how will I prove that $$ \frac{s+a}{s^2+b^2} $$ is the Laplace transform of the following equation $$ \sqrt{a^2+b^2}\sin(bt+\theta) $$ where $\theta = \arctan\left(\frac{b}{a}\right)$?
laplace transform of the given equation
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ordinary-differential-equations
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0Maybe try using the double angle formula for sine function? – 2017-02-14
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0I tried that, but I have an extra $b$ on the numerator. Please check my answer. – 2017-02-14
1 Answers
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Denote the Laplace transform by $\mathcal{L}$. Using the double-angle formula $$\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B),$$ we have that \begin{align*} \mathcal{L}(\sin(bt+\theta)) & = \cos(\theta)\mathcal{L}(\sin(bt)) + \sin(\theta)\mathcal{L}(\cos(bt)) \\ & = \cos(\theta)\left(\frac{b}{s^2 + b^2}\right) + \sin(\theta)\left(\frac{s}{s^2 + b^2}\right). \end{align*} Since $\theta=\tan^{-1}(b/a)$, $\tan\theta = b/a$ and one can show that $$\cos(\theta) = \frac{a}{\sqrt{a^2 + b^2}}, \ \ \sin(\theta) = \frac{b}{\sqrt{a^2 + b^2}}. $$ Using this and the linearity of Laplace transform, we have that \begin{align*} \mathcal{L}(\sqrt{a^2 + b^2}\sin(bt+\theta)) & = \sqrt{a^2+b^2}\mathcal{L}(\sin(bt+\theta)) \\ & = \left(\frac{\sqrt{a^2 + b^2}}{s^2 + b^2}\right)[b\cos(\theta) + s\sin(\theta)] \\ & = \left(\frac{\sqrt{a^2 + b^2}}{s^2 + b^2}\right)\left[\frac{ba}{\sqrt{a^2 + b^2}} + \frac{sb}{\sqrt{a^2 + b^2}}\right]\\ & = \frac{ba + sb}{s^2 + b^2} \\ & = \frac{b(s+a)}{s^2 + b^2}. \end{align*}