Question

help pls....Laplace transform: L{cos^3 (at)}

Answer 1

Reduce the power of the trig function first:
\[\cos^3at=\cos at\cos^2at\]
Half-angle identity:
\[\begin{align*}\cos^3at&=\cos at\left(\frac{1}{2}+\frac{1}{2}\cos2at\right)\\
&=\frac{1}{2}\cos at+\frac{1}{2}\cos at\cos 2at
\end{align*}\]
Recall the angle sum/difference identities:
\[\begin{align*}\cos(x+y)&=\cos x\cos y-\sin x\sin y\\
\cos(x-y)&=\cos x\cos y+\sin x\sin y\end{align*}\]
Adding the above equations, you have
\[\begin{align*}\cos(x+y)+\cos(x-y)&=2\cos x\cos y\\
\\\frac{\cos(x+y)+\cos(x-y)}{2}&=\cos x\cos y
\end{align*}\]
So, you have
\[\begin{align*}\cos^3at&=\frac{1}{2}\cos at+\frac{1}{2}\cos at\cos 2at\\\\
&=\frac{1}{2}\cos at+\frac{1}{2}\left(\frac{\cos3at+\cos at}{2}\right)\\\\
&=\frac{1}{2}\cos at+\frac{1}{4}\cos3at+\frac{1}{4}\cos at\\\\
&=\frac{3}{4}\cos at+\frac{1}{4}\cos3at\end{align*}\]

Answer 2

Taking the LP should be easy if you have the formula for cosine.