Question

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

40.$f\left(t\right)=10.\cos \left(t-\frac{\pi }{6}\right)$

Short answer

The Laplace transform of above function is,

$L\left\{10.\cos \left(t-\frac{\pi }{6}\right)\right\}=\frac{5}{s^2+1}\left(\sqrt{3}s+1\right)$

Step-by-step solution

The Laplace transform of above function is,

Let f be a function define for$t⩾0$.Then the Laplace transform by using trigonometry identity will be given by,

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

Is said to be Laplace transform of f, provide that integral converges.

Applying trigonometry identity

Consider the function$f\left(t\right)=10.\cos \left(t-\frac{\pi }{6}\right)$

The objective is to finddata-custom-editor="chemistry" $L\left\{f\left(t\right)\right\}$ using the trigonometry identity.

Note that, the function f is defined for$t⩾0$.

From the trigonometry identity,

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

Letsimplify it first, by using trigonometry identity formula;

So formula is $\cos \left(A-B\right)=\cos A.cosB+\sin A.\sin B$

Therefore,

$\cos \left(t-\frac{\pi }{6}\right)=\cos t.\cos \frac{\pi }{6}+\sin t.\sin \frac{\pi }{6}$

Simplification

$f\left(t\right)=10\left\{\cos t.\cos \frac{\pi }{6}+\sin t.\sin \frac{\pi }{6}\right\}$

Since$\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$and$\sin \frac{\pi }{6}=\frac{1}{2},$weputvaluesabove

Weget;

$f\left(t\right)=10.\left\{\cos t.\frac{\sqrt{3}}{2}+\sin t.\frac{1}{2}\right\}$

$f\left(t\right)=5.\left\{\sqrt{3}\cos t+\sin t\right\}$

Laplace transform by using trigonometry identity

$L\left\{f\left(t\right)\right\}=L\left\{\cos \left(t-\frac{\pi }{6}\right)\right\}=5.L\left\{\sqrt{3}\cos t+\sin t\right\}$

$=5\left[\sqrt{3}L\left\{\cos t\right\}+L\left\{\sin t\right\}\right]$

$=5\left[\sqrt{3}\frac{s}{s^2+1}+\frac{1}{s^2+1}\right]$

$=\frac{5}{s^2+1}\left(\sqrt{3}s+1\right)$

Therefore the required Laplace transform of function is,

$L\left\{10.\cos \left(t-\frac{\pi }{6}\right)\right\}=\frac{5}{s^2+1}\left(\sqrt{3}s+1\right)$