Visual walkthrough — Trig functions for angles beyond 90° — ASTC rule (All, Sin, Tan, Cos)

This is a pictures-first page. Read each step's figure before its equation.

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Step 1 — An arrow that can point anywhere

WHAT. Draw a flat sheet with two number lines crossing at a point $O$ (the origin). The line going right–left is the ==$x$-axis (left–right position). The line going up–down is the $y$-axis== (up–down position). Now draw an arrow of length $1$ starting at $O$. We can spin this arrow to point in any direction.

WHY. The old "opposite over hypotenuse" idea only lives inside a right triangle, so it only works for angles between $0°$ and $90°$. But real problems throw us $120°$, $-30°$, $750°$. We need a home where every direction is legal. A spinning arrow of fixed length is that home — its tip traces a circle of radius $1$, the unit circle (see Unit Circle).

PICTURE.

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Step 2 — The tip's shadow gives us two numbers

WHAT. Wherever the arrow points, its tip is a point we call $P$. Drop a straight line from $P$ down to the $x$-axis: the horizontal distance is the tip's left–right position; call it $x$. The vertical distance is its up–down position; call it $y$. So $P=(x,y)$.

WHY. A single arrow direction is hard to compute with. Two numbers $(x,y)$ are easy. The shadow onto each axis converts "which way it points" into "how far right" and "how far up" — quantities we can add, multiply, and put a sign on.

PICTURE.

For an acute $\theta$ this matches the old triangle rule: the right triangle $O$–foot–$P$ has hypotenuse $1$, so adjacent$/1=x=\cos \theta$ and opposite$/1=y=\sin \theta$. Old and new definitions agree.

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Step 3 — Tangent is the steepness of the arrow

WHAT. Define a third quantity as the ratio of up-position to right-position: $\tan \theta =\frac{y}{x}=\frac{\sin \theta }{\cos \theta }.$

WHY tan and not something else? We often want how steep the arrow points — its rise compared to its run. "Rise over run" is exactly $y/x$. That single number answers "for every step right, how many steps up?" — the slope of the arrow. (When $x=0$, the arrow points straight up or down, run is zero, and steepness is infinite — we handle that degenerate case in Step 7.)

PICTURE.

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Step 4 — The plane splits into four rooms (quadrants)

WHAT. The two axes cut the sheet into four corner regions. Going anticlockwise from top-right we name them QI, QII, QIII, QIV (Quadrants one to four).

WHY. As the arrow spins, its shadow $x$ and $y$ each flip between $+$ and $-$ when the arrow crosses an axis. Grouping directions into four rooms lets us track those flips cleanly — each room has one fixed sign-pattern for $(x,y)$.

PICTURE.

Room	arrow points…	$x=\cos$	$y=\sin$
QI	up-right	$+$	$+$
QII	up-left	$-$	$+$
QIII	down-left	$-$	$-$
QIV	down-right	$+$	$-$

Read the picture, not the words: left means $x<0$, down means $y<0$. That's the entire sign story.

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Step 5 — Sign of each ratio → the letters A, S, T, C fall out

WHAT. Combine Step 3 and Step 4. In each room, work out the sign of $\cos =x$, $\sin =y$, and $\tan =y/x$.

WHY. $\tan$'s sign is not free — it is forced by the other two, because a ratio of two signed numbers has a sign determined by them: $\frac{+}{+}=+$, $\frac{+}{-}=-$, $\frac{-}{-}=+$, $\frac{-}{+}=-$.

PICTURE.

Room	$\cos$	$\sin$	$\tan =\frac{\sin }{\cos }$	who is $+$
QI	$+$	$+$	$\frac{+}{+}=+$	All
QII	$-$	$+$	$\frac{+}{-}=-$	Sin
QIII	$-$	$-$	$\frac{-}{-}=+$	Tan
QIV	$+$	$-$	$\frac{-}{+}=-$	Cos

Read the "who is $+$" column anticlockwise from QI: A, S, T, C. That is the ASTC rule — derived, not memorised.

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Step 6 — The reference angle recycles an acute triangle

WHAT. From the tip $P$, the shortest tilt to the $x$-axis is an acute angle called the ==reference angle ${\theta }^{\prime }$== ("theta prime"). It is always between $0°$ and $90°$.

WHY the $x$-axis and not the $y$-axis? Because $\cos ,\sin$ were built from the horizontal foot of $P$ (Step 2). The little right triangle formed by dropping to the $x$-axis has legs of length $|x|$ and $|y|$ — the same lengths as the friendly QI triangle with angle ${\theta }^{\prime }$. So the magnitudes $|\cos \theta |$ and $|\sin \theta |$ equal $\cos {\theta }^{\prime }$ and $\sin {\theta }^{\prime }$. Only the plus/minus (Step 5) differs.

PICTURE.

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Step 7 — Degenerate arrows: straight along an axis

WHAT. What if the arrow lands exactly on an axis? Then $P$ is one of $(1,0),(0,1),(-1,0),(0,-1)$.

WHY it needs its own step. On an axis there is no triangle — one leg has length zero. The ratios still exist, but $\tan$ can break: when the run $x=0$ we would divide by zero.

PICTURE.

$\theta$	$P=(x,y)$	$\cos$	$\sin$	$\tan =y/x$
$0°$	$(1,0)$	$1$	$0$	$0/1=0$
$90°$	$(0,1)$	$0$	$1$	$1/0=$ undefined
$180°$	$(-1,0)$	$-1$	$0$	$0/-1=0$
$270°$	$(0,-1)$	$0$	$-1$	$-1/0=$ undefined

$\tan$ is undefined at $90°$ and $270°$ because the arrow points straight up/down: infinite steepness (see the vertical asymptotes in Graphs of Sin Cos Tan).

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Step 8 — Spinning past a full turn, and spinning backwards

WHAT. Two edge cases. (a) Beyond $360°$: after a full turn the arrow is back where it started, so $P$ repeats. Subtract $360°$ as many times as needed. (b) Negative angle: a minus sign means spin clockwise instead.

WHY. $P$ depends only on the arrow's final direction, not on how many loops it took to get there. This is periodicity: $\cos (\theta +360°)=\cos \theta$ and likewise for $\sin$. A clockwise turn of $30°$ ends in the same place as an anticlockwise turn of $330°$.

PICTURE.

• $\cos 750°=\cos (750°-360°-360°)=\cos 30°=\frac{\sqrt{3}}{2}.$
• $\sin (-30°)=\sin 330°$; $330°$ is QIV so $\sin <0$, reference $30°$, giving $-\sin 30°=-\frac{1}{2}.$ (Matches the identity $\sin (-\theta )=-\sin \theta$ from Trigonometric Identities.)

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The one-picture summary

Everything at once: four rooms, their sign-letters, and one worked arrow at $120°$ showing reference ${\theta }^{\prime }=60°$ and the positive sine.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a stick of length one pinned at a point, spinning like a clock hand run backwards. Its tip has a left–right number (that's cosine) and an up–down number (that's sine); their ratio, up-over-right, is how steep the stick leans (that's tangent). The flat sheet has four corner rooms. In the top-right room both numbers are positive, so all ratios are positive. Spin into the top-left and the left–right number turns negative, so only sine stays positive. Down-left, both are negative — but a negative divided by a negative is positive, so only tangent survives. Down-right, only cosine. Reading those survivors as we spin spells A-S-T-C: All Students Take Coffee. To get the actual size of any answer, tilt the stick back to its nearest friendly copy touching the sideways axis — that little acute angle is the reference angle, and it gives the same size number as a plain right triangle. If the stick points exactly up or down, it's leaning infinitely, so tangent has no value there. And spinning a whole loop, or spinning backwards, just lands the tip in a spot we already know. So: the room picks the plus-or-minus, the reference triangle picks the size, and multiplying them gives the ratio of any angle at all.

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Active recall

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Connections

• Trig Functions for Angles Beyond 90° — The ASTC Rule — the parent this page derives.
• Unit Circle — the spinning arrow of length one.
• Reference Angles — the acute triangle giving magnitude.
• Trigonometric Identities — negative-angle and shift rules.
• Radians and Degrees — same story with $\pi$ for $180°$.
• Graphs of Sin Cos Tan — quadrant signs become above/below the axis.