Derive the transform of f(t) = sin kt by using the identity sin kt = 2;(eikt-e-ikt). %3D L(sin kt) = L• 1 (eikt. Apply the Laplace transform to both sides. e 1 2 i (List the terms in the same order as they appear in the original list.)
Derive the transform of f(t) = sin kt by using the identity sin kt = 2;(eikt-e-ikt). %3D L(sin kt) = L• 1 (eikt. Apply the Laplace transform to both sides. e 1 2 i (List the terms in the same order as they appear in the original list.)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.4: Multiple-angle Formulas
Problem 40E
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![## Problem 7.1.33
**Objective:** Derive the transform of \( f(t) = \sin kt \) by using the identity:
\[
\sin kt = \frac{1}{2i} \left( e^{ikt} - e^{-ikt} \right).
\]
**Solution Steps:**
1. **Express the Laplace Transform:**
Start by expressing the Laplace transform \( \mathcal{L}(\sin kt) \) as follows:
\[
\mathcal{L}(\sin kt) = \mathcal{L} \left\{ \frac{1}{2i} \left( e^{ikt} - e^{-ikt} \right) \right\}
\]
2. **Apply the Laplace Transform to Both Terms:**
Apply the Laplace transform to each term separately:
\[
= \frac{1}{2i} \left( \mathcal{L}(e^{ikt}) - \mathcal{L}(e^{-ikt}) \right)
\]
Make sure to list the terms in the same order as they appear in the original expression.
**Note:**
This process involves using the properties of the Laplace transform and the linearity of integrals to simplify the expression by operating on each term separately.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc1f7b8ac-f3d7-4c7a-8b9c-528e950aacc6%2F9acf1b6a-9739-4d98-9891-6439f3f2c7e5%2Fg5ings_processed.png&w=3840&q=75)
Transcribed Image Text:## Problem 7.1.33
**Objective:** Derive the transform of \( f(t) = \sin kt \) by using the identity:
\[
\sin kt = \frac{1}{2i} \left( e^{ikt} - e^{-ikt} \right).
\]
**Solution Steps:**
1. **Express the Laplace Transform:**
Start by expressing the Laplace transform \( \mathcal{L}(\sin kt) \) as follows:
\[
\mathcal{L}(\sin kt) = \mathcal{L} \left\{ \frac{1}{2i} \left( e^{ikt} - e^{-ikt} \right) \right\}
\]
2. **Apply the Laplace Transform to Both Terms:**
Apply the Laplace transform to each term separately:
\[
= \frac{1}{2i} \left( \mathcal{L}(e^{ikt}) - \mathcal{L}(e^{-ikt}) \right)
\]
Make sure to list the terms in the same order as they appear in the original expression.
**Note:**
This process involves using the properties of the Laplace transform and the linearity of integrals to simplify the expression by operating on each term separately.
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