1. The three ratios
Start with a right triangle. One angle is exactly $90°$; we'll call it the right angle. Pick one of the two other angles and call it $\theta$. Every side now has a name relative to that choice:
- The hypotenuse is the longest side — the one opposite the right angle. It never changes name.
- The opposite side is the one that does not touch $\theta$. It sits across from it.
- The adjacent side is the one that touches $\theta$ and is not the hypotenuse.
With those names settled, the three functions are just ratios of pairs of those sides:
Each is a pure number — no units — that depends only on the angle $\theta$.
"Opposite" and "adjacent" are not properties of the triangle. They are properties of the triangle plus a chosen angle. Pick the other non-right angle and the labels swap: yesterday's opposite is today's adjacent.
2. SOH-CAH-TOA
The mnemonic everyone learns. It's nine letters that pack the three definitions:
| Chunk | Stands for | Ratio |
|---|---|---|
| SOH | Sine = Opposite over Hypotenuse | $\sin\theta = \dfrac{O}{H}$ |
| CAH | Cosine = Adjacent over Hypotenuse | $\cos\theta = \dfrac{A}{H}$ |
| TOA | Tangent = Opposite over Adjacent | $\tan\theta = \dfrac{O}{A}$ |
Lean on it whenever you stare at a triangle and forget which side goes on top. The structure is also worth noticing: every ratio is one of $\{O, A\}$ divided by the larger of the relevant pair. Tangent is the odd one out — no hypotenuse — and unsurprisingly, it can become arbitrarily large (because as $\theta \to 90°$, $A \to 0$).
A useful identity drops out for free: $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$. Cancel the hypotenuse in $\dfrac{O/H}{A/H}$ and you're left with $O/A$. The three functions aren't independent — tangent is fully determined by the other two.
3. Why ratios, not lengths
Here is the conceptual move that makes all of trigonometry possible: the ratios depend only on the angle, not on the size of the triangle. Two right triangles with the same $\theta$ have the same $\sin\theta$, the same $\cos\theta$, and the same $\tan\theta$ — even if one fits in your palm and the other spans a football field.
The reason is similar triangles. Two triangles are similar when their angles match; when that happens, corresponding sides are all scaled by the same factor $k$:
$$ \frac{O'}{H'} = \frac{k \cdot O}{k \cdot H} = \frac{O}{H} $$The $k$ cancels. Whatever ratio you compute in the small triangle, you'll compute again in the big one. Scale is irrelevant.
That cancellation is what justifies treating $\sin$, $\cos$, $\tan$ as functions of the angle alone:
$$ \sin : \theta \longmapsto \frac{\text{opposite}}{\text{hypotenuse}} $$Feed in an angle, get out a number. There's no ambiguity about which triangle you used, because every right triangle with that angle gives the same answer.
Sine, cosine, and tangent are not properties of a triangle. They are properties of an angle.
Once you accept that $\sin 30°$ is a single fixed number — not "the sine of some 30° triangle" — you can tabulate it once and use it forever. That's how a single calculator key can answer a question about any right triangle in the world.
4. Special angles
At three angles — 30°, 45°, 60° — the ratios come out as exact algebraic expressions instead of decimal noise. They show up so often that it's worth deriving them once and remembering the picture, not the values.
The 45-45-90 triangle
Start with a square of side $1$ and cut it along a diagonal. You get two congruent right triangles, each with legs $1$ and $1$ and a hypotenuse you can compute with the Pythagorean theorem:
$$ h^2 = 1^2 + 1^2 = 2 \quad\Longrightarrow\quad h = \sqrt{2} $$Both non-right angles are $45°$ by symmetry. So:
$$ \sin 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \qquad \cos 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \qquad \tan 45° = \frac{1}{1} = 1 $$The 30-60-90 triangle
Start with an equilateral triangle of side $2$ and drop an altitude from one vertex. The altitude bisects the opposite side, creating two right triangles. Each has a hypotenuse of $2$, a short side of $1$ (half the original side), and a long leg you find by Pythagoras:
$$ \ell^2 + 1^2 = 2^2 \quad\Longrightarrow\quad \ell = \sqrt{3} $$The angle at the top vertex is half of $60°$, so $30°$; the angle at the base is still $60°$. Reading the ratios off this triangle:
$$ \sin 30° = \frac{1}{2}, \quad \cos 30° = \frac{\sqrt{3}}{2}, \quad \tan 30° = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$ $$ \sin 60° = \frac{\sqrt{3}}{2}, \quad \cos 60° = \frac{1}{2}, \quad \tan 60° = \sqrt{3} $$Notice the symmetry: $\sin 30° = \cos 60°$, $\cos 30° = \sin 60°$. That's not coincidence — it's because $30°$ and $60°$ are complementary, and "opposite" for one angle is "adjacent" for the other.
Table of exact values
Including the endpoints $0°$ and $90°$, which you can read off by squashing the triangles flat:
| $\theta$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |
|---|---|---|---|
| $0°$ | $0$ | $1$ | $0$ |
| $30°$ | $\tfrac{1}{2}$ | $\tfrac{\sqrt{3}}{2}$ | $\tfrac{\sqrt{3}}{3}$ |
| $45°$ | $\tfrac{\sqrt{2}}{2}$ | $\tfrac{\sqrt{2}}{2}$ | $1$ |
| $60°$ | $\tfrac{\sqrt{3}}{2}$ | $\tfrac{1}{2}$ | $\sqrt{3}$ |
| $90°$ | $1$ | $0$ | undefined |
For the $\sin$ column, write $\sqrt{0}/2,\ \sqrt{1}/2,\ \sqrt{2}/2,\ \sqrt{3}/2,\ \sqrt{4}/2$. That gives $0,\ \tfrac{1}{2},\ \tfrac{\sqrt{2}}{2},\ \tfrac{\sqrt{3}}{2},\ 1$ in order. The cosine column is the same sequence reversed.
5. Using them to solve right triangles
"Solving a triangle" means: given some of its parts (sides and angles), find the rest. With a right triangle and one extra piece of information beyond the right angle, $\sin$/$\cos$/$\tan$ are usually enough.
Case A: an angle and a side, find another side
Identify which side you have and which you want — opposite, adjacent, or hypotenuse — relative to the known angle. Pick the function whose ratio links those two sides, then solve.
Example: known angle $\theta = 35°$, known hypotenuse $h = 10$. Find the opposite side $o$.
The function that links opposite and hypotenuse is sine: $\sin\theta = o/h$. Rearrange:
$$ o = h \sin\theta = 10 \sin 35° \approx 10 \times 0.5736 \approx 5.74 $$Case B: two sides, find an angle
Form the ratio you can compute, then invert. The inverse functions $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ (sometimes written $\arcsin$, $\arccos$, $\arctan$) take a ratio and return the angle that produces it.
Example: opposite $= 4$, adjacent $= 7$. Find $\theta$.
$$ \tan\theta = \frac{4}{7} \quad\Longrightarrow\quad \theta = \tan^{-1}\!\left(\frac{4}{7}\right) \approx 29.7° $$$\sin^{-1}$ and $\cos^{-1}$ only accept inputs between $-1$ and $1$, because $\sin$ and $\cos$ can only output values in that range. If your ratio comes out as $1.3$, you've made an arithmetic error — there is no angle that produces it.
The two-step pattern
Whatever the specific case, the recipe is the same:
- Label. Mark the known angle and the sides as opposite, adjacent, hypotenuse.
- Pick. Choose the function whose ratio uses exactly the sides you know and the one you want.
- Solve. Algebra or inverse function — whichever isolates the unknown.
6. The reciprocal trio: sec, csc, cot
Once you have $\sin$, $\cos$, $\tan$, you have three more for free — their reciprocals. They get their own names because they appear often enough in formulas (especially in calculus) that always writing "$1/\cos\theta$" gets tiresome.
The pairing is a little perverse — secant is the reciprocal of cosine, not sine — but it's the convention everywhere. The "co-" prefix on cosecant and cotangent matches the function they pair with under complementary angles, not the one they're reciprocals of.
You can do almost everything in introductory trigonometry with just $\sin$, $\cos$, $\tan$. The reciprocal trio mostly earns its keep later, when you start doing identities and calculus. Recognize them when they appear; don't bother memorizing values.