Visual walkthrough — Co-function identities

This walkthrough goes deeper than the parent note. If a word like "opposite" or "hypotenuse" is unfamiliar, peek at Right Triangle Trigonometry (SOH-CAH-TOA) first — but you shouldn't need to; we rebuild it here.

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Step 1 — Draw the stage: a right triangle

WHAT. We draw a triangle with three corners, and we make one corner a perfect square corner — a $90°$ angle. We name the corners $A$, $B$, $C$, and we put the square corner at $C$.

WHY. Every co-function identity lives inside a right triangle. The square corner is the anchor: it is the reason the other two angles are forced to cooperate (Step 2). Without it, none of this works.

PICTURE. In the figure, the small square drawn inside corner $C$ is the standard mark for "this is exactly $90°$". The two slanted corners $A$ and $B$ are the ones we will study.

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Step 2 — The two slanted angles must add to $90°$

WHAT. Name the angle at corner $A$ with the Greek letter $\theta$ (say "theta" — it just means "the angle we're focusing on"). Name the angle at corner $B$ with $\varphi$ ("phi" — its partner). We now prove $\theta +\varphi =90°$.

WHY. This is the hidden engine of every co-function identity. Two angles that add to $90°$ are called complementary (see Complementary and Supplementary Angles). We must earn this fact before using it.

$\underset{\text{angle at }A}{\underbrace{\theta }}+\underset{\text{angle at }B}{\underbrace{\varphi }}+\underset{\text{angle at }C}{\underbrace{90°}}=\underset{\text{every triangle}}{\underbrace{180°}}$

Every term explained: the three angles of any triangle always total $180°$ (a straight line's worth of turning). One of ours is $90°$. Subtract it from both sides:

$\theta +\varphi =180°-90°=90°⟹\boxed{\varphi =90°-\theta }$

The boxed line is the star: $\varphi$ is not a free number — it is exactly the leftover after $\theta$ is taken from $90°$. That is what "$90°-\theta$" means physically.

PICTURE. Watch the two coloured arcs: as $\theta$ (blue) grows, $\varphi$ (orange) shrinks by the same amount, so their sum stays pinned at $90°$.

Recall Quick check

If $\theta =30°$, what is $\varphi$? $\varphi =90°-30°=60°$ ::: and indeed $30°+60°+90°=180°$. ✔

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Step 3 — Name the three sides from $\theta$'s point of view

WHAT. Stand at corner $A$ and look across the triangle as angle $\theta$ would. Three sides get names:

• the side across from you (not touching $\theta$) is the opposite, length $a$;
• the side along the floor next to you (touching $\theta$, but not the long slanted one) is the adjacent, length $b$;
• the longest side, always across from the square corner, is the hypotenuse, length $h$.

WHY. Sine and cosine are defined using these three lengths. We can't talk about $\sin \theta$ until we've labelled what "opposite" and "hypotenuse" point to. The labels depend on which corner you stand at — that dependence is the whole secret of this page.

PICTURE. Blue = opposite ($a$), green = adjacent ($b$), gray = hypotenuse ($h$), all tagged from $\theta$'s corner.

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Step 4 — Now re-name the same sides from $\varphi$'s point of view

WHAT. Don't redraw anything. Just walk to corner $B$ and re-ask "which side is opposite me now, which is adjacent?" The physical sides never moved — only your viewpoint did.

WHY. This is the crux. The single most important observation on the whole page: the side that was opposite $\theta$ is adjacent to $\varphi$, and the side that was adjacent to $\theta$ is opposite $\varphi$. Standing at the other corner swaps the two roles.

PICTURE. Same triangle, two viewpoints side by side. Follow side $a$: labelled "opposite" from $A$'s desk (left), it is labelled "adjacent" from $B$'s desk (right). Nothing about the triangle changed — only the word attached to $a$.

$\underset{\text{opp of }\theta }{\underbrace{a}}=\underset{\text{adj of }\varphi }{\underbrace{a}},\underset{\text{adj of }\theta }{\underbrace{b}}=\underset{\text{opp of }\varphi }{\underbrace{b}}$

Both lines say the obvious — $a$ equals $a$, $b$ equals $b$ — but the labels underneath them are what carry the magic.

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Step 5 — Write $\cos \varphi$ and watch it become $\sin \theta$ (and the twin)

WHAT. Apply the cosine definition, but at corner $B$ this time.

WHY. Cosine at $B$ needs "adjacent-to-$\varphi$ over hypotenuse". From Step 4, adjacent-to-$\varphi$ is the length $a$. So:

$\cos \varphi =\frac{\text{adj of }\varphi }{h}=\frac{a}{h}$

Now look back at Step 3: $\sin \theta =\frac{a}{h}$ too. Two different expressions, one identical fraction. Setting them equal:

$\underset{a/h\text{ from }\theta }{\underbrace{\sin \theta }}=\underset{a/h\text{ from }\varphi }{\underbrace{\cos \varphi }}$

Finally substitute the boxed result of Step 2, $\varphi =90°-\theta$:

$\boxed{\sin \theta =\cos (90°-\theta )}$

Term by term: on the left, $\theta$'s sine (its tallness). On the right, the cosine of $\theta$'s complement. They are equal because they are the same side over the same hypotenuse, seen from two chairs.

The twin, derived (not just asserted). Now do the mirror computation — write $\sin \varphi$ at corner $B$ and $\cos \theta$ at corner $A$. From Step 4, the side opposite $\varphi$ is the length $b$ (the very side that is adjacent to $\theta$):

$\sin \varphi =\frac{\text{opp of }\varphi }{h}=\frac{b}{h},\cos \theta =\frac{\text{adj of }\theta }{h}=\frac{b}{h}$

Same fraction $b/h$ again, so $\sin \varphi =\cos \theta$. Substituting $\varphi =90°-\theta$:

$\boxed{\cos (90°-\theta )=\sin \theta }\text{is the }a/h\text{ line, and}\boxed{\sin (90°-\theta )=\cos \theta }\text{is this new }b/h\text{ line.}$

So both core identities come from the same triangle — one from tracking side $a$, the other from tracking side $b$. No hand-waving "trade the roles"; each has its own fraction.

PICTURE. The fraction $a/h$ is highlighted once in blue (as $\sin \theta$) and once in orange (as $\cos \varphi$); the fraction $b/h$ is highlighted in green (as both $\cos \theta$ and $\sin \varphi$).

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Step 6 — Get all the others by dividing (no new triangle)

WHAT. We now have both core identities from Step 5: $\sin (90°-\theta )=\cos \theta ,\cos (90°-\theta )=\sin \theta .$

WHY. Tangent, cotangent, secant, cosecant are all built from sine and cosine (we wrote their definitions in Step 3). So we never need geometry again — just algebra.

$\tan (90°-\theta )=\frac{\sin (90°-\theta )}{\cos (90°-\theta )}=\frac{\cos \theta }{\sin \theta }=\cot \theta$

Reading it left to right: $\tan$ is by definition $\sin /\cos$. Feed in the complement; the numerator $\sin (90°-\theta )$ becomes $\cos \theta$ and the denominator $\cos (90°-\theta )$ becomes $\sin \theta$. The result $\frac{\cos \theta }{\sin \theta }$ is exactly the Step 3 definition of $\cot \theta$.

$\sec (90°-\theta )=\frac{1}{\cos (90°-\theta )}=\frac{1}{\sin \theta }=\csc \theta$

Same trick with reciprocals: $\sec =1/\cos$ by definition, the denominator flips to $\sin \theta$, and $\frac{1}{\sin \theta }$ is the Step 3 definition of $\csc \theta$.

PICTURE. A dependency fan: the one geometric fact at the root branches into all six identities.

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Step 7 — Beyond the triangle: the unit circle & angles in every quadrant

WHAT. A triangle only holds angles between $0°$ and $90°$. But $\sin$ and $\cos$ are defined for any angle — even $150°$, $210°$, $300°$ — using the unit circle: a circle of radius $1$ centred at the origin. Spin a ray by angle $\theta$ from the positive $x$-axis; where it hits the circle is the point $(\cos \theta ,\sin \theta )$. So $\cos \theta$ is the point's $x$-coordinate and $\sin \theta$ its $y$-coordinate — and coordinates can be negative.

WHY. We must show the identity survives outside the first quadrant, where signs flip. The clean way: $90°-\theta$ is a reflection across the line $y=x$. Reflecting a point across $y=x$ swaps its coordinates: $(x,y)\to (y,x)$. Since the point for $\theta$ is $(\cos \theta ,\sin \theta )$, the point for $90°-\theta$ is $(\sin \theta ,\cos \theta )$. Reading off that point's coordinates:

$\cos (90°-\theta )=\underset{x\text{-coord after swap}}{\underbrace{\sin \theta }},\sin (90°-\theta )=\underset{y\text{-coord after swap}}{\underbrace{\cos \theta }}$

This argument uses no triangle — it works for every $\theta$, and the signs take care of themselves because the swap moves a negative coordinate to a negative coordinate.

PICTURE. Take $\theta =150°$ (second quadrant, where $\cos$ is negative). Its point is $(-0.866,0.5)$. Reflect across $y=x$ (dashed line) to reach the point for $90°-150°=-60°$, namely $(0.5,-0.866)$ — coordinates swapped. Check: $\cos (-60°)=0.5=\sin 150°$ ✔ and $\sin (-60°)=-0.866=\cos 150°$ ✔. Negative signs and all, the identity holds.

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Step 8 — Special value & the $\theta =45°$ symmetry (a feature, not a bug)

WHAT. Walk $\theta$ from $0°$ up to $90°$ and watch $\sin \theta$ against $\cos (90°-\theta )$.

WHY. A good identity must survive the limits and have a clean symmetric point. At $\theta =0°$ the triangle collapses (no height); at $\theta =90°$ it collapses the other way (no width). The formula must still give the right answer — and Step 7's unit circle already guarantees it does, since the reflection argument never needed a genuine triangle.

$\theta$	complement $90°-\theta$	$\sin \theta$	$\cos (90°-\theta )$	agree?
$0°$	$90°$	$0$	$\cos 90°=0$	✔
$30°$	$60°$	$0.5$	$\cos 60°=0.5$	✔
$45°$	$45°$	$0.707$	$\cos 45°=0.707$	✔
$90°$	$0°$	$1$	$\cos 0°=1$	✔

PICTURE. As $\theta$ slides from $0°$ to $90°$, the blue $\sin \theta$ curve and the orange $\cos (90°-\theta )$ curve lie exactly on top of each other — including the flat endpoints. The identity is a single curve wearing two names.

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The one-picture summary

Everything above collapses into one image: one triangle, two viewpoints, one shared fraction $a/h$. From $\theta$'s chair that fraction is $\sin \theta$; from $\varphi$'s chair the very same fraction is $\cos \varphi =\cos (90°-\theta )$. And Step 7 upgraded this to all angles via the reflection across $y=x$. That's the whole proof.

Recall Feynman retelling — the whole walkthrough in plain words

Draw a slide with a perfect square corner at the bottom. Two kids sit at the two slanted corners. Because a triangle's turning always adds to $180°$ and one corner already ate $90°$, the two kids share the last $90°$ — whatever one kid's angle is, the other's is the leftover. Now here's the trick: the side that is in front of kid $\theta$ is beside kid $\varphi$. So when kid $\theta$ measures "how tall the far corner looks" (that's sine = opposite/hypotenuse), kid $\varphi$ measures the exact same side but calls it "how wide it looks" (that's cosine = adjacent/hypotenuse). Same side, same long slide, so the same number — one kid names it sine, the other names it cosine. Divide or flip that pair (using $\cot =\cos /\sin$, $\csc =1/\sin$) and you get tan/cot and sec/csc for free. For angles bigger than $90°$ the triangle runs out of room, so we spin a ray on a circle of radius $1$ instead; the point for $90°-\theta$ is just the point for $\theta$ flipped across the diagonal $y=x$, which swaps $x$ and $y$ — and swapping the coordinates is exactly what swaps cosine and sine, even when they're negative. Right in the middle, at $45°$, the angle equals its own partner, so both kids read off the very same value.

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Connections

• Co-function Identities — the parent topic this page derives.
• Right Triangle Trigonometry (SOH-CAH-TOA) — where opposite/adjacent/hypotenuse come from.
• Complementary and Supplementary Angles — the $\theta +\varphi =90°$ backbone.
• Pythagorean Identities — pairs with this to collapse ${\sin }^2+{\cos }^2$.
• Reference Angles and Supplementary Identities — the $180°-\theta$ sibling & per-quadrant signs.
• Angle Sum and Difference Formulas — the general rule this is a special case of.
• Radian Measure — for the $\frac{\pi }{2}$ form.