Visual walkthrough — Inverse trig functions — arcsin, arccos, arctan — domain, range, graphs

We assume only this: a function is a machine that takes one number in and gives one number out. Everything else — sine, one-to-one, reflection — we build below.

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Step 1 — What sine actually is, drawn on a circle

WHAT. Draw a circle of radius $1$ centred at the origin. Pick an angle $\theta$ measured anticlockwise from the positive $x$-axis. Mark the point $P$ where that angle's ray hits the circle. The height of $P$ above the horizontal axis is what we call $\sin \theta$.

WHY a circle and not a triangle? A triangle only shows angles between $0$ and a quarter turn ($\frac{\pi }{2}$). The circle lets $\theta$ keep spinning — past the top, round the back — so we can see every angle's sine, including negative ones. That freedom is exactly what will cause trouble in Step 3.

PICTURE. Look at the red vertical segment: that length is $\sin \theta$. As the point climbs, the segment grows; past the top it shrinks again.

Here $\theta$ is the angle (the input) and the height is the output — a plain signed length on the circle, not a fraction of anything. Keep that direction in mind: angle → height. The inverse will run height → angle.

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Step 2 — Unrolling the circle into the sine wave

WHAT. Now spin $\theta$ steadily and plot the height against $\theta$ on ordinary axes: horizontal axis $=\theta$, vertical axis $=\sin \theta$. The circle's up-and-down height traces the familiar wave.

WHY unroll it? To invert a machine we need to see it as a graph — a curve on $x$–$y$ axes — because the whole trick of inverting (Step 5) is a reflection of that curve. The circle can't be reflected across $y=x$; a graph can.

PICTURE. Each dot on the circle (left) sends its height across to a dot on the wave (right). The wave crosses zero at $0,\pi ,2\pi$ and peaks at $+1$ then dips to $-1$, forever.

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Step 3 — The disaster: one height, many angles

WHAT. Draw a single horizontal line at height $y=0.5$. Count how many times it hits the wave.

WHY this is fatal for "going backwards". An inverse must answer the question "which angle gave this height?" with exactly one angle. But the line at $0.5$ hits the wave infinitely often: $\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{\pi }{6}+2\pi ,\dots$ So "the angle" is not well-defined. This is what one-to-one means and why sine fails it.

PICTURE. The dashed line at $0.5$ pierces the wave at many crossings (red dots). Any machine claiming to undo sine would have to output all of them — impossible for a function.

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Step 4 — The fix: keep only the rising slice

WHAT. Erase everything except the piece of the wave from $\theta =-\frac{\pi }{2}$ to $\theta =+\frac{\pi }{2}$. On this slice the wave rises steadily from $-1$ up to $+1$ without ever turning back.

WHY this exact slice? We need the simplest continuous piece that is (a) strictly increasing — so every height is hit once — and (b) covers the whole output range $[-1,1]$, and (c) is symmetric about $0$ so it treats positive and negative inputs fairly. The rising branch $[-\frac{\pi }{2},\frac{\pi }{2}]$ is the unique simplest choice. Now the horizontal line at $0.5$ hits exactly once.

PICTURE. The kept slice is solid; the discarded rest is faint grey. The line at $0.5$ now meets the curve at a single green dot — uniqueness restored.

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Step 5 — Flipping the slice to birth arcsin

WHAT. Take that rising slice and reflect it across the diagonal line $y=x$. The reflected curve is arcsin.

WHY reflect across $y=x$? Reflecting a point across $y=x$ swaps its coordinates: $(a,b)\mapsto (b,a)$. If the point $(\theta ,\sin \theta )$ sat on sine (angle in, height out), its mirror $(\sin \theta ,\theta )$ has height in, angle out — which is precisely the inverse's job. So the mirror line is not decoration; it is the act of reversing input and output. (See Reflection of graphs across y=x.)

PICTURE. The lavender slice (sine) and its coral mirror (arcsin) sit either side of the dashed diagonal. Notice the axes swap roles: what was horizontal reach $[-1,1]$ is now horizontal input, and the vertical spread $[-\frac{\pi }{2},\frac{\pi }{2}]$ is now the output.

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Step 6 — Cosine needs a different slice (why arccos ≠ arcsin)

WHAT. Repeat Steps 3–5 for cosine. Try the same slice $[-\frac{\pi }{2},\frac{\pi }{2}]$ first — and watch it fail.

WHY it fails there. Since cosine is the width (the horizontal coordinate), on $[-\frac{\pi }{2},\frac{\pi }{2}]$ it is a hill: it climbs to $1$ at $\theta =0$ then falls, so $\cos (-a)=\cos (a)$. A horizontal line hits it twice — not one-to-one. We must slide to the next simplest monotone slice: $[0,\pi ]$, where cosine falls steadily from $+1$ to $-1$. Reflecting that slice gives arccos.

PICTURE. Top: cosine on $[-\frac{\pi }{2},\frac{\pi }{2}]$ with a line cutting it twice (red — rejected). Bottom: cosine on $[0,\pi ]$, strictly falling, cut once (green — accepted), then its mirror = arccos.

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Step 7 — arctan and the walls it can never touch

WHAT. Tangent is $\tan \theta =\frac{\sin \theta }{\cos \theta }$ — height over width on the circle. As $\theta \to \frac{\pi }{2}$, the width $\cos \theta \to 0$, so the ratio explodes to $+\infty$. Plot $\tan \theta$ on $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ and reflect.

WHY this exact slice? As with sine, we hunt the simplest monotone piece. On $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ tangent is strictly increasing: its slope is $\frac{d}{d\theta }\tan \theta =\frac{1}{{\cos }^2\theta }$, which is always positive (a square in the denominator can never be negative), so the curve only ever climbs — never turns back. That means every ratio is hit exactly once. It also sweeps every real number, from $-\infty$ up to $+\infty$, so no output is missed. Any other slice would either straddle an asymptote (breaking monotonicity — the curve would leap from $+\infty$ back to $-\infty$) or fail to cover all reals. So $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ is forced, just like the sine slice was.

WHY open interval, not closed? At exactly $\pm \frac{\pi }{2}$ the width is $0$ and dividing by $0$ is undefined — the curve shoots up a vertical wall (asymptote) without ever arriving. Reflecting, those vertical walls become horizontal ceilings at $y=\pm \frac{\pi }{2}$ that arctan approaches but never reaches. Hence the range is open: $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.

PICTURE. Left: tangent rising from $-\infty$ to $+\infty$ between two vertical dashed walls. Right: its mirror, arctan, flattening toward two horizontal dashed ceilings it never touches.

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Step 8 — The snap-back: why $\arcsin (\sin x)$ isn't always $x$

WHAT. Feed an angle outside the safe zone, say $x=\frac{3\pi }{4}$, through sine and then arcsin.

WHY it changes. Sine sends $\frac{3\pi }{4}$ to height $\frac{\sqrt{2}}{2}$. Arcsin must then return an angle inside $[-\frac{\pi }{2},\frac{\pi }{2}]$ with that same height — and that angle is $\frac{\pi }{4}$, not $\frac{3\pi }{4}$. The inverse "snaps" the answer into its range. It only leaves $x$ untouched when $x$ was already in the safe zone.

PICTURE. On the circle, $\frac{3\pi }{4}$ (outside zone) and $\frac{\pi }{4}$ (inside zone) share the same height — a dashed horizontal line links them. Arcsin always chooses the one inside the shaded safe zone.

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The one-picture summary

All three inverses are the same idea three times: chop the trig curve to its simplest monotone slice, then reflect across $y=x$. The slice you keep becomes the inverse's range; the outputs it covered become the inverse's domain.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a dot going round a circle. Its height is sine and its width is cosine; that height, plotted as the dot spins, makes a wave. We wanted to run the machine backwards — hand it a height and get the angle — but the wave repeats, so one height matches endlessly many angles. No good. So we keep only a small rising piece of the wave where each height happens once, and we tip that piece over the diagonal mirror $y=x$, which swaps "in" and "out". The tipped piece is arcsin. Cosine is a hill on the same little interval, so it fails the one-hit test and gets sent to a falling slice $[0,\pi ]$ instead — that's arccos, always positive. Tangent is height-over-width; near the sides the width hits zero and the value flies off to infinity along walls it never touches, so when we tip it over, those walls become ceilings arctan forever approaches but never reaches — an open range. And if you ever feed sine an angle from outside its safe slice, arcsin politely hands back the equal-height angle that is inside. Same recipe, three flavours.

Recall Quick self-test

Why is arccos never negative? ::: Its slice $[0,\pi ]$ lies entirely above zero; cosine wasn't one-to-one on a zero-symmetric slice, so it was forced onto the top half. Why is arctan's range open at $\pm \pi /2$? ::: Those are asymptotes — tangent's vertical walls become arctan's horizontal ceilings, approached but never reached. What single geometric act turns a trig slice into its inverse? ::: Reflection across the line $y=x$, which swaps input and output coordinates.

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Connections

• Parent: Inverse trig functions
• Unit circle and reference angles
• One-to-one functions and invertibility
• Reflection of graphs across y=x
• Trigonometric identities
• Solving trigonometric equations
• Derivatives of inverse trig functions

Concept Map