Q:

Find ℒ{f(t)} by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 16cos(t−π/6)ℒ{f(t)} = ?

Accepted Solution

A:
Answer: [tex]L\{f(t)\}=\frac{8(\sqrt3s+1)}{s^2+1}[/tex]Step-by-step explanation:Given : [tex]f(t)=16\cos (t-\frac{\pi}{6})[/tex]To find : ℒ{f(t)} by first using a trigonometric identity ?Solution : First we solve the function,[tex]f(t)=16\cos (t-\frac{\pi}{6})[/tex]Applying trigonometric identity, [tex]\cos (A-B)=\cos A\cos B+\sin A\sin B[/tex][tex]f(t)=16(\cos t\cos (\frac{\pi}{6})+\sin t\sin(\frac{\pi}{6})[/tex][tex]f(t)=16(\frac{\sqrt3}{2}\cos t+\frac{1}{2}\sin t)[/tex][tex]f(t)=\frac{16}{2}(\sqrt3\cos t+\sin t)[/tex][tex]f(t)=8(\sqrt3\cos t+\sin t)[/tex]We know, [tex]L(\cos at)=\frac{s}{s^2+a^2}[/tex] and [tex]L(\sin at)=\frac{a}{s^2+a^2}[/tex]Applying Laplace in function, [tex]L\{f(t)\}=8\sqrt3L(\cos t)+8L(\sin t)[/tex] [tex]L\{f(t)\}=8\sqrt3(\frac{s}{s^2+1})+8(\frac{1}{s^2+1})[/tex] [tex]L\{f(t)\}=\frac{8\sqrt3s+8}{s^2+1}[/tex] [tex]L\{f(t)\}=\frac{8(\sqrt3s+1)}{s^2+1}[/tex]Therefore, The Laplace transformation is [tex]L\{f(t)\}=\frac{8(\sqrt3s+1)}{s^2+1}[/tex]