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

The inverse machine is only ever allowed to give back an angle inside its safe zone.

• arcsin: the interval from $-\frac{\pi }{2}$ up to $+\frac{\pi }{2}$, including both endpoints. Symmetric around $0$.
• arctan: the same interval $-\frac{\pi }{2}$ to $+\frac{\pi }{2}$ but open — the endpoints are asymptotes, never reached.
• arccos: sits higher up, from $0$ to $\pi$, including both endpoints — never negative. In the figure the three bars are stacked: arcsin (blue) and arctan (green, hollow endpoints) share the same left-right span centred on $0$; the arccos (yellow) bar is shifted entirely to the right of $0$, running from $0$ to $\pi$. Every problem below is really the question: "which angle inside my safe zone gives this ratio?"

Input check: $\frac{1}{2}\in [-1,1]$ ✅, so arccos is allowed to eat it. WHAT we want: the angle $y$ in arccos's safe zone $[0,\pi ]$ with $\cos y=\frac{1}{2}$. In figure s01 this is the yellow bar — the answer must land somewhere on it. WHY that zone: so the answer is unique — cosine equals $\frac{1}{2}$ at many angles, but only one sits in $[0,\pi ]$. On the unit circle, $\cos 60°=\frac{1}{2}$, and $60°=\frac{\pi }{3}$ is inside the yellow bar $[0,\pi ]$. ✅ $\arccos \left(\frac{1}{2}\right)=\frac{\pi }{3}.$

WHAT: the angle $y\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ with $\tan y=0$ — the green bar in s01 (hollow ends: the asymptotes are off-limits). $\tan y=\frac{\sin y}{\cos y}$ is zero exactly when $\sin y=0$. Inside the open safe zone, the only such angle is $y=0$ — dead centre of the green bar. $\arctan (0)=0.$

Input check: $-1\in [-1,1]$ ✅ (it's the very edge — the most negative sine allowed). WHAT: the angle $y\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ with $\sin y=-1$ — the blue bar in s01. Sine reaches its lowest value $-1$ at the bottom of the unit circle, $y=-\frac{\pi }{2}$, which is the (closed) left endpoint of the blue bar. Allowed. $\arcsin (-1)=-\frac{\pi }{2}.$

Input check: $-\frac{\sqrt{3}}{2}\approx -0.87\in [-1,1]$ ✅. WHAT: angle in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ (the blue bar of s01) with sine $-\frac{\sqrt{3}}{2}$. WHY the sign steers us: negative sine means we're in the lower half of the blue bar (angles between $-\frac{\pi }{2}$ and $0$). WHY arcsin is odd — derive it, don't assume it. We want to show $\arcsin (-x)=-\arcsin x$. Let $\theta =\arcsin x$, so $\sin \theta =x$ with $\theta \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$. Since sine is an odd function ($\sin (-\theta )=-\sin \theta$), $\sin (-\theta )=-\sin \theta =-x.$ Also, if $\theta \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ then $-\theta$ is in that same interval (the blue bar is symmetric about $0$), so $-\theta$ is a valid arcsin output. Therefore $\arcsin (-x)=-\theta =-\arcsin x$. Done — the oddness of sine plus the symmetry of the safe zone forces the inverse to be odd. Apply it with reference angle $\frac{\pi }{3}$ (since $\arcsin \frac{\sqrt{3}}{2}=\frac{\pi }{3}$): $\arcsin \left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi }{3}.$

Input check: $-\frac{\sqrt{2}}{2}\approx -0.71\in [-1,1]$ ✅. WHAT: angle in $[0,\pi ]$ (the yellow bar) with cosine $-\frac{\sqrt{2}}{2}$. WHY not $-\frac{\pi }{4}$: that's outside the yellow bar $[0,\pi ]$. Negative cosine means the angle is past $\frac{\pi }{2}$ (second-quadrant part of the zone). WHERE the identity $\arccos (-x)=\pi -\arccos x$ comes from — derive it, don't memorise it. Let $\theta =\arccos x$, so $\cos \theta =x$. Now look at the angle $\pi -\theta$. On the unit circle, rotating to $\pi -\theta$ is the mirror image of $\theta$ across the vertical axis: the point keeps the same height (same $\sin$) but its horizontal coordinate flips sign. Since the horizontal coordinate is the cosine, $\cos (\pi -\theta )=-\cos \theta =-x.$ Also, if $\theta \in [0,\pi ]$ then $\pi -\theta \in [0,\pi ]$ too, so $\pi -\theta$ is a valid arccos output. Therefore $\arccos (-x)=\pi -\theta =\pi -\arccos x$. That's the whole justification. Apply it with reference angle $\frac{\pi }{4}$ (since $\arccos \frac{\sqrt{2}}{2}=\frac{\pi }{4}$): $\arccos \left(-\frac{\sqrt{2}}{2}\right)=\pi -\frac{\pi }{4}=\frac{3\pi }{4}.$

Input check: arctan eats any real, so $-\sqrt{3}$ is fine. WHAT: angle in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ (the green bar) with $\tan =-\sqrt{3}$. WHY arctan is odd — same derivation pattern. We want $\arctan (-x)=-\arctan x$. Let $\theta =\arctan x$, so $\tan \theta =x$ with $\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$. Tangent is an odd function ($\tan (-\theta )=-\tan \theta$), so $\tan (-\theta )=-\tan \theta =-x.$ And $-\theta$ stays inside $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ (the green bar is symmetric about $0$), so it's a valid arctan output. Hence $\arctan (-x)=-\theta =-\arctan x$. Apply it with reference angle $\frac{\pi }{3}$ ($\tan \frac{\pi }{3}=\sqrt{3}$); negative tangent lands in the lower half of the green bar: $\arctan (-\sqrt{3})=-\frac{\pi }{3}.$

Naive guess: $\frac{5\pi }{6}$. Check the yellow bar: $\frac{5\pi }{6}\approx 2.62$ is inside $[0,\pi ]$ (which ends at $\pi \approx 3.14$). So here the undo works directly. (Input check on the re-inversion: $\cos \frac{5\pi }{6}=-\frac{\sqrt{3}}{2}\approx -0.87\in [-1,1]$ ✅ — a legal arccos input.) $\arccos \left(\cos \frac{5\pi }{6}\right)=\frac{5\pi }{6}.$ (Included on purpose: sometimes the naive answer is correct — always check the zone rather than assume.)

Naive guess: $\frac{5\pi }{6}$ — but $\frac{5\pi }{6}>\frac{\pi }{2}$, so it lies off the right end of the blue bar, outside arcsin's zone. The answer must snap back. Method: compute the ratio, then re-invert. $\sin \frac{5\pi }{6}=\frac{1}{2}$ (input check: $\frac{1}{2}\in [-1,1]$ ✅). Now $\arcsin \frac{1}{2}=\frac{\pi }{6}$, and $\frac{\pi }{6}$ is inside the blue bar. $\arcsin \left(\sin \frac{5\pi }{6}\right)=\frac{\pi }{6}.$ Geometrically: $\frac{5\pi }{6}$ and $\frac{\pi }{6}$ are mirror angles about $\frac{\pi }{2}$, so they share the same sine; the inverse returns the one inside its zone.

Naive guess: $\frac{3\pi }{4}$ — outside the green bar $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$. Snap back needed. $\tan \frac{3\pi }{4}=-1$. Then $\arctan (-1)=-\frac{\pi }{4}$, which lives in the zone. WHY it shifts by $\pi$: tangent repeats every $\pi$ (its period), so $\frac{3\pi }{4}$ and $\frac{3\pi }{4}-\pi =-\frac{\pi }{4}$ have the same tangent; arctan keeps the in-zone one. $\arctan \left(\tan \frac{3\pi }{4}\right)=-\frac{\pi }{4}.$

Input check: $\frac{3}{5}=0.6\in [-1,1]$ ✅, so arcsin accepts it. Set up: let $\theta =\arcsin \frac{3}{5}$, so $\sin \theta =\frac{3}{5}$ and $\theta \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$. Draw the triangle (figure s02 above): opposite $=3$, hypotenuse $=5$. By Pythagoras the adjacent side is $\sqrt{5^2-3^2}=\sqrt{16}=4$. Sign check: on $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ cosine is $\ge 0$, so we take the positive root. $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{4}{5}.$

Input check: $\frac{5}{13}\approx 0.38\in [-1,1]$ ✅, a legal arccos input. Set up: $\varphi =\arccos \frac{5}{13}$, so $\cos \varphi =\frac{5}{13}$, $\varphi \in [0,\pi ]$. Since $\cos \varphi >0$, $\varphi$ is in the first-quadrant part of the zone, where everything is positive. Triangle: adjacent $=5$, hypotenuse $=13$, opposite $=\sqrt{13^2-5^2}=\sqrt{144}=12$. $\tan \varphi =\frac{\text{opposite}}{\text{adjacent}}=\frac{12}{5}.$

Set up: $\theta =\arctan x$, so $\tan \theta =x=\frac{x}{1}$: opposite $=x$, adjacent $=1$. Hypotenuse: $\sqrt{x^2+1^2}=\sqrt{1+x^2}$ (always positive). Sign across all cases: on $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, cosine is $>0$, so hypotenuse-based ratios keep the sign of the opposite side $x$ — this makes the formula valid for negative $x$ too (both sides flip sign together). $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{x}{\sqrt{1+x^2}}.$ At $x=2$: $\frac{2}{\sqrt{1+4}}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$.

Input checks: $\frac{3}{5}\in [-1,1]$ ✅ (legal arccos input) and $\frac{5}{13}\in [-1,1]$ ✅ (legal arcsin input). Name the pieces: let $\alpha =\arccos \frac{3}{5}$ (so $\cos \alpha =\frac{3}{5}$, $\sin \alpha =\frac{4}{5}$, since $\alpha \in [0,\pi ]$ with positive cosine ⇒ first quadrant ⇒ $\sin \ge 0$). Let $\beta =\arcsin \frac{5}{13}$ (so $\sin \beta =\frac{5}{13}$, $\cos \beta =\frac{12}{13}$, positive since $\beta \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$). Why the sine-addition formula: we want $\sin$ of a sum, and $\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$ turns it into known pieces. $\sin (\alpha +\beta )=\frac{4}{5}\cdot \frac{12}{13}+\frac{3}{5}\cdot \frac{5}{13}=\frac{48}{65}+\frac{15}{65}=\frac{63}{65}.$

Where the arctan addition formula comes from. Let $\alpha =\arctan a$, $\beta =\arctan b$, so $\tan \alpha =a$, $\tan \beta =b$. The tangent-addition identity (from Trigonometric identities) says $\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{a+b}{1-ab}.$ Taking arctan of both sides gives a candidate angle $\arctan \frac{a+b}{1-ab}$ — but arctan can only answer inside $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, while the true sum $\alpha +\beta$ can live anywhere in $(-\pi ,\pi )$. So we must check which "branch" the true sum falls in. Full case split (for $a,b>0$, so $\alpha ,\beta \in (0,\frac{\pi }{2})$ and $\alpha +\beta \in (0,\pi )$):

• $ab<1$: then $1-ab>0$, the candidate is positive and $\alpha +\beta <\frac{\pi }{2}$, so $\arctan a+\arctan b=\arctan \frac{a+b}{1-ab}$ directly.
• $ab=1$: the denominator $1-ab=0$, so $\tan (\alpha +\beta )$ blows up — the sum is exactly $\frac{\pi }{2}$ (the asymptote). Then $\arctan a+\arctan b=\frac{\pi }{2}$. (E.g. $\arctan 1+\arctan 1$ has $ab=1$: check, $\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}$. ✅)
• $ab>1$: then $1-ab<0$, the candidate comes out negative, but the true sum exceeds $\frac{\pi }{2}$. Since arctan returned an angle a full $\pi$ too low, we add $\pi$: $\arctan a+\arctan b=\pi +\arctan \frac{a+b}{1-ab}$.

Combine $\arctan 2+\arctan 3$: here $ab=6>1$, so we're in the third case. Candidate: $\frac{2+3}{1-6}=\frac{5}{-5}=-1$, and $\arctan (-1)=-\frac{\pi }{4}$. Add $\pi$: $\arctan 2+\arctan 3=\pi +\left(-\frac{\pi }{4}\right)=\frac{3\pi }{4}.$ Add $\arctan 1=\frac{\pi }{4}$: $\frac{\pi }{4}+\frac{3\pi }{4}=\pi .$ Sanity: each arctan is positive and less than $\frac{\pi }{2}$; their sum $\pi$ is reasonable. ✅ A classic identity.

The undo-law holds only inside the range: $\arcsin (\sin x)=x$ iff $x\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$. Note: the inner $\sin x$ always lands in $[-1,1]$, so arcsin's input rule is never violated — the failure is never about illegal inputs, it is purely about the output being forced back into the blue-bar range. Full case analysis (why it fails outside), walking the figure s03 below:

• Central segment $x\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ (blue curve on the red dashed $y=x$ line): $x$ is already in the safe zone, so the machine returns it unchanged. Here $\arcsin (\sin x)=x$. ✅ In the figure this is the green highlighted diagonal segment.
• Right of centre $x\in \left[\frac{\pi }{2},\frac{3\pi }{2}\right]$: sine is symmetric about $x=\frac{\pi }{2}$, so $\sin x=\sin (\pi -x)$ and $\pi -x$ lands in the zone. Output $=\pi -x$ — a downward line. It equals $x$ only at the single point $x=\frac{\pi }{2}$ (where the tent peaks), not on the interior. In s03 this is the blue curve peeling below the red line and sloping down.
• Left of centre $x\in \left[-\frac{3\pi }{2},-\frac{\pi }{2}\right]$ (the mirror-image negative side): by the same symmetry, sine reflects about $x=-\frac{\pi }{2}$: $\sin x=\sin (-\pi -x)$, and $-\pi -x$ lands in the zone. Output $=-\pi -x$ — an upward line, equal to $x$ only at the single point $x=-\frac{\pi }{2}$ (the tent's trough). In s03 this is the blue curve rising above the red line on the left.
• General $x$ (the periodic pattern): everything repeats with period $2\pi$ because $\sin$ does. Shifting any of the above pieces by a multiple of $2\pi$ gives the same behaviour, producing a symmetric zig-zag "tent wave" (blue in s03). This tent touches the line $y=x$ (red dashed) at exactly one segment — the central one — and merely crosses it at the isolated peak/trough points elsewhere. The full solution set: $\boxed{\arcsin (\sin x)=x⟺x\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]}$ (The isolated touch-points at $x=\pm \frac{\pi }{2}$ are already the endpoints of this interval, so nothing extra is added.) The figure below makes the "only the central segment lies on $y=x$" claim visible, and its left–right symmetry confirms the negative side behaves exactly like the positive side.