# Get Cos[2 x] in “Solve”

My goal is to get an analytical solution of x ( $\lambda$ in the problem), when $\epsilon$ (f[x] in the problem) is maximum (use inverse form).

I found that if I can use trigonometric identity to change Sin[x] and Cos[x] to Sin[2x] or Cos[2x], then I can achieve my goal. Here is the code that I wrote, but I found it is not very good, because for the third line I copy and paste from the evaluation cell.

So, can anyone think of an easier way to change $\epsilon$ in terms of 2$\lambda$, or can provide another way to achieve the goal?

f[x_] = (R ω^2 Sin[x] Cos[x])/(g - ω^2 (Cos[x])^2 R);

f'[x] // FullSimplify;

Solve[2 R ω^2 (R ω^2 + (-2 g + R ω^2) Cos[2 x]) == 0, Cos[2 x]][[1]] // FullSimplify;

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You don't have to copy anything, try der = f'[x] // FullSimplify Solve[der == 0, Cos[2 x]][[1]] // FullSimplify –  Kuba Apr 16 at 6:31

Another way is to use Maximize rather than solving for the zero of the derivative.

f[x_] = (r ω^2 Sin[x] Cos[x])/(g - ω^2 (Cos[x])^2 r);

as = {r > 0, ω > 0, g > 0, r ω^2 < g};


You can see that

f[x] // TrigReduce
(* -((r ω^2 Sin[2 x])/(-2 g + r ω^2 + r ω^2 Cos[2 x])) *)


therefore you can make a simple substitution and, assuming the angle is very small,

sol = Simplify[Maximize[(f[x] // TrigReduce) /. {Cos[2 x] -> y, Sin[2 x] -> Sqrt[1 - y^2]},
y],
Assumptions -> as]
(* {(r ω^2)/(2 Sqrt[g (g - r ω^2)]), {y -> (r ω^2)/(2 g - r ω^2)}} *)

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Thanks a lot. Your way is exactly what I want to get. –  Lawerance Apr 16 at 7:13