Find L{f(t)} by first using a trigonometric identity. (Write your answer as a function of s.) f(t) = 18 cos t - L{f(t)} =
Find L{f(t)} by first using a trigonometric identity. (Write your answer as a function of s.) f(t) = 18 cos t - L{f(t)} =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Problem Statement:**
Find \(\mathcal{L}\{f(t)\}\) by first using a trigonometric identity. (Write your answer as a function of \(s\).)
\[
f(t) = 18 \cos \left( t - \frac{\pi}{6} \right)
\]
\[
\mathcal{L}\{f(t)\} = \_\_\_\_\_\_\_
\]
**Explanation:**
To solve the problem, you need to find the Laplace transform of the function \(f(t) = 18 \cos \left( t - \frac{\pi}{6} \right)\). Start by using the trigonometric identity for cosine of a difference:
\[
\cos(a - b) = \cos a \cos b + \sin a \sin b
\]
Applying this identity:
\[
\cos \left( t - \frac{\pi}{6} \right) = \cos t \cos \frac{\pi}{6} + \sin t \sin \frac{\pi}{6}
\]
Then substitute the values for \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\) and \(\sin \frac{\pi}{6} = \frac{1}{2}\):
\[
\cos \left( t - \frac{\pi}{6} \right) = \cos t \cdot \frac{\sqrt{3}}{2} + \sin t \cdot \frac{1}{2}
\]
Now,
\[
f(t) = 18 \left( \cos t \cdot \frac{\sqrt{3}}{2} + \sin t \cdot \frac{1}{2} \right)
\]
Simplifying, we get:
\[
f(t) = 9\sqrt{3} \cos t + 9 \sin t
\]
The Laplace transform of \(\cos(at)\) is \(\frac{s}{s^2 + a^2}\) and for \(\sin(at)\) is \(\frac{a}{s^2 + a^2}\). Therefore, we calculate:
\[
\mathcal{L}\{f(t)\} = 9\sqrt{3} \cdot \frac{s}{s^2 + 1} + 9 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F59c1a09d-455b-4e35-b247-874cd3f14c27%2F47c669a9-a42c-4b31-8348-0f9b58bc4862%2Fr66t1pp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \(\mathcal{L}\{f(t)\}\) by first using a trigonometric identity. (Write your answer as a function of \(s\).)
\[
f(t) = 18 \cos \left( t - \frac{\pi}{6} \right)
\]
\[
\mathcal{L}\{f(t)\} = \_\_\_\_\_\_\_
\]
**Explanation:**
To solve the problem, you need to find the Laplace transform of the function \(f(t) = 18 \cos \left( t - \frac{\pi}{6} \right)\). Start by using the trigonometric identity for cosine of a difference:
\[
\cos(a - b) = \cos a \cos b + \sin a \sin b
\]
Applying this identity:
\[
\cos \left( t - \frac{\pi}{6} \right) = \cos t \cos \frac{\pi}{6} + \sin t \sin \frac{\pi}{6}
\]
Then substitute the values for \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\) and \(\sin \frac{\pi}{6} = \frac{1}{2}\):
\[
\cos \left( t - \frac{\pi}{6} \right) = \cos t \cdot \frac{\sqrt{3}}{2} + \sin t \cdot \frac{1}{2}
\]
Now,
\[
f(t) = 18 \left( \cos t \cdot \frac{\sqrt{3}}{2} + \sin t \cdot \frac{1}{2} \right)
\]
Simplifying, we get:
\[
f(t) = 9\sqrt{3} \cos t + 9 \sin t
\]
The Laplace transform of \(\cos(at)\) is \(\frac{s}{s^2 + a^2}\) and for \(\sin(at)\) is \(\frac{a}{s^2 + a^2}\). Therefore, we calculate:
\[
\mathcal{L}\{f(t)\} = 9\sqrt{3} \cdot \frac{s}{s^2 + 1} + 9 \
![The given mathematical function is:
\[ f(t) = 18 \cos \left( t - \frac{\pi}{6} \right) \]
This function describes a cosine wave with an amplitude of 18, a horizontal shift (or phase shift) of \(\frac{\pi}{6}\) to the right, and no vertical shift. The variable \(t\) typically represents time or an angle, depending on the context of the problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F59c1a09d-455b-4e35-b247-874cd3f14c27%2F47c669a9-a42c-4b31-8348-0f9b58bc4862%2Fukrvoed_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The given mathematical function is:
\[ f(t) = 18 \cos \left( t - \frac{\pi}{6} \right) \]
This function describes a cosine wave with an amplitude of 18, a horizontal shift (or phase shift) of \(\frac{\pi}{6}\) to the right, and no vertical shift. The variable \(t\) typically represents time or an angle, depending on the context of the problem.
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