It is obvious from the definition of $f(x)=\sin(x)$ using the unit circle of radius $1$ that the range of that function is the set $[-1,1]$. But also there are approaches where the sine is defined using its Taylor series expansion:
$\sin(x)=\sum_{i=0}^{\infty}\dfrac {{(-1)^i}{x^{2i+1}}}{(2i+1)!}$
Now, what I would like of you to show me, how we can show that the range of the sine function is $[-1,1]$ if we define it by its Taylor series expansion?
Cheers!
Edit: I was thinking of this, we could somehow show by using the Taylor series expansion for sine and cosine that it holds that $(\sin x)^2+(\cos x)^2=1$. So to solve the question it should be sufficient to show that there exists some $x_1$ such that $\sin (x_1)=-1$ and some $x_2$ such that $\sin (x_2)=1$ and now the solution will follow form the fact that sine is continuous, so by the intermediate value theorem it would take any value between $[-1,1]$. But how to show the existence of such $x_1$ and $x_2$?
Best Answer
The formulas $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$ and $\cos x = \frac{e^{ix}+e^{-ix}}{2}$ follow quite straightforwardly from the Taylor expansions for each function. Now, $\sin^2x +\cos^2x = \frac{e^{2ix}-2+e^{-2ix}}{-4}+\frac{e^{2ix}+2+e^{-2ix}}{4}=1$.
Now, $\cos(0)=1$ is immediate from the series, as are the relations $\sin '(x)=\cos x$ and $\cos' (x)=-\sin (x)$. So, $\sin (x)$ is increasing at $x=0$, but $\sin (\pi 3/2)<0$ by direct evaluation of the series at $x=\pi 3/2$. So, $\sin (x)$ attains a maximum at some $y>0$, thus $\cos(y)=0$, and thus $\sin(y)=1$. Now, the formula $\sin(-x)=-\sin(x)$ is immediate from the series expansion, and so $\sin (-y)=-1$.
This proof is along the lines you were asking for. Note also that the above does not really contain any complex analysis. The part involving $i$ can be replaced by a slightly more cumbersome expression. Basically, the $e^{ix}$ business is just a slick way to hide some combinatorics. You can unpack it if you like.