Find the Laplace transform L{sin2tcost2t}.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![**Problem Statement:**
Find the Laplace transform \( \mathcal{L} \{ \sin 2t \cos 2t \} \).
**Solution:**
To find the Laplace transform of the given function \( \sin 2t \cos 2t \), we can use trigonometric identities to simplify the expression. Specifically, we can use the product-to-sum identities, which state:
\[ \sin A \cos B = \frac{1}{2} [\sin (A+B) + \sin (A-B)] \]
For the given problem, \( A = 2t \) and \( B = 2t \), which gives:
\[ \sin 2t \cos 2t = \frac{1}{2} [\sin (2t + 2t) + \sin (2t - 2t)] \]
\[ = \frac{1}{2} [\sin 4t + \sin 0] \]
\[ = \frac{1}{2} \sin 4t \]
Now we need to find \( \mathcal{L} \left\{ \frac{1}{2} \sin 4t \right\} \).
The Laplace transform of \( \sin kt \) is given by:
\[ \mathcal{L} \{\sin kt\} = \frac{k}{s^2 + k^2} \]
In this case, \( k = 4 \), so:
\[ \mathcal{L} \{ \sin 4t \} = \frac{4}{s^2 + 16} \]
Since:
\[ \mathcal{L} \left\{ \frac{1}{2} \sin 4t \right\} = \frac{1}{2} \mathcal{L} \{ \sin 4t \} \]
Therefore:
\[ \mathcal{L} \left\{ \frac{1}{2} \sin 4t \right\} = \frac{1}{2} \cdot \frac{4}{s^2 + 16} = \frac{2}{s^2 + 16} \]
Thus:
\[ \mathcal{L} \{ \sin 2t \cos 2t \} = \frac{2}{s^2 +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1d700aa-072a-4ff7-be1d-cdca0f10aa9d%2F25534722-f680-49bd-8984-06ceae7b7366%2Fwmpi329_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the Laplace transform \( \mathcal{L} \{ \sin 2t \cos 2t \} \).
**Solution:**
To find the Laplace transform of the given function \( \sin 2t \cos 2t \), we can use trigonometric identities to simplify the expression. Specifically, we can use the product-to-sum identities, which state:
\[ \sin A \cos B = \frac{1}{2} [\sin (A+B) + \sin (A-B)] \]
For the given problem, \( A = 2t \) and \( B = 2t \), which gives:
\[ \sin 2t \cos 2t = \frac{1}{2} [\sin (2t + 2t) + \sin (2t - 2t)] \]
\[ = \frac{1}{2} [\sin 4t + \sin 0] \]
\[ = \frac{1}{2} \sin 4t \]
Now we need to find \( \mathcal{L} \left\{ \frac{1}{2} \sin 4t \right\} \).
The Laplace transform of \( \sin kt \) is given by:
\[ \mathcal{L} \{\sin kt\} = \frac{k}{s^2 + k^2} \]
In this case, \( k = 4 \), so:
\[ \mathcal{L} \{ \sin 4t \} = \frac{4}{s^2 + 16} \]
Since:
\[ \mathcal{L} \left\{ \frac{1}{2} \sin 4t \right\} = \frac{1}{2} \mathcal{L} \{ \sin 4t \} \]
Therefore:
\[ \mathcal{L} \left\{ \frac{1}{2} \sin 4t \right\} = \frac{1}{2} \cdot \frac{4}{s^2 + 16} = \frac{2}{s^2 + 16} \]
Thus:
\[ \mathcal{L} \{ \sin 2t \cos 2t \} = \frac{2}{s^2 +
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