I was told that it can be shown that $|\sin(x)| \leq 1$ by its definition
$\sin (z):= \frac{1}{2i} \bigl( (\exp(iz)-\exp(-iz)\bigr) $
I am aware that as soon as I choose $x \in \mathbb{R}$ and substitute it into the equation, the equation itself becomes real $\forall x \in \mathbb{R}$
My approach:
$ |\sin (z)| \leq 1 \implies |\sin z|^2 \leq 1 \implies \sin z \cdot \overline{ \sin(z)} \leq 1$ by definition. Since I want to show this for $x \in \mathbb{R}$ it applies that $z= \overline{z}$ and I substitute $x$ into the definition. $$ \frac{1}{2i}((\exp(ix)-\exp(-ix))\cdot \left(-\frac{1}{2i}((\exp(-ix)-\exp(ix))\right) \\ \implies \frac{1}{4}( \exp(0)-\exp(2ix)-\exp(-2ix)\exp(0)) \\ \iff \frac{1}{4}(2-\exp(2ix)-\underbrace{\exp(-2ix)}_{ \in ]0,1]})$$ So this seems to be pretty near to the result I want to proof, however, $\exp(2ix)$ can get sufficiently large for big $x$ so I doubt that my approach is correct.
Any hints, ideas, corrections to keep me going?
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There's a comment saying that $|\sin(x)| \leq 1$ follows by $ \sin^2(x) + \cos^2(x) =1 $ which I don't see why.