functions.

Subtract the functions.

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

 

Hence the $\sin \left(\theta \right)$ is, $\frac{e^{\left(\theta i\right)}-e^{\left(-\theta i\right)}}{2i}$.

Therefore, 

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