Trigonometry

The point $(\cos\theta, \sin\theta)$ lies on the unit circle $x^2+y^2=1$ by definition — and that equation is the Pythagorean theorem applied to the right triangle with legs $|\cos\theta|$, $|\sin\theta|$ and hypotenuse $1$. This single identity is the source of almost every other trigonometric identity.

Let $P = (\cos\theta, \sin\theta)$ be the point on the unit circle obtained by rotating $(1,0)$ counter-clockwise through angle $\theta$. Drop a perpendicular from $P$ to the horizontal axis at $Q = (\cos\theta, 0)$. The segment $OP$ is the radius of the unit circle, so $|OP|=1$.

In the right triangle $\triangle OPQ$, the horizontal leg $OQ$ has length $|\cos\theta|$ and the vertical leg $QP$ has length $|\sin\theta|$. By the classical Pythagorean theorem, $|OQ|^2+|QP|^2=|OP|^2$, giving $|\cos\theta|^2+|\sin\theta|^2=1^2$, which simplifies to $\sin^2\theta+\cos^2\theta=1$ for every real $\theta$.

Drop the triangle into a circle of radius $R$ through all three vertices. The inscribed angle theorem says the angle $A$ subtending side $a$ is half the central angle for $a$, which gives $a = 2R\sin A$ directly — the same argument for each side produces the common ratio $2R$.

Inscribe $\triangle ABC$ in its circumcircle of radius $R$ and center $O$. Draw the diameter $BD$ through vertex $B$ so that $|BD|=2R$, then connect $D$ to $C$. By Thales' theorem, since $BD$ is a diameter, the inscribed angle $\angle BCD=90°$.

When angle $A$ is acute, $\angle A$ and $\angle BDC$ both subtend arc $BC$ on the same side, so $\angle BDC=A$. In right triangle $\triangle BCD$ we have $\sin A=\sin(\angle BDC)=\dfrac{|BC|}{|BD|}=\dfrac{a}{2R}$, giving $\dfrac{a}{\sin A}=2R$. When $A$ is obtuse, $\angle BDC=180°-A$ and $\sin(180°-A)=\sin A$, yielding the same equation. Repeating for sides $b$ and $c$ gives $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$.

This is exactly the Pythagorean theorem plus a correction term: place $C$ at the origin with one side along the x-axis, so the opposite vertex is at $(a\cos C, a\sin C)$; the distance formula to the point $(b,0)$ expands to $c^2=a^2+b^2-2ab\cos C$. When $C=90°$, $\cos C = 0$ and it collapses back to $a^2+b^2=c^2$ — the law of cosines is the Pythagorean theorem for triangles that aren't right triangles.

Place vertex $C$ at the origin $(0,0)$ and align side $CB$ of length $a$ along the positive horizontal axis, so $B=(a,0)$. Since $CA$ has length $b$ and makes angle $C$ with the horizontal axis, the coordinates of $A$ are $A=(b\cos C, b\sin C)$.

The length $c$ of side $AB$ equals the Euclidean distance between $A$ and $B$. Squaring and expanding: $c^2 = (b\cos C - a)^2 + (b\sin C)^2 = b^2\cos^2 C - 2ab\cos C + a^2 + b^2\sin^2 C = a^2 + b^2(\cos^2 C+\sin^2 C) - 2ab\cos C$. Applying the Pythagorean identity $\cos^2 C+\sin^2 C=1$ yields $c^2=a^2+b^2-2ab\cos C$.

Both sides solve the same differential equation $f'(\theta) = i f(\theta)$ with $f(0)=1$ (differentiate the right side using $\frac{d}{d\theta}\cos\theta=-\sin\theta$, $\frac{d}{d\theta}\sin\theta=\cos\theta$ to check), so by uniqueness of solutions they must be equal. Setting $\theta=\pi$ gives the famous special case $e^{i\pi}+1=0$, linking five fundamental constants in one equation.

Define $g(\theta) = e^{-i\theta}(\cos\theta + i\sin\theta)$ for all real $\theta$. Differentiating by the product rule gives $g'(\theta) = -i e^{-i\theta}(\cos\theta+i\sin\theta) + e^{-i\theta}(-\sin\theta+i\cos\theta)$.

Expanding the first term using $i^2=-1$: $-i(\cos\theta+i\sin\theta) = -i\cos\theta-i^2\sin\theta = \sin\theta-i\cos\theta$. Adding this to $(-\sin\theta+i\cos\theta)$ gives $0$, so $g'(\theta)=0$ for all $\theta\in\mathbb{R}$. Therefore $g$ is constant and $g(\theta)=g(0)=e^{0}(\cos 0+i\sin 0)=1$, which rearranges to $e^{i\theta}=\cos\theta+i\sin\theta$.