Question

In Problems 37–40 find $L\left\{f\left(t\right)\right\}$by first using a trigonometric identity.

37.$f\left(t\right)=\sin 2t\cos 2t$

Short answer

The Laplace transform of given function is,

$L\left\{\sin 2t.\cos 2t\right\}=\left(\frac{2}{s^2+16}\right),s>4$

Step-by-step solution

Theorem

Here we determine the Laplace transform of the given function by comparing it to general form as provided in theorem 7.1.1.

$L\left\{\sin at\right\}=\frac{a}{s^2+a^2},s>a$

Solution

Consider the function$f\left(t\right)=\sin 2t.\cos 2t$

The objective is to find $L\left\{f\left(t\right)\right\}$using the theorem.

From the linearity property of the Laplace transform, we have

$$\begin{gathered} f\left(t\right)=\sin 2t.\cos 2t \\ f\left(t\right)=\frac{1}{2}\left(2\sin 2t.\cos 2t\right) \\ f\left(t\right)=\frac{1}{2}\left(\sin 4t\right)............................\left(\because \sin 2\theta =2\sin \theta \cos \theta \right) \\ L\left\{f\left(t\right)\right\}=\frac{1}{2}L\left\{\sin 4t\right\} \end{gathered}$$

From the theorem7.1.1,

$L\left\{\sin at\right\}=\frac{a}{s^2+a^2},s>a$, Where role="math" localid="1663917444012" $a=0,1,2,3,............$

$$\begin{gathered} \left\{\sin 2t.\cos 2t\right\}=\frac{1}{2}\left(\frac{4}{s^2+4^2}\right) \\ L\left\{\sin 2t.\cos 2t\right\}=\left(\frac{2}{s^2+16}\right),s>4 \end{gathered}$$

Therefore the required Laplace transform of function is,

$L\left\{\sin 2t.\cos 2t\right\}=\left(\frac{2}{s^2+16}\right),s>4$