What's the maximum and minimum of $\sin Z$ and $\cos Z$ where $Z$ is a complex number? I see that a lot of people try solving something  like $\sin Z=2$. So that breaks the the maximum of $\sin x$. Now I wonder if there is a maximum and minimum of $\sin Z$ and $\cos Z$ if $Z$ is a complex number.
 A: By Liouville's theorem, a bounded entire function must be constant. Since sine and cosine are entire and obviously non-constant, they must be unbounded. Hence they have no largest value. Note: by "largest" we mean largest modulus.
References: https://en.wikipedia.org/wiki/Liouville%27s_theorem_(complex_analysis)
and https://en.wikipedia.org/wiki/Entire_function.
EDIT: They still have a "smallest" value, namely zero (this is not true for all non-constant entire functions, as noted in the comments below).
A: I'll be elementary here.
Since,
for all complex $x$,
$e^{ix}
=\cos(x)+i\sin(x)
$
and
$e^{-ix}
=\cos(x)-i\sin(x)
$,
$\sin(x)
=\dfrac{e^{ix}-e^{-ix}}{2i}
=-i\dfrac{e^{ix}-e^{-ix}}{2}
=-i\sinh(ix)
$.
This holds for all complex $x$.
Letting $x = -iy$,
so $ix = y$,
$\sin(-iy)
=-i\sinh(y)
$.
Letting
$y \to \infty$,
$\sinh(y) \to \infty$
so
$|\sin(iy)|
\to \infty$.
A: Note that $\sin z=sin(x+iy)=\sin x \cosh y+i\cos x\sinh y$
Taking absolute value to have the meaning of maximum and minimum, $|\sin z|=\sqrt{\sin^2 x\cosh^2 y+\cos^2 x\sinh^2 y}=\sqrt{\sin^2 x+\sinh^2  y}=\sqrt{\sin^2 x+(\frac{e^y-e^{-y}}{2})^2}$
which is unbounded along the imaginary axis in both directions. Similarly you can proceed for $\cos z$. The minimum value for both the functions can be easily seen to be zero at origin for $|\sin z|$ and at $z=\pi/2$ for $\cos z$.
