Finding the maximum of a equation

Question

Hi guys I have the following function

$R_t = \exp(-2 k_x^2 \sin^2(\frac{t}{2})) k_x^2 \theta(t) \bigg[ -\sin(2t) - \sin(t) (1 - k_x^2 \sin^2(t)) \bigg]$

and I want to find the maximum value of it.

I tried the following method in Mathematica.

R = k^2 HeavisideTheta[t] Exp[ -2 k^2 Sin[(t)/2]^2 ](-Sin[2t] - Sin[t](1-k^2 Sin[t]^2));
criticalPoints = Solve[D[R,t] == 0 && 0 <= t <= 2 Pi, t];
maxValue = MaxValue[R, t] /. criticalPoints;

However, it seems it does not work with Solve. I tried Findinstance and Reduce which are even more confusing.

I appreciate it if you could help me.

Answer 1

This can be done as follows (The distribution HeavisideTheta is of no importance here. ).

f[k_?NumericQ] := Maximize[{Exp[-2 k^2 Sin[(t)/2]^2] (-Sin[2 t] - 
Sin[t] (1 - k^2 Sin[t]^2)), t >= 0 && t <= 2*Pi && k > 0}, t]
f[1]

{E^(-2 Sin[2 ArcTan[Sqrt[ Root[3 - 144 # + 612 #^2 - 816 #^3 + 946 #^4 - 816 #^5 + 612 #\ ^6 - 144 #^7 + 3 #^8& , 4, 0]]]]^2) (-Sin[4 ArcTan[Sqrt[ Root[3 - 144 # + 612 #^2 - 816 #^3 + 946 #^4 - 816 #^5 + 612 #\ ^6 - 144 #^7 + 3 #^8& , 4, 0]]]] + Sin[4 ArcTan[Sqrt[ Root[3 - 144 # + 612 #^2 - 816 #^3 + 946 #^4 - 816 #^5 + 612 #\ ^6 - 144 #^7 + 3 #^8& , 4, 0]]]]^3 - Sin[8 ArcTan[Sqrt[ Root[3 - 144 # + 612 #^2 - 816 #^3 + 946 #^4 - 816 #^5 + 612 #\ ^6 - 144 #^7 + 3 #^8& , 4, 0]]]]), {t -> 4 ArcTan[Sqrt[ Root[3 - 144 # + 612 #^2 - 816 #^3 + 946 #^4 - 816 #^5 + 612 #\ ^6 - 144 #^7 + 3 #^8& , 4, 0]]]}}

f[1.]

{1.10567, {t -> 5.68091}}

Plot[f[k] // First, {k, 0, 2}]

Answer 2

If "k" is an input for this function then, this may be helpfull

k=1;
 f = Maximize[{Exp[-2 k^2 Sin[(t)/2]^2] 
(-Sin[2 t] - Sin[t] (1 - k^2 Sin[t]^2)), 
 t >= 0 && t <= 2*Pi}, t] // N

The output is

{1.10567, {t -> 5.68091}}