Proofs and Derivations

The unit-circle point $(\cos\theta, \sin\theta)$ lies at distance $1$ from the origin, and the distance formula squares to $\cos^2\theta + \sin^2\theta = 1$ — the Pythagorean theorem wearing trig notation. Dividing by $\cos^2\theta$ yields $1 + \tan^2\theta = \sec^2\theta$. $\blacksquare$

In plain terms. It is the Pythagorean theorem on a radius-1 circle. Every "Pythagorean identity" is this one line, re-divided.

Example. $\sin\theta = \tfrac35$ (Quadrant I) forces $\cos\theta = \tfrac45$: no triangle needed.

Place the triangle with $C$ at the origin and side $a$ along the $x$-axis: $B = (a, 0)$ and $A = (b\cos C, b\sin C)$. The distance $AB = c$ squares to $c^2 = (b\cos C - a)^2 + (b\sin C)^2 = a^2 + b^2 - 2ab\cos C$ after applying $\sin^2 + \cos^2 = 1$. $\blacksquare$

In plain terms. Drop the triangle onto coordinates and use the distance formula once — the identity mops up.

Example. Sides $5, 7$ with a $60^\circ$ angle between: $c^2 = 25 + 49 - 70 \cdot \tfrac12 = 39$.

The area of a triangle is $\tfrac12 ab\sin C$ — computed from ANY vertex. Equating the three versions, $\tfrac12 ab \sin C = \tfrac12 bc \sin A = \tfrac12 ca \sin B$, and dividing through by $\tfrac12 abc$ gives $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$. $\blacksquare$

In plain terms. One area, three formulas for it — setting them equal is the whole proof.

Example. A $30^\circ$ angle opposite side $4$: every other side is $\dfrac{\text{its opposite’s sine}}{\sin 30^\circ} \cdot 4 \cdot \tfrac12 \ldots$ — i.e. $\tfrac{a}{\sin A}$ is the constant $8$.

Expanding $(1 + x)^n$ means choosing, from each of the $n$ factors, either the $1$ or the $x$. A term $x^k$ appears once for every way to pick WHICH $k$ factors contribute an $x$ — that is $\binom{n}{k}$ ways. Hence $(1+x)^n = \sum_k \binom{n}{k}x^k$. $\blacksquare$

In plain terms. The coefficient counts choices, so it must be "n choose k" — the theorem is combinatorics, not algebra.

Example. The $x^2$ coefficient in $(1+x)^5$ is $\binom52 = 10$.

Let $m = \log_b x$, $n = \log_b y$, so $x = b^m$, $y = b^n$. Then $xy = b^{m+n}$ says exactly $\log_b(xy) = m + n$, and $x^p = b^{mp}$ says $\log_b(x^p) = p\log_b x$. Logs ARE exponents, so exponent laws translate one-for-one. $\blacksquare$

In plain terms. Every log rule is an exponent rule read backwards.

Example. $\log_2 8 + \log_2 4 = 3 + 2 = 5 = \log_2 32$.

From $x = b^{\log_b x}$, take $\log_c$ of both sides and pull the exponent down (power rule): $\log_c x = \log_b x \cdot \log_c b$. Divide: $\log_b x = \dfrac{\log_c x}{\log_c b}$. $\blacksquare$

In plain terms. One log in any base is two logs in the base your calculator likes.

Example. $\log_2 5 = \dfrac{\ln 5}{\ln 2} \approx 2.32$.