How can the sine function be derived/proven?
The definition for $\sin(x)$ is of course given as $\frac{\text{opposite}}{\text{hyoptenuse}}$ of a right-angled triangle, which solving for $x$ can be had from the Maclaurin series:
$$\sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}$$
Which can be simplified to (using euler's formula, which is also derived from maclaurin series):
$$\sin(x) = \large \frac{e^{ix}-e^{-ix}}{2i}$$
But since the use of maclaurin series assumes the derivative of the sin function to be known, which requires knowing the function, the proof becomes circular. How to prove that the identity holds without resorting to this circularity?
More than likely the derivation of the formula is going to use the pythagorean theorem:
$$a^2+b^2=c^2$$
Where $a$ and $b$ are the sides of the right-angled triangle and $c$ is the hypotenuse. From this it can be derived that the ratio between circumference of a circle and it's radius is (denoted by $\pi$):
$$\large \pi=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} = 3.14...$$
Which can be used to evaluate an angle (in units of radians):
$$\theta = s/r$$
$s=$ arc length, $r =$ radius.
And the equation for characterizing a unit-circle, also derived from pythagorian theorem:
$$x^2+y^2 = 1$$
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$\begingroup$You can geometrically prove that $\frac{d}{dx}\sin(x)=\cos(x)$ as seen here. You can similarly prove that $\frac{d}{dx}\cos(x)=-\sin(x)$, which gives rise to the differential equation representation of $\sin(x)$. This differential equation has all the info you need about $\sin(x)$ to get the other things.
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