Chapter 10.3, Problem 46E

(a)

To determine

Verify Euler’s identity for $\cos \theta$ using Euler’s formula.

Answer to Problem 46E

Euler’s identity for $\cos \theta$ has been verified.

Explanation of Solution

Given information:

  $\cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}$

According to Euler’s formula,

  $e^{i\theta }=\cos \theta +i\sin \theta$

Also,

  $e^{-i\theta }=\cos \left(-\theta \right)+i\sin \left(-\theta \right)=\cos \theta -i\sin \theta$

Thus,

  $e^{i\theta }+e^{-i\theta }=2\cos \theta$

Rewrite the above equation:

  $2\cos \theta =e^{i\theta }+e^{-i\theta }$

Divide both sides by 2:

  $\cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}$

Thus,

Euler’s identity for $\cos \theta$ has been verified.

(b)

To determine

Verify Euler’s identity for $\sin \theta$ using Euler’s formula.

Answer to Problem 46E

Euler’s identity for $\sin \theta$ has been verified.

Explanation of Solution

Given information:

  $\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}$

According to Euler’s formula,

  $e^{i\theta }=\cos \theta +i\sin \theta$

Also,

  $e^{-i\theta }=\cos \left(-\theta \right)+i\sin \left(-\theta \right)=\cos \theta -i\sin \theta$

Thus,

  $e^{i\theta }-e^{-i\theta }=2i\sin \theta$

Rewrite the above equation:

  $2i\sin \theta =e^{i\theta }-e^{-i\theta }$

Divide both sides by 2i:

  $\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}$

Thus,

Euler’s identity for $\sin \theta$ has been verified.