Sine and cosine are the coordinates of a point on the unit circle. Four more functions are built from them. They are not new information, just useful ratios that show up often enough to deserve names.
Key idea
$$\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \csc\theta = \frac{1}{\sin\theta}.$$ If $(x, y)$ is any point on the terminal side at distance $r = \sqrt{x^2 + y^2}$ from the origin: $\sin\theta = \frac{y}{r}$, $\cos\theta = \frac{x}{r}$, $\tan\theta = \frac{y}{x}$, $\csc\theta = \frac{r}{y}$, $\sec\theta = \frac{r}{x}$, $\cot\theta = \frac{x}{y}$.
Why it works
A point $(x, y)$ at distance $r$ is the unit-circle point scaled up by $r$. Dividing by $r$ brings it back: $\left(\frac{x}{r}, \frac{y}{r}\right) = (\cos\theta, \sin\theta)$. Tangent is the slope of the terminal side, $\frac{y}{x}$, and the "co" and reciprocal functions follow. A function is undefined wherever its denominator is $0$: tangent and secant when $x = 0$, cotangent and cosecant when $y = 0$.
Worked examples
Worked example: All six from a point
The terminal side of $\theta$ passes through $(-5, 12)$. Find all six trig values.
- $r = \sqrt{25 + 144} = 13$.
- $\sin\theta = \frac{12}{13}$, $\cos\theta = -\frac{5}{13}$, $\tan\theta = \frac{12}{-5} = -\frac{12}{5}$.
- Reciprocals: $\csc\theta = \frac{13}{12}$, $\sec\theta = -\frac{13}{5}$, $\cot\theta = -\frac{5}{12}$.
Answer: $\frac{12}{13}, -\frac{5}{13}, -\frac{12}{5}, \frac{13}{12}, -\frac{13}{5}, -\frac{5}{12}$
Worked example: At special angles
Find $\sec\frac{\pi}{3}$ and $\csc\pi$.
- $\cos\frac{\pi}{3} = \frac{1}{2}$, so $\sec\frac{\pi}{3} = \frac{1}{1/2} = 2$.
- $\sin\pi = 0$, so $\csc\pi = \frac{1}{0}$, which is undefined.
Answer: $\sec\frac{\pi}{3} = 2$; $\csc\pi$ is undefined
Common mistake
Pairing secant with sine because both start with "s". Each reciprocal pair has exactly one "co": sine goes with cosecant, cosine with secant, tangent with cotangent. Check at $0$: $\sec 0$ should be $\frac{1}{\cos 0} = 1$; if you had used sine you would be dividing by $0$.