Question bank — Inverse trig functions — arcsin, arccos, arctan — domain, range, graphs

False. It only holds when $x$ already lies in $[-\pi /2,\pi /2]$; outside that the output "snaps back" into the range, e.g. $\arcsin (\sin 3\pi /4)=\pi /4\ne 3\pi /4$.

True. Here we feed a valid ratio in, get an angle inside the range, and sine of that angle returns the ratio — this direction never leaves the safe zone, so it always undoes cleanly.

False. Its range is $[0,\pi ]$ by construction, so the smallest possible output is $0$ and it climbs only to $\pi$; a negative answer is impossible.

False. $\pi /2$ is a horizontal asymptote: $\arctan x\to \pi /2$ as $x\to \infty$ but never reaches it, which is exactly why the range is the open interval.

True. Sine is odd and its inverse inherits that, so $\arcsin (-x)=-\arcsin x$; geometrically its graph has half-turn symmetry about the origin.

False. Instead $\arccos (-x)=\pi -\arccos x$, so it is symmetric about the point $(0,\pi /2)$, not the origin — feeding in $-x$ reflects the answer through the middle of $[0,\pi ]$.

False. It holds for the whole domain $x\in [-1,1]$, because the cofunction argument $\pi /2-\arcsin x$ always lands inside $[0,\pi ]$, the valid arccos range.

False. That is the domain of $\arcsin$ and $\arccos$. $\tan$ spans all real numbers, so $\arctan$ accepts any real input.

False. You must first restrict sine to $[-\pi /2,\pi /2]$; reflecting the whole wavy sine would fail the vertical-line test and not be a function.

True. It is the reflection of the decreasing cosine branch on $[0,\pi ]$, so as $x$ rises from $-1$ to $1$ the angle falls from $\pi$ down to $0$.

The input $2$ is outside $[-1,1]$; no real angle has sine $2$, so $\arcsin (2)$ is undefined (a sine ratio can never exceed $1$).

$\cos (-\pi /3)=+1/2$, not $-1/2$, and $-\pi /3\notin [0,\pi ]$ anyway. The correct answer is $2\pi /3$, found in the second quadrant where cosine is negative.

"Blows up" means approaches without arriving. $\arctan$ gets arbitrarily close to $\pi /2$ but the range is open, so it never actually equals $\pi /2$.

They do not. Cosine is even on $[-\pi /2,\pi /2]$ (fails one-to-one), forcing arccos onto $[0,\pi ]$, whereas arcsin uses $[-\pi /2,\pi /2]$.

$100°\notin [-90°,90°]$, so the answer cannot be $100°$. Since $\sin 100°=\sin 80°$ and $80°$ is in range, the result is $80°$.

Only the $+$ root is valid: $\arcsin x$ lands in $[-\pi /2,\pi /2]$ where cosine is $\ge 0$, so $\cos (\arcsin x)=\sqrt{1-x^2}$ with no negative option.

$3\pi /4\notin (-\pi /2,\pi /2)$. Because $\tan$ repeats every $\pi$, $\tan (3\pi /4)=-1$ and $\arctan (-1)=-\pi /4$, so the answer is $-\pi /4$.

Because the original trig functions are not one-to-one — many angles share a ratio — so without a restriction the "inverse" would have to return several angles at once, which no function may do.

On $[-\pi /2,\pi /2]$ cosine is even ($\cos (-a)=\cos a$), so it repeats outputs and cannot be inverted; $[0,\pi ]$ is the simplest interval where cosine is strictly monotone and still covers every value in $[-1,1]$ once.

Because inverting swaps inputs and outputs: if $(a,b)$ lies on $f$ then $(b,a)$ lies on $f^{-1}$, and swapping coordinates is exactly a reflection across $y=x$.

Sine reaches its extremes $\pm 1$ at real angles $\pm \pi /2$ (so those endpoints are attainable), but $\tan$ only approaches $\pm \infty$ near $\pm \pi /2$ without ever hitting them, so those angle-endpoints are excluded.

Sine is odd, and reflecting an odd function across $y=x$ keeps it odd, so $\arcsin (-x)=-\arcsin x$; cosine is even, and its range shift to $[0,\pi ]$ turns the symmetry into $\arccos (-x)=\pi -\arccos x$ instead.

With $\theta =\arcsin x$ we have $\cos (\pi /2-\theta )=\sin \theta =x$, and the angle $\pi /2-\theta$ automatically lies in $[0,\pi ]$, so it is $\arccos x$, forcing the two to sum to $\pi /2$.

$\arcsin (1)=\pi /2$, the closed upper boundary of arcsin's range — attainable precisely because sine actually reaches $1$ there.

$\arccos (-1)=\pi$, the top of arccos's range; this is where the decreasing branch bottoms out in cosine value while topping out in angle.

$\arctan (0)=0$; zero sits comfortably in the open interval's interior, and $\tan 0=0$ makes it the natural anchor point of the whole curve.

On the closed lower edge: it is attained because $\sin (-\pi /2)=-1$, so unlike arctan's asymptotes this endpoint genuinely belongs to the range.

It approaches $-\pi /2$ from above but never reaches it, mirroring the upper asymptote — this is the open lower end of the range.

$\cos (-\pi /3)=1/2$ and $\arccos (1/2)=\pi /3$; since arccos can never return a negative, $-\pi /3$ is impossible, so the answer is $\pi /3$.

Yes: $\arcsin (0)=0$ and $\arccos (0)=\pi /2$, summing to $\pi /2$ — the identity holds right across the domain including this midpoint.