Inverse trig functions — arcsin, arccos, arctan — domain, range, graphs: Worked examples

You have met the definitions in the parent note. This page is the practice arena. We list every kind of problem inverse trig can throw at you, then knock each one down with a fully worked example — you forecast first, we build the answer step by step.

Before we start, a one-line reminder of the only three "safe zones" (ranges) — every answer must land inside its zone:

Recall The three safe zones (memorise the picture, not the words)

$arcsin$ : answers live in $[- \frac{π}{2}, \frac{π}{2}]$ — right half of the circle.
$arccos$ : answers live in $[0, π]$ — top half of the circle.
$arctan$ : answers live in $(- \frac{π}{2}, \frac{π}{2})$ — right half, endpoints never reached.

If you ever forget why these zones exist, revisit One-to-one functions and invertibility — a function can only be reversed where each output has exactly one input.

The scenario matrix

Every problem below belongs to one of these cells. Together the worked examples cover all of them.

Cell	What makes it tricky	Example that hits it
A. Positive input, standard	plain lookup on the increasing branch	Ex 1
B. Negative input, sign rules	odd (sin, tan) vs. reflection ( $arccos (- x) = π - arccos x$ )	Ex 2
C. Zero / boundary input	$x = 0, \pm 1$ — where graphs start, end, or touch asymptotes	Ex 3
D. Out-of-range "snap back"	$arcsin (sin θ)$ , $arccos (cos θ)$ in every quadrant + multi-turn	Ex 4, Ex 4b, Ex 4c
E. Nested / composite identity	$tan (arccos x)$ — build an expression, both signs of $x$	Ex 5
F. Limiting behaviour	$x \to + \infty$ and $x \to - \infty$ for $arctan$ ; both asymptotes	Ex 6
G. Real-world word problem	angle of elevation from a distance	Ex 7
H. Exam-style twist	solve an equation using an inverse, then check all cases	Ex 8

Cell A — positive input, standard lookup

Cell B — negative input, the sign rules

Here the reflection rule matters. $arccos$ is not odd — feeding a negative flips it a different way than $arcsin$ or $arctan$ do. To cover the full cell we compute a negative input through all three functions in one shot.

The figure below is a unit circle (radius-1 circle centred at the origin). Its horizontal axis reads off $cos$ , its vertical axis reads off $sin$ ; angles are measured anticlockwise from the positive $cos$ -axis. Three coloured radius-arrows show where each answer lands: teal = $arcsin (- \frac{1}{2})$ in the lower-right (a negative angle, correctly in arcsin's zone), orange = $arccos (- \frac{1}{2})$ in the upper-left (arccos's zone is the top half, so it climbs up, never negative), and plum = $arctan (- 1)$ also in the lower-right (arctan is odd, so a negative input gives a negative angle). The plum dashed vertical line marks the input value $x = - \frac{1}{2}$ used by arcsin/arccos.

Figure — Inverse trig functions — arcsin, arccos, arctan — domain, range, graphs

Worked example Ex 2. Compute

$arcsin (- \frac{1}{2})$ , $arccos (- \frac{1}{2})$ , and $arctan (- 1)$ — three negative inputs, three different sign behaviours. Forecast: $arcsin$ of a negative should be negative, and $arctan$ of a negative should be negative (both odd). But is $arccos$ of a negative also negative? Predict all three.

$arcsin$ — use oddness. $arcsin (- x) = - arcsin x$ because sine is an odd function, so its inverse is odd too. Why this step? Oddness turns a negative-input problem into a positive one we already know.
$arcsin \frac{1}{2} = \frac{π}{6}$ (since $sin 30° = \frac{1}{2}$ ), so $arcsin (- \frac{1}{2}) = - \frac{π}{6}$ . Why this step? $- \frac{π}{6}$ is in $[- \frac{π}{2}, \frac{π}{2}]$ — the teal arrow in the lower-right of the figure. Valid.
$arccos$ — use the reflection rule. $arccos (- x) = π - arccos x$ . This is NOT $- arccos x$ : negatives are illegal outputs for $arccos$ . Why this step? The safe zone $[0, π]$ has no negatives, so a "sign flip" can't work; the correct move is $π - (\cdot)$ , mirroring across $\frac{π}{2}$ .
$arccos \frac{1}{2} = \frac{π}{3}$ , so $arccos (- \frac{1}{2}) = π - \frac{π}{3} = \frac{2 π}{3}$ — the orange arrow, top-left, still inside $[0, π]$ . Why this step? Confirms arccos climbs into the second quadrant instead of going negative.
$arctan$ — use oddness (like arcsin). $arctan (- x) = - arctan x$ because tangent is odd. Why this step? $arctan$ shares arcsin's symmetric zone $(- \frac{π}{2}, \frac{π}{2})$ , so its negative-input rule is a plain sign flip, unlike arccos.
$arctan 1 = \frac{π}{4}$ (slope $1$ is a $45°$ line), so $arctan (- 1) = - \frac{π}{4}$ — the plum arrow, lower-right. Why this step? A slope of $- 1$ is a line tipping the other way; the angle is the mirror of $\frac{π}{4}$ below the axis.

Answers: $arcsin (- \frac{1}{2}) = - \frac{π}{6}$ , $arccos (- \frac{1}{2}) = \frac{2 π}{3}$ , $arctan (- 1) = - \frac{π}{4}$ .

Verify: $sin (- \frac{π}{6}) = - \frac{1}{2}$ ✅, $cos \frac{2 π}{3} = - \frac{1}{2}$ ✅, $tan (- \frac{π}{4}) = - 1$ ✅. All three land in their zones; only arccos avoids negatives.

Cell C — zero and boundary inputs

The endpoints $x = 0, \pm 1$ are where the graphs start, end, or touch an asymptote. Get these and you own the shape of every curve — see Reflection of graphs across y=x.

Worked example Ex 3. Evaluate the full boundary set:

$arcsin 0, arcsin 1, arcsin (- 1), arccos 1, arccos (- 1), arctan 0$ . Forecast: these are the "corner" values. Sketch each on the unit circle in your head first. Note $arcsin (- 1)$ — don't forget the bottom endpoint.

$arcsin 0$ : which $y \in [- \frac{π}{2}, \frac{π}{2}]$ has $sin y = 0$ ? Only $y = 0$ . Why this step? Zero sine happens at $0, π, 2 π, \dots$ ; only $0$ is in the zone.
$arcsin 1$ : $sin y = 1$ at $y = \frac{π}{2}$ — the top of the safe zone (a closed endpoint, so it's allowed). Why this step? Unlike $arctan$ 's open ends, $arcsin$ includes $\pm \frac{π}{2}$ , so $arcsin (\pm 1)$ is defined.
$arcsin (- 1)$ : $sin y = - 1$ at $y = - \frac{π}{2}$ — the bottom endpoint of the zone. Why this step? By oddness, $arcsin (- 1) = - arcsin (1) = - \frac{π}{2}$ ; it is the mirror image of the top corner.
$arccos 1$ : which $y \in [0, π]$ has $cos y = 1$ ? At $y = 0$ . Why this step? Cosine peaks at $1$ when the angle is $0$ — the right edge of the top half.
$arccos (- 1)$ : cosine is $- 1$ at $y = π$ (the far left of the circle). Why this step? This is the other endpoint of the $[0, π]$ range — the arccos graph ends here.
$arctan 0$ : $tan y = 0$ at $y = 0$ . Why this step? Tangent (slope) is flat only at $0$ inside the open zone.

Answers: $arcsin 0 = 0$ , $arcsin 1 = \frac{π}{2}$ , $arcsin (- 1) = - \frac{π}{2}$ , $arccos 1 = 0$ , $arccos (- 1) = π$ , $arctan 0 = 0$ .

Verify: $sin 0 = 0, sin \frac{π}{2} = 1, sin (- \frac{π}{2}) = - 1, cos 0 = 1, cos π = - 1, tan 0 = 0$ ✅.

Cell D — out-of-range "snap back"

This is the classic trap: an inverse cannot return an angle outside its zone, even if you feed it the sine (or cosine) of one. To be exhaustive we cover an angle from quadrant II (Ex 4), one from quadrant III (Ex 4b), and a quadrant-IV multi-turn angle (Ex 4c) — so every position of the inner angle is demonstrated.

The figure is again a unit circle (horizontal axis $= cos$ , vertical axis $= sin$ ). The orange arrow points to the raw inner angle $\frac{5 π}{6}$ (outside arcsin's zone); the teal arrow points to the snapped-back answer $\frac{π}{6}$ (inside the zone). The plum dashed horizontal line at height $\frac{1}{2}$ shows that both arrows have the same $sin$ -value — that shared height is exactly why arcsin returns the lower one.

Worked example Ex 4. (Quadrant II) Compute

$arcsin (sin \frac{5 π}{6})$ . Forecast: the tempting (wrong) answer is $\frac{5 π}{6}$ . Why can't it be?

Check the inner angle against the zone. Is $\frac{5 π}{6} \in [- \frac{π}{2}, \frac{π}{2}]$ ? No — $\frac{5 π}{6} = 150° > 90°$ (quadrant II). Why this step? $arcsin (sin θ) = θ$ only when $θ$ is already inside the zone. Here it isn't, so we must reduce.
Collapse to a ratio. $sin \frac{5 π}{6} = sin (π - \frac{5 π}{6}) = sin \frac{π}{6} = \frac{1}{2}$ . Why this step? Sine of an obtuse angle equals sine of its supplement (the two equal-height dots on the plum line). We keep only the ratio.
Ask arcsin the clean question. $arcsin \frac{1}{2} = \frac{π}{6}$ , which IS in the zone. Why this step? The inverse always returns the one angle inside $[- \frac{π}{2}, \frac{π}{2}]$ giving that ratio — it "snaps back."

Answer: $arcsin (sin \frac{5 π}{6}) = \frac{π}{6}$ .

Verify: $sin \frac{5 π}{6} = \frac{1}{2}$ and $sin \frac{π}{6} = \frac{1}{2}$ produce the same ratio ✅; $\frac{π}{6}$ is in range while $\frac{5 π}{6}$ is not.

Worked example Ex 4b. (Quadrant III) Compute

$arccos (cos \frac{7 π}{6})$ — the arccos snap-back (different zone!). Forecast: naive answer $\frac{7 π}{6}$ . But $arccos$ can only return $[0, π]$ , and $\frac{7 π}{6} > π$ . Which angle in $[0, π]$ comes out?

Check the inner angle against the zone. Is $\frac{7 π}{6} \in [0, π]$ ? No — $\frac{7 π}{6} = 210° > 180°$ (quadrant III). Why this step? $arccos (cos θ) = θ$ only if $θ$ already sits in $[0, π]$ . It doesn't, so reduce.
Collapse to a ratio. $cos \frac{7 π}{6} = - cos \frac{π}{6} = - \frac{3}{2}$ . (In quadrant III cosine is negative, reference angle $\frac{7 π}{6} - π = \frac{π}{6}$ .) Why this step? We keep only the cosine value; the "which angle" information is deliberately discarded.
Ask arccos the clean question. Which $y \in [0, π]$ has $cos y = - \frac{3}{2}$ ? Using $arccos (- x) = π - arccos x$ : $π - \frac{π}{6} = \frac{5 π}{6}$ . Why this step? Negative cosine forces a second-quadrant answer inside $[0, π]$ — arccos snaps up, not down.

Answer: $arccos (cos \frac{7 π}{6}) = \frac{5 π}{6}$ .

Verify: $cos \frac{7 π}{6} = - \frac{3}{2} = cos \frac{5 π}{6}$ ✅; $\frac{5 π}{6} \in [0, π]$ while $\frac{7 π}{6}$ is not. (General rule: for $θ \in [π, 2 π]$ , $arccos (cos θ) = 2 π - θ$ ; here $2 π - \frac{7 π}{6} = \frac{5 π}{6}$ ✅.)

Worked example Ex 4c. (Quadrant IV, multi-turn) Compute

$arctan (tan \frac{9 π}{4})$ and $arcsin (sin \frac{11 π}{6})$ — spin past $2 π$ and dip into quadrant IV. Forecast: $\frac{9 π}{4}$ is more than a full turn, and $\frac{11 π}{6}$ sits in quadrant IV (below the axis). Both must snap into their zones.

Multi-turn first: peel off full circles. $tan$ repeats every $π$ , so $tan \frac{9 π}{4} = tan (\frac{9 π}{4} - 2 π) = tan \frac{π}{4} = 1$ . Why this step? Subtracting whole periods leaves the ratio unchanged; $\frac{π}{4}$ is already inside $(- \frac{π}{2}, \frac{π}{2})$ .
Ask arctan the clean question. $arctan 1 = \frac{π}{4}$ , in the zone. So $arctan (tan \frac{9 π}{4}) = \frac{π}{4}$ . Why this step? Shows multi-turn inputs are no harder — reduce to one period, then snap.
Quadrant IV next. $\frac{11 π}{6} = 330°$ is in quadrant IV, where sine is negative: $sin \frac{11 π}{6} = - sin \frac{π}{6} = - \frac{1}{2}$ (reference angle $2 π - \frac{11 π}{6} = \frac{π}{6}$ ). Why this step? Quadrant IV gives a negative ratio, so the snapped answer will be negative — the opposite sign story from Ex 4.
Ask arcsin the clean question. $arcsin (- \frac{1}{2}) = - \frac{π}{6}$ , inside $[- \frac{π}{2}, \frac{π}{2}]$ . So $arcsin (sin \frac{11 π}{6}) = - \frac{π}{6}$ . Why this step? Confirms snap-back works below the axis too, completing all four quadrants.

Answers: $arctan (tan \frac{9 π}{4}) = \frac{π}{4}$ , $arcsin (sin \frac{11 π}{6}) = - \frac{π}{6}$ .

Verify: $tan \frac{9 π}{4} = 1 = tan \frac{π}{4}$ ✅; $sin \frac{11 π}{6} = - \frac{1}{2} = sin (- \frac{π}{6})$ ✅. Both answers sit in their zones.

Cell E — nested / composite identity

Sometimes the answer isn't a number but an algebraic expression. We turn the inverse into a triangle — but a triangle only handles the acute case, so we must also treat $x < 0$ separately, where the angle is obtuse. This connects directly to Trigonometric identities.

The figure is a right triangle carrying the angle $θ = arccos x$ at the lower-left corner, drawn for the acute case $x > 0$ . Its adjacent side (teal, along the bottom) has length $x$ ; its hypotenuse (plum) has length $1$ ; its opposite side (orange, vertical) has length $1 - x^{2}$ by Pythagoras. Reading $tan = opposite / adjacent$ straight off this triangle gives the formula.

Worked example Ex 5. Simplify

$tan (arccos x)$ for all valid $x \in [- 1, 1]$ , $x \neq = 0$ . Forecast: for $x > 0$ the answer should be positive and shrink to $0$ at $x = 1$ . What sign should it take when $x < 0$ ? Predict before reading.

Name the angle. Let $θ = arccos x$ , so $cos θ = x$ with $θ \in [0, π]$ . Why this step? Naming the inverse lets us build a triangle (for $x > 0$ ) or reason by quadrant (for $x < 0$ ).
Acute case $x \in (0, 1]$ : draw the triangle. Adjacent $= x$ , hypotenuse $= 1$ , opposite $= 1 - x^{2}$ (figure). Then $tan θ = \frac{1 - x ^{2}}{x}$ , positive. Why this step? Here $θ \in [0, \frac{π}{2})$ where $tan \geq 0$ , so the $+$ root and positive value are correct.
Obtuse case $x \in [- 1, 0)$ : check the quadrant. Now $cos θ = x < 0$ forces $θ \in (\frac{π}{2}, π]$ , the second quadrant, where $tan θ < 0$ . Why this step? The triangle picture no longer applies (the angle is obtuse); we must decide the sign from the quadrant instead.
Fix the sign. The magnitude is still $\frac{1 - x ^{2}}{∣ x ∣}$ , but since $x < 0$ , $\frac{1 - x ^{2}}{x}$ is already negative — so the single expression $\frac{1 - x ^{2}}{x}$ works for both signs. Why this step? Dividing the (always non-negative) $1 - x^{2}$ by $x$ itself carries the correct sign automatically, so one formula covers the whole domain.

Answer: $tan (arccos x) = \frac{1 - x ^{2}}{x}$ for all $x \in [- 1, 1]$ , $x \neq = 0$ (positive for $x > 0$ , negative for $x < 0$ ).

Verify: at $x = \frac{1}{2}$ : $\frac{3 /2}{1/2} = 3 = tan \frac{π}{3}$ , and $arccos \frac{1}{2} = \frac{π}{3}$ ✅. At $x = - \frac{1}{2}$ : $\frac{3 /2}{- 1/2} = - 3 = tan \frac{2 π}{3}$ , and $arccos (- \frac{1}{2}) = \frac{2 π}{3}$ ✅ (negative, as forecast). At $x = 1$ : $0 = tan 0$ ✅.

Cell F — limiting behaviour

$arctan$ never reaches $\pm \frac{π}{2}$ ; it approaches them from both directions. This is the two-asymptote story, and it foreshadows Derivatives of inverse trig functions (the slope flattens to $0$ far out on either side).

The figure plots $y = arctan x$ (teal S-curve) against $x$ . Two orange dashed horizontal lines at $y = \pm \frac{π}{2}$ are the asymptotes: the curve hugs the upper one as $x \to + \infty$ and the lower one as $x \to - \infty$ , but touches neither.

Worked example Ex 6. Find

$x \to + \infty lim arctan x$ and $x \to - \infty lim arctan x$ , plus $arctan (1000)$ and $arctan (- 1000)$ numerically. Forecast: as $x$ grows hugely positive the line becomes nearly vertical upward; as $x$ grows hugely negative it becomes nearly vertical downward. Guess both ceilings.

Interpret geometrically. $arctan x$ is the angle whose slope (rise/run) is $x$ . As $x \to + \infty$ the line is nearly vertical with positive steepness. Why this step? Tangent = slope; an infinite positive slope means a vertical line, i.e. angle $\to + 90°$ .
Positive limit. The angle approaches but never equals $+ \frac{π}{2}$ (upper dashed line in the figure). Why this step? A truly vertical line has undefined slope, so the range is open at $+ \frac{π}{2}$ .
Negative limit. As $x \to - \infty$ the slope is huge but negative, so the line tips the other way: angle $\to - \frac{π}{2}$ (lower dashed line). Why this step? $arctan$ is odd, so its behaviour at $- \infty$ mirrors that at $+ \infty$ across $0$ — the graph has TWO horizontal asymptotes.
Numerical feel. $arctan (1000) \approx 1.5698$ rad (just under $\frac{π}{2} \approx 1.5708$ ); $arctan (- 1000) \approx - 1.5698$ rad (just above $- \frac{π}{2}$ ). Why this step? Concrete numbers make "approaches from both sides" tangible.

Answers: $x \to + \infty lim arctan x = \frac{π}{2}$ and $x \to - \infty lim arctan x = - \frac{π}{2}$ (neither reached); $arctan (\pm 1000) \approx \pm 1.5698$ .

Verify: $\frac{π}{2} - arctan (1000) > 0$ and $arctan (- 1000) + \frac{π}{2} > 0$ , confirming the values stay strictly inside both ceilings ✅.

Cell G — real-world word problem

Worked example Ex 7. You stand

$30 m$ from the base of a tower. Your eye-line to the top rises $40 m$ . What is the angle of elevation? Forecast: the rise ( $40$ ) beats the run ( $30$ ), so the angle should be more than $45°$ . Guess a number.

Set up the ratio. Angle of elevation $α$ satisfies $tan α = \frac{opposite}{adjacent} = \frac{40}{30} = \frac{4}{3}$ . Why this step? Elevation is a slope: vertical rise over horizontal distance — exactly a tangent.
Invert with arctan. $α = arctan (\frac{4}{3})$ . Why arctan and not arcsin? We know the two legs (opposite and adjacent), not the hypotenuse. $tan$ uses those two legs, so its inverse is the right tool.
Evaluate. $arctan (1.333 \dots) \approx 0.9273 rad$ . Convert to degrees: $0.9273 \times \frac{180}{π} \approx 53.13°$ . Why this step? A positive input gives a first-quadrant answer, comfortably in $(- \frac{π}{2}, \frac{π}{2})$ ; degrees make the answer intuitive.
Sanity-check the size. $53.13° > 45°$ , matching the forecast that a rise bigger than the run tilts past $45°$ . Why this step? Guarding against calculator-mode errors (radians vs degrees) by comparing to the $45°$ landmark.

Answer: $α = arctan \frac{4}{3} \approx 0.9273 rad \approx 53.13°$ .

Verify: $tan (53.13°) = \frac{sin}{cos} \approx \frac{0.8}{0.6} = 1.333 = \frac{4}{3}$ ✅, and $53.13° > 45°$ as forecast ✅. (This is the famous $3$ – $4$ – $5$ triangle.)

Cell H — exam-style twist

Worked example Ex 8. Solve

$2 arcsin x = \frac{π}{3}$ for $x$ , then state which values of $x$ are legal. Forecast: halving the equation gives an $arcsin$ we can undo. But watch the domain of $x$ .

Isolate the inverse. Divide by $2$ : $arcsin x = \frac{π}{6}$ . Why this step? We want $arcsin x$ alone so we can apply $sin$ to both sides.
Undo with sine. Apply $sin$ : $x = sin \frac{π}{6} = \frac{1}{2}$ . Why this step? $sin (arcsin x) = x$ is always valid for $x \in [- 1, 1]$ (this direction never leaves the zone), so applying sine is safe.
Domain check. For $arcsin x$ to exist we need $x \in [- 1, 1]$ ; and $\frac{π}{6}$ must lie in the range $[- \frac{π}{2}, \frac{π}{2}]$ — it does. So $x = \frac{1}{2}$ is the unique solution. Why this step? Inverse-trig equations can hide illegal or missing solutions; a range/domain audit catches them. See Solving trigonometric equations.

Answer: $x = \frac{1}{2}$ .

Verify: $2 arcsin \frac{1}{2} = 2 \cdot \frac{π}{6} = \frac{π}{3}$ ✅.

Connections

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