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Chapter 11: Problem 2

Express \(\cos \omega t\) in terms of exponential functions.

Short Answer

Expert verified
Question: Express the cosine function \(\cos \omega t\) in terms of exponential functions using Euler's formula. Answer: \(\cos \omega t = \frac{1}{2}(e^{i\omega t} + e^{-i\omega t})\)

Step by step solution

01

Recall Euler's Formula

Euler's formula states that \(e^{ix} = \cos x + i \sin x\). This formula connects complex exponentials with trigonometric functions.
02

Write the complex conjugate of Euler's formula

The complex conjugate of Euler's formula is obtained by replacing \(i\) with \(-i\) in the formula: \(e^{-ix} = \cos x - i \sin x\)
03

Add the two equations

Now, add the Euler's formula and its complex conjugate together: \(e^{ix} + e^{-ix} = (\cos x + i \sin x) + (\cos x - i \sin x)\) This simplifies to: \(e^{ix} + e^{-ix} = 2\cos x\)
04

Express the cosine function in terms of exponential functions

Finally, solve for \(\cos x\) by dividing both sides by 2: \(\cos x = \frac{1}{2}(e^{ix} + e^{-ix})\)
05

Apply the result to \(\cos \omega t\)

Now, we can replace \(x\) with \(\omega t\) to express \(\cos \omega t\) in terms of exponential functions: \(\cos \omega t = \frac{1}{2}(e^{i\omega t} + e^{-i\omega t})\) So, \(\cos \omega t\) can be expressed as \(\frac{1}{2}(e^{i\omega t} + e^{-i\omega t})\) in terms of exponential functions.

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