maximise $\sin^2(x) +\sinh^2(y)$ subject to $x^2+y^2=1$ I want to maximise $\sin^2(x) +\sinh^2(y)$ subject to $x^2+y^2 =1$
I tried to use Lagrange multipliers but got nowhere. Then, I tried to parametrise $x=cos(t)$, $y=sin(t)$. 
I obtained  $$\cos(t) (-\sin(2\cos(t)) +\sinh(2\sin(t))) =0 $$ 
The solutions I am after are there, namely $\cos (t) =0$ gives $t= 2 \pi n \pm \frac{\pi}{2}$. However, I can't figure out how to deal with the other bracket. I want to differentiate again and show that those solutions give minimum, but I am stuck. I feel like there is an easier way. 
 A: Lagrange multipliers still works. We get the following system of equations
$$\begin{cases}\sin(2x) = \lambda (2x)\\ \sinh(2y) = \lambda (2y)\\ x^2+y^2=1\end{cases}$$
Now geometrically consider the intersection of $\lambda z$ with $\sin(z)$ and $\sinh(z)$. Changing lambda changes the slope and notice that for $\lambda > 1$, the first equation will only ever have one intersection (at $z=0$), but the other equation will have three intersections, at $z=\pm$ some arbitrary value specified by the choice of $\lambda$, and $z=0$.
But recognize that the last equation says that if one value is fixed to $0$, the other must be $\pm 1$, so finding $\lambda$ is no longer necessary since we have shown that it is possible to choose a $\lambda$ to allow $z=2\cdot(\pm 1)$ to be a solution.
A similar thing happens with $\lambda < 1$, but with the roles of the equations reversed. So we get the following points to check from Lagrange multipliers:
$$(0,\pm 1)\hspace{16pt}(\pm 1, 0) \implies f(\pm 1, 0) = \sin^2(1)\hspace{16 pt} f(0,\pm 1) = \sinh^2(1)$$
and we can conclude that the function is minimized at $(\pm 1, 0)$ and maximized at $(0,\pm 1)$
A: The parametrized function is
$$f(t) = \sin^2(\cos t) +\sinh^2(\sin t)$$
Set $f'(t) = 0$ for the extrema,
$$-\sin(2\cos t)\sin t + \sinh(2\sin t)\cos t=0$$
which has solutions at $t = \pm \frac\pi2,0,\pi$. Given that $f(t)$ is bounded, it is straightforward to verify that the maximum value is at
$$f(\pm\frac\pi2) = \sinh^2(1)$$
while the minimum is at $f(0)=f(\pi)=\sin^2(1)$.
A: For $x, y \in \Bbb R$ we have
$$ 
\sin^2(x) + \sinh^2(y) = |\sin(x+iy)|^2 \, ,
$$
see for example How to show $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$.
From the power series expansion we get for all $z \in \Bbb C$
$$ 
|\sin(z)| = \left| \sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{(2n+1)!}\right|
\le  \sum_{n=0}^\infty \frac{ |z|^{2n+1}}{(2n+1)!} = \sinh(|z|)
$$
with equality for purely imaginary $z$.
It follows that for $x^2+y^2=1$
$$
\sin^2(x) + \sinh^2(y) = |\sin(x+iy)|^2 \le \sinh^2 (|x+iy|) = \sinh^2(1) \, ,
$$
with equality for $(x, y) = (0, \pm 1)$.
