I would like to analytically derive the maximum of $\frac{\sin{(\sin{t}})}{{\cos{(\cos{t})}}}$ on $t\in [0,2\pi]$. I feel like we wouldn't have a closed form, but still I gave it a try.
My attempt:
I thought this problem can be seen as optimizing the $2$-variable function $f(x,y) = \sin{x}/\cos{y}$ on $g(x,y)=0$ where $g(x,y)=x^2+y^2-1$. I used the Lagrange Multiplier method to solve this problem.
Since $f_x = \cos{x}/\cos{y}, f_y=\sin{x}\sin{y}/\cos^2{y}$, we need to solve the following:
$\begin{cases} \frac{\cos{x}}{\cos{y}}-\lambda \cdot 2x=0, \\ \frac{\sin{x}\sin{y}}{\cos{y}}-\lambda \cdot 2y=0, \\ x^2+y^2-1=0.\end{cases}$
By first two equations, we have $2xy\lambda = \frac{y\cos{x}}{\cos{y}}=\frac{x\sin{x}\sin{y}}{\cos^2{y}}$ and therefore we have $y\cos{x}\cos{y}=x\sin{x}\sin{y}$ .
Now we need to solve
$\begin{cases} y\cos{x}\cos{y}=x\sin{x}\sin{y}, \\ x^2+y^2=1.\end{cases}$
$x=\pm1,y=0$ are the trivial solution of this equation. According to Wolfram|Alpha, this seems to have few more solutions, but I couldn't find a way to find them.
Any ideas?
In my humble opinion, you make the problem more difficult than it is intr0ducing two variables and Lagrange multiplier.
We have $$f(t)=\sin (\sin (t)) \sec (\cos (t))$$ We do not need to work on $t\in [0,2\pi]$ since $$f(\pi -t)+f(t+\pi )=0 \qquad \text{and} \qquad f\left(\frac{\pi }{2}-t\right)-f\left(t+\frac{\pi }{2}\right)=0$$ So, we shall only consider $$f(x)=\sin (\cos (x)) \sec (\sin (x))\qquad \text{for} \qquad x\in \Big[-\frac \pi 2,+\frac \pi 2\Big]$$ Now $$f'(x)=\cos (x) \tan (\sin (x)) \sin (\cos (x)) \sec (\sin (x))-\sin (x) \cos (\cos (x)) \sec (\sin (x))$$ We can see that $f'(0)=0$ but the second derivative test show that this is a minimum $$f(0)=\sin (1)~~<~~1\qquad \text{for} \qquad f''(0)=\sin (1)-\cos (1)~~>~~0$$
Composing the Taylor series around $x=0$, we have $$f'(t)= (s-c)x-\frac{5 c+2 s}{6} x^3 +\frac{14 c-53 s}{120} x^5 +\frac{897 c-163 s}{5040}x^7+$$ $$\frac{8639 c+17484 s}{362880}x^9+O\left(x^{11}\right)\tag 1$$ where $s=\sin(1)$ and $c=\cos(1)$
Dividing now by $x$, then $(1)$ is approximated by a polynomial in $x^2$; as written here, it ca be solved with radicals. Using the expansions to $O(x^{2n+1}$ and considering only the postive root, the successive approximations give $$x_{(1)}=0.641982\qquad x_{(2)}=0.59834\qquad x_{(3)}=0.601188\qquad x_{(4)}=0.602014$$
We could have continued and quickly converge to the solution $x=0.602034$ which, as shown below, would have been very quickly ontained using Newton method strating with $x_0=x_{(1)}$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.641982 \\ 1 & 0.605799 \\ 2 & 0.602072 \\ 3 & 0.602034 \end{array} \right)$$
To have a nice looking result say that $x \sim \frac \pi 5$ which makes $$f\left(\frac{\pi }{5}\right) \sim \sec \left(\frac{ \sqrt[4]{5}}{2 \sqrt{2}}\sqrt{\sqrt{5}-1}\right)\sin \left(\frac{1}{4} \left(1+\sqrt{5}\right)\right)=0.869544$$ while a full optimization would give a maximum value $0.869785$ for $x=0.602034$.