$|\sin z|^2+|\cos z|^2=1$ I have a task here:
Show that $|\cos z|^{2}+|\sin z|^{2} = 1$ if and only if $z\in \mathbb R$.
I've tried to different ways.
1:
Since $z\in \mathbb R$ $z^2\geq0$
$-1\leq \cos z\leq1\leftrightarrow |\cos z|\leq1$
$(\cos z)^2\leq(1)^2\leftrightarrow |\cos z|^2\leq1^2$
$(\cos z)^2\leq1\leftrightarrow |\cos z|^2\leq1$ Which also means that $|\cos z|^2=(\cos z)^2$
$-1\leq \sin z\leq1\leftrightarrow |\sin z|\leq1$
$(\sin z)^2\leq(1)^2\leftrightarrow |\sin z|^2\leq1^2$
$(\sin z)^2\leq1\leftrightarrow |\sin z|^2\leq1$ Which also means that $|\sin z|^2=(\sin z)^2$
$|\cos z|^{2}+|\sin z|^{2} = (\cos z)^2+(\sin z)^2=1  \blacksquare$
2:
$|\cos z|^{2}+|\sin z|^{2} = |\frac{e^{iz}+e^{-iz}}{2}|^2+|\frac{e^{iz}-e^{-iz}}{2i}|^2=||\frac{e^{iz}}{2}|+|\frac{e^{-iz}}{2}||^2+||\frac{e^{iz}}{2i}|+|\frac{-e^{-iz}}{2i}||^2$
$|\frac{e^{iz}}{2}|^2 + 2|\frac{e^{iz}}{2}||\frac{e^{-iz}}{2}|+|\frac{e^{-iz}}{2}|^2+|\frac{e^{iz}}{2i}|^2 + 2|\frac{e^{iz}}{2i}||\frac{-e^{-iz}}{2i}|+|\frac{e^{-iz}}{2i}|^2$
$=|\frac{e^{iz}}{2}|^2 -|\frac{e^{iz}}{2}|^2+2|\frac{e^{iz}}{2}||\frac{e^{-iz}}{2}|+2|\frac{e^{iz}}{2i}||\frac{-e^{-iz}}{2i}|+|\frac{e^{-iz}}{2}|^2-|\frac{-e^{-iz}}{2}|^2=2|\frac{e^{iz}}{2}||\frac{e^{-iz}}{2}|+2|\frac{e^{iz}}{2i}||\frac{-e^{-iz}}{2i}|$
$=2|\frac{e^0}{4}|+2|\frac{-e^{0}}{-4}|=\frac{1}{2}+\frac{1}{2}=1 \blacksquare$
Is these two proofs sufficent?
 A: The definition of $\cos z$ and $\sin z$ is
$$
\cos z=\frac{e^{iz}+e^{-iz}}{2}
\qquad
\sin z=\frac{e^{iz}-e^{-iz}}{2i}
$$
Note that $\bar{e^{w}}=e^{\bar{w}}$, so
$$
\overline{\cos z}=\frac{e^{-i\bar{z}}+e^{i\bar{z}}}{2}=\cos\bar{z}
$$
and, similarly, $\overline{\sin z}=\sin\bar{z}$. This also follows from the power series definition.
Now $\lvert\cos z\rvert^2=\cos z\cos\bar{z}$ and $\lvert\sin z\rvert^2=\sin z\sin\bar{z}$ and the addition formula gives
$$
\lvert\cos z\rvert^2+\lvert\sin z\rvert^2=
\cos z\cos\bar{z}+\sin z\sin\bar{z}=\cos(z-\bar{z})
$$
Let $w=z-\bar{z}$; then the equation $\cos w=1$ is equivalent to
$$
\frac{e^{iw}+e^{-iw}}{2}=1
$$
or
$$
e^{iw}+\frac{1}{e^{iw}}=2
$$
that becomes
$$
e^{2iw}+1=2e^{iw}
$$
and therefore $e^{iw}=1$, so $iw=2k\pi i$ and finally $w=2k\pi$, for some integer $k$.
From $z-\bar{z}=2k\pi$ we get $\bar{z}-z=2k\pi$, so $\bar{z}=z$.
A: By the addition formula and the definition of the hyperbolic functions,
$$\cos z=\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y$$ and
$$\sin z=\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y.$$
From this, after simplifications
$$|\cos z|^2+|\sin z|^2=\cosh^2y+\sinh^2y=\cosh 2y.$$
And obviously,
$$\cosh 2y=1\iff y=0.$$

If the use of the hyperbolic functions is disallowed, we have
$$\cos iy=\frac{e^{-y}+e^y}2,$$ a real function, let $c(y)$, and
$$\sin iy=\frac{e^{-y}-e^y}{2i},$$ an imaginary function, let $i\,s(y)$.
By the same argument,
$$|\cos z|^2+|\sin z|^2=c^2(y)+s^2(y)=\frac{e^{-2y}+2+e^{2y}+e^{-2y}-2+e^{2y}}4=\frac{e^{2y}+e^{-2y}}2.$$
Now, with $t:=e^{2y}$
$$t+\frac1t=2\iff t=1\iff y=0.$$
A: Imagine a right angle triangle with sides a, b and c, with c as hypotenuse, whereby $a^2 + b^2 = c^2$ and
$sin z = \frac {a}{c}$
$cos z = \frac {b}{c}$
Thus, $(sinz)^2 + (cosz)^2)$
$=(\frac {a}{c})^2 + {\frac {b}{c}})^2$
$= \frac {a^2+b^2}{c^2}$
$=\frac {c^2}{c^2}$
$=1$
