Trig ratios of standard angles — 0°, 30°, 45°, 60°, 90° (derive, don't memorize blindly): Visual walkthrough

This page rebuilds the entire table from the parent note — the standard-angle table — starting from a single right triangle and two ideas: a ratio, and the Pythagorean Theorem. By the last picture you will be able to redraw every value with a pencil.

We never assume you know what "sine" is. We build it.

Step 0 — What a "trig ratio" even means

WHY divide sides at all? Because a ratio of sides doesn't care how big you draw the triangle — only its shape (its angles) sets the ratio. So each of these numbers is a fingerprint of the angle $θ$ , nothing else. That is the whole reason trigonometry works.

In the figure: $θ$ sits at the bottom-left corner. Trace the red opposite side (across from $θ$ ), the orange adjacent side (under $θ$ ), and the violet hypotenuse (the slope). Every value below is one of these divided by another.

Step 1 — Draw the 45° triangle from equality

WHAT. Build the triangle for $θ = 45°$ .

WHY 45° first? Because $45°$ is the balanced case: the two non-right angles are equal ( $45° + 45° + 90° = 180°$ ). Equal angles force equal opposite sides — the two legs must be the same length. That symmetry hands us the triangle for free, with no measuring.

PICTURE. Make both legs length $1$ . This is an isosceles right triangle (two equal legs). See Special Triangles.

$hypotenuse h^{2} = one leg 1^{2} + other leg 1^{2} = 2 \Rightarrow h = 2$

Here $h^{2}$ is the square built on the slanted side; the two $1$ 's are the squares on the legs. The Pythagorean Theorem says the big square equals the two small ones added — so $h = 2$ (the number that squares to $2$ , about $1.414$ ).

Step 2 — Read off the 45° values

WHAT. Plug the sides into the Step 0 definitions.

WHY. We now have all three sides ( $opp = 1$ , $adj = 1$ , $hyp = 2$ ), so the ratios are just division.

$sin 45° = \frac{1 opp}{hyp 2} = \frac{2}{2}, cos 45° = \frac{1 adj}{hyp 2} = \frac{2}{2}, tan 45° = \frac{1 opp}{adj 1} = 1$

Why is $\frac{1}{2}$ rewritten as $\frac{2}{2}$ ? That is Rationalization: multiply top and bottom by $2$ so no root sits underneath — $\frac{1}{2} \cdot \frac{2}{2} = \frac{2}{2}$ . Same number, tidier for later addition. See Exact vs Approximate Values.

PICTURE. $tan 45° = 1$ means the slope rises exactly as fast as it runs — a perfect diagonal. The figure in Step 1 already shows this: equal legs, $45°$ slant.

Step 3 — Build one equilateral triangle for BOTH 30° and 60°

WHAT. Draw an equilateral triangle (all sides equal, all angles $60°$ ) with each side $= 2$ .

WHY size 2? So that when we cut it in half the pieces come out as whole numbers, not fractions. It is a convenience, nothing more.

WHY does one triangle give two angles? Because dropping a straight line from the top vertex down to the base does two things at once: it splits the $60°$ top angle into two $30°$ angles, and (by the triangle's mirror Symmetry in Trigonometry) it cuts the base exactly in half. One clean cut manufactures a $30 - 60 - 90$ triangle.

PICTURE.

The dashed violet line is the cut. On the left half: the base became $1$ (half of $2$ ), the slanted side stays $2$ , and the height is still unknown — call it $b$ .

Step 4 — Find the height with Pythagoras

WHAT. Solve for the height $b$ of the half-triangle.

WHY. We know two sides of the half-triangle ( $base = 1$ , $hyp = 2$ ) and need the third to read any ratio.

$height^{2} b^{2} + half-base^{2} 1^{2} = full side^{2} 2^{2} \Rightarrow b^{2} = 4 - 1 = 3 \Rightarrow b = 3$

$b^{2}$ is the square on the tall side; $1^{2}$ the square on the short base; $2^{2}$ the square on the slant. Subtract to isolate the height: $b = 3 \approx 1.732$ .

PICTURE.

The completed $30 - 60 - 90$ triangle: short side $1$ sits opposite the small $30°$ angle; long side $3$ sits opposite the bigger $60°$ angle; hypotenuse $2$ across from the right angle. The small angle always faces the small side — that single fact keeps you from ever mixing up $30°$ and $60°$ .

Step 5 — Read off 30° and 60°

WHAT. Apply the definitions to the same triangle, once looking from the $30°$ corner, once from the $60°$ corner.

WHY the same triangle twice? Because "opposite" and "adjacent" swap depending on which corner you stand at. Stand at $30°$ : the short side ( $1$ ) is opposite. Stand at $60°$ : now the tall side ( $3$ ) is opposite. One triangle, two viewpoints.

$sin 30° = \frac{1 opp of 30°}{hyp 2} = \frac{1}{2}, cos 30° = \frac{3 adj of 30°}{2} = \frac{3}{2}, tan 30° = \frac{1}{3} = \frac{3}{3}$

$sin 60° = \frac{3 opp of 60°}{2} = \frac{3}{2}, cos 60° = \frac{1 adj of 60°}{2} = \frac{1}{2}, tan 60° = \frac{3}{1} = 3$

PICTURE.

Notice the swap: $sin 30° = cos 60°$ and $cos 30° = sin 60°$ . That is not a coincidence — $30°$ and $60°$ are Complementary Angles (they sum to $90°$ ), and swapping viewpoints swaps opposite with adjacent, which swaps sine with cosine.

Step 6 — The edge cases: 0° and 90° by squashing the triangle

WHAT. Get $0°$ and $90°$ where no fat triangle exists.

WHY a limit and not a triangle? At exactly $0°$ or $90°$ the triangle collapses flat — one side shrinks to nothing. So instead of building, we watch what happens as $θ$ slides toward the extreme, using the Unit Circle (a circle of radius $1$ ). A point on it at angle $θ$ sits at position $(cos θ, sin θ)$ : its horizontal reach is cosine, its height is sine.

PICTURE.

At $0°$ the point rests at $(1, 0)$ : no height, full width. So $sin 0° = 0$ , $cos 0° = 1$ , and $tan 0° = \frac{0}{1} = 0$ .
At $90°$ the point is at $(0, 1)$ : full height, no width. So $sin 90° = 1$ , $cos 90° = 0$ , and $tan 90° = \frac{1}{0}$ is undefined — you cannot divide by zero, and geometrically the slope is straight up (infinitely steep).

$tan 90° = \frac{s i n 90°}{c o s 90°} = \frac{1}{0} \Rightarrow undefined (vertical line, no finite slope)$

The one-picture summary

Everything above lives in this one frame: the two source triangles on the left ( $45 - 45 - 90$ and $30 - 60 - 90$ ), the unit circle for the flat cases on the right, and the finished table underneath. If you can redraw the two triangles, you own every entry.

Angle	$sin$	$cos$	$tan$
$0°$	$0$	$1$	$0$
$30°$	$\frac{1}{2}$	$\frac{3}{2}$	$\frac{3}{3}$
$45°$	$\frac{2}{2}$	$\frac{2}{2}$	$1$
$60°$	$\frac{3}{2}$	$\frac{1}{2}$	$3$
$90°$	$1$	$0$	undefined

Recall Feynman retelling — say it out loud

A trig ratio is just one side of a right triangle divided by another. Because a ratio ignores size, it only depends on the angle — so it's a code for the angle. For $45°$ I draw a triangle with two equal sides of $1$ ; Pythagoras gives the slant $2$ , and dividing gives $\frac{2}{2}$ for both sine and cosine, and $1$ for tangent. For $30°$ and $60°$ I draw an equilateral triangle of side $2$ , chop it down the middle, and get a triangle with sides $1$ , $3$ , $2$ . Standing at the small angle, the small side is opposite, so $sin 30° = \frac{1}{2}$ . Standing at the big angle, the tall side is opposite, so $sin 60° = \frac{3}{2}$ . Small angle faces small side — that's how I never confuse them. For $0°$ and $90°$ I picture a dot on a circle of radius $1$ : its height is sine, its width is cosine. At $0°$ it's flat right (sine $0$ , cosine $1$ ); at $90°$ it's straight up (sine $1$ , cosine $0$ ), and tangent blows up because we'd divide by zero.

Recall

Why does a ratio of two sides only depend on the angle, not the triangle's size? ::: Scaling a triangle multiplies every side by the same factor, which cancels in a ratio — so shape (angle) alone sets the value. On the 30-60-90 triangle, why is $sin 30° = \frac{1}{2}$ and not $\frac{3}{2}$ ? ::: The small $30°$ angle faces the short side (length $1$ ); opposite over hypotenuse is $\frac{1}{2}$ . Why is $tan 90°$ undefined? ::: It equals $sin 90°/ cos 90° = 1/0$ ; division by zero is undefined and the line is vertical (infinite slope). Where does $2$ come from in the $45°$ case? ::: Pythagoras on legs $1$ and $1$ : hypotenuse $= 1^{2} + 1^{2} = 2$ .