Is there a proof show that : $\cos(z)$ and $\sin(z)$ are images of unbounded functions? if we knew that :cos and sin are bounded function $\mathbb{R}$ for any real 
number $x$ .
let $z $ be a complex variable , Is there a proof show that :
$\cos(z)$ and $\sin(z)$ are images of  unbounded functions ?
Any kind of help is appreciated.
 A: In the spirit of giving an explicit computation accessible to good high-school students:
If one accepts that
$$
\left.
\begin{gathered}
\exp(iz) = \cos z + i\sin z, \\
\exp(z + w) = \exp(z) \exp(w),
\end{gathered}\right\}\quad\text{for all complex $z$ and $w$,}
\tag{1}
$$
then unboundedness of both $\cos$ and $\sin$ along the imaginary axis follows from Yves Daoust's comment.

More generally, the real and imaginary parts of the circular functions are given by:
  $$
\left.
\begin{aligned}
\cos(x + iy) &= \cos x \cosh y - i\sin x \sinh y, \\
\sin(x + iy) &= \sin x \cosh y + i\cos x \sinh y.
\end{aligned}\right\}\quad\text{for all real $x$, $y$.}
$$
  Since the hyperbolic functions $\cosh$ and $\sinh$ are unbounded, $\cos$ and $\sin$ are also unbounded.

Proof: By Euler's formula,
$$
\left.
\begin{aligned}
\exp(iz)  &= \cos z + i\sin z, \\
\exp(-iz) &= \cos z - i\sin z,
\end{aligned}\right\}\quad\text{for all complex $z$.}
\tag{2}
$$
Adding and subtracting, we find
$$
\left.
\begin{aligned}
\cos z &= \frac{\exp(iz) + \exp(-iz)}{2}, \\
\sin z &= \frac{\exp(iz) - \exp(-iz)}{2i},
\end{aligned}\right\}\quad\quad\text{for all complex $z$.}
\tag{3}
$$
Setting $z = x + iy$ with $x$ and $y$ real in (3), so that $iz = ix - y$ and $-iz = -ix + y$, the law of exponents and Euler's formula give
\begin{align*}
\cos(x + iy) &= \tfrac{1}{2} \bigl[\exp(ix - y) + \exp(-ix + y)\bigr] \\
  &= \tfrac{1}{2}\bigl[e^{-y}(\cos x + i\sin x) + e^{y}(\cos x - i\sin x)\bigr] \\
  &= \cos x \bigl[\tfrac{1}{2}(e^{y} + e^{-y})\bigr] - i\sin x \bigl[\tfrac{1}{2}(e^{y} - e^{-y})\bigr] \\
  &= \cos x \cosh y - i\sin x \sinh y,
\end{align*}
and
\begin{align*}
\sin(x + iy) &= \tfrac{1}{2i} \bigl[\exp(ix - y) - \exp(-ix + y)] \\
  &= -\tfrac{i}{2} \bigl[e^{-y}(\cos x + i\sin x) - e^{y}(\cos x - i\sin x)\bigr] \\
  &= \sin x \bigl[\tfrac{1}{2}(e^{y} + e^{-y})\bigr] + i\cos x \bigl[\tfrac{1}{2}(e^{y} - e^{-y})\bigr] \\
  &= \sin x \cosh y + i\cos x \sinh y.
\end{align*}
