Laplace Transform of Sine/Proof 5
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Theorem
Let $\sin$ denote the real sine function.
Let $\laptrans f$ denote the Laplace transform of a real function $f$.
Then:
- $\laptrans {\sin at} = \dfrac a {s^2 + a^2}$
where $a \in \R_{>0}$ is constant, and $\map \Re s > 0$.
Proof
From Laplace Transform of Second Derivative:
- $(1): \quad \laptrans {\map {f} t} = s^2 \laptrans {\map f t} - s \, \map f 0 - \map {f'} 0$
under suitable conditions.
Then:
\(\ds \map f t\) | \(=\) | \(\ds \sin a t\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {f'} t\) | \(=\) | \(\ds a \cos a t\) | |||||||||||
\(\ds \map {f} t\) | \(=\) | \(\ds -a^2 \sin a t\) | ||||||||||||
\(\ds \map f 0\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \map {f'} 0\) | \(=\) | \(\ds a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \laptrans {-a^2 \sin a t}\) | \(=\) | \(\ds s^2 \laptrans {\sin a t} - s \times 0 - a\) | from $(1)$, substituting for $\map f t$, $\map {f'} 0$ and $\map f 0$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds -a^2 \laptrans {\sin a t}\) | \(=\) | \(\ds s^2 \laptrans {\sin a t} - a\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \laptrans {\sin a t}\) | \(=\) | \(\ds \dfrac a {s^2 + a^2}\) | rearranging |
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Laplace Transform of Derivative: $16$