I have one trigonometric function which is simplified as

f[t_, RV_, H_] := 0.000308148 H Cos[0.172439 RV Sin[2 \[Pi] t]]

It is an input to one ODE which I want to solve with Laplace transform where RV and H are just two arbitrary parameters.

My question is how to find the Laplace transform of this function. One idea I am trying to do is using the following rules to transform the identities.

Rules = {p0_ Cos[ p1_*Sin[p2_*u_]] -> 
   p0 Sum[((-1)^k (p1 Sin[p2 u])^(2 k))/(2 k)!, {k, 0, Num}]}

Then apply the Laplace transform, which gives

LaplaceTransform[f[t, RV, H] /. Rules, t, s] // FullSimplify

(0.000308148 H (99746.9 + 1.37804 RV^4 + 789.568 s^2 + s^4 +    RV^2 (-741.499 - 1.1739 s^2)))/(s (157.914 + s^2) (631.655 + s^2))

Is there any other way, since I have several of these functions and I do not know exactly how to apply these rules repeatedly to functions like sin(sin(sin...))) and write them such that I can take the Laplace transform.

  • $\begingroup$ The following will do the job: LaplaceTransform[f[t, RV, H], t, w] $\endgroup$ Dec 14 '20 at 16:30
  • $\begingroup$ @DanielHuber, Not really in Wolfram 12.1; it just returns the expression. $\endgroup$
    – hesamaero
    Dec 14 '20 at 19:05
  • $\begingroup$ I tried it in 12.1 and it worked. Did you define f[t, RV, H] beforehand? $\endgroup$ Dec 14 '20 at 19:12
  • $\begingroup$ @DanielHuber In[7]:= LaplaceTransform[f[t, RV, H], t, w] Out[7]= -((0.129421 H)/w) + 0.000308148 H LaplaceTransform[ Cos[0.0523599 - 0.0862193 RV - 0.172439 RV Sin[4.71239 - 6.28319 t]], t, w] - 0.0153853 H LaplaceTransform[ Sin[0.0523599 - 0.0862193 RV - 0.172439 RV Sin[4.71239 - 6.28319 t]], t, w] + 1/(39.4784 + w^2) 0.032 (5.0832 - 0.587785 w Cos[2 ArcTan[(2 [Pi])/w]] - 3.69316 Sin[2 ArcTan[(2 [Pi])/w]])```, it just return parts of expression without taking any transformation. $\endgroup$
    – hesamaero
    Dec 14 '20 at 19:29
  • 2
    $\begingroup$ The problem is: MMA can not transform: Cos[Sin[t]] Probably, this transform is not representable by standard functions. I do not know enough about Laplace transforms to give a definitive answer. $\endgroup$ Dec 14 '20 at 20:07

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