Question

find the Laplace transformation of cos^3(2t​

Answer 1

Let $\bold{f(t)=cos³(2t)}$

Using the identity $\boxed{\bold{cos²(x)=\dfrac{1+cos(2x)}{2}}}$ ,

it follows that $\bold{cos³(x)=\dfrac{1+cos(2x)}{2}⋅cos(x)}$.

In our case, the x=2t , so we can re-write our function as:

$\bold{f(t)=\dfrac{1+cos(4t)}{2}⋅cos(2t)}$

$\bold{f(t)=\dfrac{cos(2t)}{2}+\dfrac{cos(4t)cos(2t)}{2} }$

Now, using the identity: $\bold{cos(a)⋅cos(b)=\dfrac{1}{2}(cos(a−b)+cos(a+b)) }$

where a=4t and b=2t , we obtain:

$\bold{\dfrac{1}{2}cos(4t)⋅cos(2t) =\dfrac{1}{4}(cos(2t)+cos(6t))}$

$\bold{=\dfrac{cos(2t)}{4}+\dfrac{cos(6t)}{4} }$

So, to sum it up, our function is now:

$\bold{f(t)=\dfrac{cos(2t)}{2}+\dfrac{cos(2t)}{4}+\dfrac{cos(6t)}{4}}$

$\bold{f(t)=\dfrac{3}{4}cos(2t)+\dfrac{1}{4}cos(6t) }$