Complementary angle relationships — sin(90−θ) = cos θ etc.: Visual walkthrough

This page builds the cofunction identities from nothing but a triangle. No formula is used before you can see it. By the end you will watch sine turn into cosine just by tilting your head to look at the other corner — and then see the same thing on the Unit Circle for every angle, not just triangle-sized ones.

We rely on ideas from Right Triangle Trigonometry (what sine/cosine mean) and later peek at the Unit Circle.

Step 1 — Build the stage: a right triangle and its two angles

WHAT. Draw a triangle with one square (90°) corner. The two remaining corners are slanted — call them the acute angles (each smaller than 90°).

WHY. Every trig ratio in this chapter is defined on a right triangle, so before we can speak of $sin$ or $cos$ we must have the triangle in front of us. This is the "stage" every symbol will stand on.

PICTURE. The pink corner is our angle $θ$ (theta — just a name for "the angle we care about"). The blue corner is the third one.

Figure — Complementary angle relationships — sin(90−θ) = cos θ etc.

Step 2 — The two slanted corners must add to 90°

WHAT. Add up the angles: $90° + θ + (blue angle) = 180°$ . Subtract the $90°$ from both sides. What remains is $θ + (blue angle) = 90°$ , so the blue angle is exactly $90° - θ$ .

WHY. This is the engine of the whole page. It tells us the two slanted corners are complementary — they share $90°$ between them. Everything else is just re-reading the same triangle.

PICTURE. Watch the pink and blue wedges: together they fill the $90°$ that is not used by the square corner.

$pink corner θ + blue corner (90° - θ) = what the two slants share 90°$

Recall Why does the blue corner equal 90° − θ?

Angles of a triangle sum to 180°; one corner eats 90°, so the other two must share the remaining 90°. If one is θ, the other is 90° − θ. ::: The right angle "uses up" 90°, leaving 90° for the two slanted corners to split.

Step 3 — Name the sides from θ's point of view

WHAT. Standing at the pink corner $θ$ , label the sides:

hypotenuse $h$ — always the long side across from the right angle;
opposite $a$ — the side that does not touch $θ$ ;
adjacent $b$ — the side (not the hypotenuse) that does touch $θ$ .

WHY. Sine and cosine are ratios of sides, but "opposite" and "adjacent" only mean something once you fix which corner you are standing at. We fix $θ$ first.

PICTURE. Yellow = hypotenuse $h$ , pink = opposite $a$ (across the room from $θ$ ), blue = adjacent $b$ (leaning on $θ$ ).

$sin θ = \frac{a opposite θ}{hypotenuse h} cos θ = \frac{b adjacent θ}{hypotenuse h}$

Step 4 — Walk to the other corner; the labels swap

WHAT. Now stand at the blue corner, whose angle is $90° - θ$ . Nothing about the triangle moved — but "opposite" and "adjacent" are re-read from this corner:

side $b$ is now opposite the blue corner (it's across from it);
side $a$ is now adjacent to the blue corner (it touches it);
$h$ is still the hypotenuse (it never changes — it's fixed to the right angle).

WHY. This is the only new fact we need. The side that was "adjacent to $θ$ " is "opposite to $90° - θ$ ", and vice-versa. The swap is a fact of geometry, not algebra.

PICTURE. The same pink and blue sides, but the "opposite/adjacent" tags have jumped to the other side of each colour.

Step 5 — Write sine of the blue angle, and watch it become cosine

WHAT. Apply the same definition of sine, but at the blue corner $90° - θ$ :

$sin (90° - θ) = \frac{opposite the blue corner}{h} = \frac{b}{h}$

Now recall from Step 3 that $\frac{b}{h}$ is literally the definition of $cos θ$ . So:

$sin (90° - θ) = \frac{b}{h} = cos θ$

WHY. We didn't invent a new rule — we used the one definition of sine twice, once at each corner, and the side-swap from Step 4 did all the work.

PICTURE. The fraction $b / h$ is highlighted twice: once labelled " $sin$ of blue", once labelled " $cos$ of $θ$ " — same fraction, two names.

$blue corner’s sine sin (90° - θ) = same fraction \frac{b}{h} = pink corner’s cosine cos θ$

Step 6 — The mirror image: cosine of the blue angle becomes sine

WHAT. Now do the cosine of the blue corner. Cosine is "adjacent over hypotenuse". From the blue corner the adjacent side is $a$ (Step 4), so:

$cos (90° - θ) = \frac{adjacent the blue corner}{h} = \frac{a}{h}$

And $\frac{a}{h}$ is literally the definition of $sin θ$ from Step 3. Therefore:

$cos (90° - θ) = \frac{a}{h} = sin θ$

WHY. This is the perfect mirror of Step 5 and deserves its own picture: it shows the swap runs both ways. Sine of the complement is cosine, and cosine of the complement is sine — the two statements are one symmetry seen from two sides.

PICTURE. The fraction $a / h$ is highlighted twice: once labelled " $cos$ of blue", once labelled " $sin$ of $θ$ ".

$blue corner’s cosine cos (90° - θ) = same fraction \frac{a}{h} = pink corner’s sine sin θ$

Step 7 — The other four identities, straight from their side-pictures

WHAT. In Step 3 we defined all six ratios on the triangle, so we can read the remaining four directly off the swapped sides — no black-box algebra. From the blue corner, tangent = opposite/adjacent $= b / a$ :

$tan (90° - θ) = \frac{opp blue}{adj blue} = \frac{b}{a} = cot θ$

because $\frac{b}{a}$ was defined as $cot θ$ (adj/opp) at the pink corner in Step 3. The reciprocals flip along:

$csc (90° - θ) = \frac{h}{b} = sec θ, sec (90° - θ) = \frac{h}{a} = csc θ, cot (90° - θ) = \frac{a}{b} = tan θ$

WHY. Every one of these is just "same swapped sides, different pairing." Because we gave each ratio a picture in Step 3, none of them arrive as unexplained algebra. See Trigonometric Identities for the full family tree.

PICTURE. A little map: the two proven core identities in the middle, arrows out to the four that inherit the same side-swap.

Step 8 — The general picture: the same identity on the unit circle for ALL θ

WHAT. The triangle only exists for $0° < θ < 90°$ . To cover negative angles, obtuse angles, and everything past a full turn, we move to the Unit Circle: a circle of radius $1$ where the point at angle $θ$ has coordinates $(cos θ, sin θ)$ — the horizontal coordinate is cosine, the vertical coordinate is sine, for any angle.

WHY. On the circle, " $90° - θ$ " has a clean geometric meaning: it is the reflection of the angle across the diagonal line $y = x$ . Reflecting across $y = x$ literally swaps the x- and y-coordinates of a point. Since x is cosine and y is sine, swapping the coordinates swaps sine and cosine — and this is true no matter which quadrant the point lands in.

PICTURE. The angle $θ$ (pink point) and the angle $90° - θ$ (blue point) are mirror images across the dashed diagonal $y = x$ . Read off: the blue point's height (its sine) equals the pink point's width (its cosine).

$point at (90° - θ) : (its x cos (90° - θ), its y sin (90° - θ)) = (swapped sin θ, swapped cos θ)$

Checking every quadrant with signs

WHAT. Let's confirm the coordinate-swap gives the right signs in each region, using $ϕ = 90° - θ$ as the reflected angle.

$θ$	$90° - θ$ lands in	$cos θ$ sign	should equal $sin (90° - θ)$	matches?
$30°$ (Q I)	$60°$ , Q I	$+$	$sin 60° = +$	✓
$120°$ (Q II)	$- 30°$ , Q IV	$-$	$sin (- 30°) = -$	✓
$200°$ (Q III)	$- 110°$ , Q III	$-$	$sin (- 110°) = -$	✓
$- 40°$ (Q IV)	$130°$ , Q II	$+$	$sin 130° = +$	✓

WHY. In each row the reflection across $y = x$ moves the point to a new quadrant, but because reflection only swaps the coordinates (it never changes their values), $cos θ$ and $sin (90° - θ)$ always carry the same number with the same sign. No quadrant escapes the rule.

PICTURE. All four cases plotted: each pink point and its blue reflection sit symmetrically about $y = x$ , in whatever quadrants they fall.

Step 9 — Degenerate edges: θ = 0° and θ = 90°

WHAT. What about the exact endpoints, where the triangle has collapsed to a line? The circle handles these too.

PICTURE. As $θ \to 0°$ the pink point sits at $(1, 0)$ ; its reflection $90° - θ \to 90°$ sits at $(0, 1)$ .

$θ = 0° : sin (90° - 0°) = sin 90° = 1 = ? cos 0° = 1 ✓$ $θ = 90° : sin (90° - 90°) = sin 0° = 0 = ? cos 90° = 0 ✓$

The one-picture summary

Everything above is one triangle read from two corners — and one point reflected across $y = x$ . This final figure stacks both views: the complementary corners with colour-coded sides, plus the unit-circle reflection that extends it to all $θ$ .

Recall Feynman retelling — say it to a 12-year-old

Draw a right triangle. It has one square corner and two slanted corners. Because a triangle's angles add to 180° and the square corner takes 90°, the two slanted corners split the other 90° between them — so if one is θ, the other must be 90° − θ. Now, "sine" just means "the far side divided by the long slanted side," and "cosine" means "the near side divided by the long slanted side." Here's the trick: the side that is near to one slanted corner is far from the other. So when you walk from one corner to the other, near becomes far and far becomes near — cosine turns into sine and sine turns into cosine. That's the identity sin(90°−θ) = cos θ, and its mirror cos(90°−θ) = sin θ. Tangent, cotangent, secant and cosecant just tag along because they're all built out of the same swapped sides. The triangle only works for angles between 0° and 90°, so for negative or big angles we draw a circle of radius 1 instead: the point at angle θ has cosine across and sine up, and the angle 90°−θ is just that point flipped across the slanted mirror line y = x — flipping swaps across-and-up, which is exactly swapping cosine and sine. Same answer, every angle, every quadrant. And it's all in degrees or radians — just don't mix them.

Recall One-line memory hook

Cofunctions are "the same triangle seen from the other corner" — or "the same point reflected across y = x." ::: Opposite ↔ adjacent (or x ↔ y) swap ⇒ sin ↔ cos swap, for every θ.