Define and sketch the inverse trigonometric functions $\arcsin$, $\arccos$ and $\arctan$, including their domains and ranges

What this dot point is asking

NESA wants you to know how the inverse trigonometric functions are defined as inverses of suitably restricted trig functions, to state and sketch their principal-value graphs (with domains and ranges), and to evaluate or simplify expressions involving them. The crucial idea, and the one most students rush past, is that $\sin$, $\cos$ and $\tan$ are not one-to-one, so an inverse only exists once we cut each function down to a single rising or falling branch. Everything else, the ranges, the graphs, the "why $\arcsin(\sin x)$ is not always $x$" trap, follows from that single decision.

The answer

Why we have to restrict the domain first

A function has an inverse only if it is one-to-one: each output must come from exactly one input. But $\sin x$, $\cos x$ and $\tan x$ are periodic, so a single output value comes from infinitely many inputs. Geometrically, a horizontal line cuts the sine curve over and over.

To get an inverse we choose a principal branch: a largest interval on which the function is strictly increasing or strictly decreasing (hence one-to-one) and still hits every output value once. For sine, the natural choice is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

The inverse is then the reflection of that branch in the line $y = x$ (reflecting any function in $y = x$ swaps the roles of $x$ and $y$, which is exactly what taking an inverse does). Reflecting the restricted sine branch produces the $\arcsin$ curve.

Reading the reflected curve on its own gives the graph you must be able to sketch from memory: the domain is what was the range of the restricted sine ($[-1, 1]$ on the $x$-axis) and the range is what was the domain ($\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ on the $y$-axis).

The vertical tangents at the endpoints $x = \pm 1$ are the mirror image of sine's horizontal tangents at its peak and trough: reflection turns a flat tangent into an upright one. That is why $\arcsin$ rises ever more steeply as $x \to \pm 1$.

$\arcsin$ (also written $\sin^{-1}$)

Restrict $\sin x$ to $x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, where it increases from $-1$ to $1$. The inverse $\arcsin$ then maps $[-1, 1]$ onto $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$. The graph passes through $\left(-1, -\frac{\pi}{2}\right)$, $(0, 0)$ and $\left(1, \frac{\pi}{2}\right)$; it is an odd function (it has $180^\circ$ rotational symmetry about the origin), strictly increasing, with vertical tangents at the endpoints.

$\arccos$ (also written $\cos^{-1}$)

Restrict $\cos x$ to $x \in [0, \pi]$, where it decreases from $1$ to $-1$. The inverse $\arccos$ maps $[-1, 1]$ onto $[0, \pi]$.

The graph passes through $(-1, \pi)$, $\left(0, \frac{\pi}{2}\right)$ and $(1, 0)$, is strictly decreasing, and has vertical tangents at the endpoints. Unlike $\arcsin$, it is neither even nor odd: it has point symmetry about $\left(0, \frac{\pi}{2}\right)$, which is the geometric content of the identity $\arccos(-x) = \pi - \arccos x$.

$\arctan$ (also written $\tan^{-1}$)

Restrict $\tan x$ to $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, where it increases from $-\infty$ to $+\infty$. The inverse $\arctan$ maps all of $\mathbb{R}$ onto $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.

The graph passes through the origin, is odd, strictly increasing, and approaches the horizontal asymptotes $y = \pm \frac{\pi}{2}$ without ever reaching them. The asymptotes are the reflection of tangent's vertical asymptotes at $x = \pm \frac{\pi}{2}$: reflecting in $y = x$ turns a vertical asymptote into a horizontal one. Note the domain is all real numbers, so unlike $\arcsin$ and $\arccos$ there is no "out of domain" restriction to check on the argument of $\arctan$.

The identities you are expected to know

The complementary identity links $\arcsin$ and $\arccos$:

It is true because the two angles are the acute angles of the same right triangle (one has sine $x$, the other has cosine $x$), and they add to a right angle. For symmetric (negated) arguments:

For positive $x$, the reciprocal-argument identity for $\arctan$ is

Composing a trig function with its inverse

Applying a function then its inverse cancels, provided you start in the right set:

The reverse order only returns $x$ when $x$ is already inside the principal range:

For an $x$ outside that interval you must first reduce $\sin x$ using periodicity or a reflection identity, then take $\arcsin$ of the resulting in-range value. This is the single most-tested subtlety on the topic.

Composing across different functions: the right-triangle move

To simplify something like $\tan(\arccos x)$ or $\cos(\arcsin x)$, set $\theta$ equal to the inverse-trig part, draw a right triangle (or use a Pythagorean identity) for the known ratio, and read off the ratio you want, keeping the sign correct for the principal range. The worked examples below do this twice.

How exam questions ask about inverse trig functions

• "Find the exact value of $\arcsin(\ldots)$ / $\arccos(\ldots)$ / $\arctan(\ldots)$": identify the unique angle in the principal range with that ratio. State the range you are working in to justify the sign.
• "State the domain (and range) of $f(x) = \arcsin(\text{expression})$": set the inside between $-1$ and $1$ and solve; for $\arctan$ the domain is all reals. The range of a transformed inverse function shifts and scales the base range.
• "Sketch $y = \arccos x$ (or a transformation of it)": draw the base curve through its three key points, then apply shifts and stretches. Mark endpoints and asymptotes.
• "Evaluate $\arcsin(\sin \theta)$ for a given $\theta$": a principal-range trap. Reduce the inner value first if $\theta$ is outside the range.
• "Simplify $\cos(\arcsin x)$ / $\tan(\arccos x)$": the right-triangle move, watching the sign.
• "Show that $\arctan a + \arctan b = \ldots$": the $\arctan$ addition formula, with the $\pm\pi$ correction when $ab > 1$.

Edge cases worth knowing

• Endpoints are included for $\arcsin$ and $\arccos$ but the range of $\arctan$ is open. $\arcsin(\pm 1) = \pm\frac{\pi}{2}$ and $\arccos(\mp 1) = 0$ or $\pi$ are defined, but $\arctan$ never actually outputs $\pm\frac{\pi}{2}$ (those are asymptotes).
• $\arccos$ is not odd. The reflex $\arccos(-x) = \pi - \arccos x$ (not $-\arccos x$) trips up students who assume all inverse trig functions behave like $\arcsin$.
• Transformed graphs shift the range, not just the curve. For $y = 2\arctan x$ the range becomes $(-\pi, \pi)$ with asymptotes at $y = \pm\pi$; for $y = \arcsin x + \frac{\pi}{2}$ the range becomes $[0, \pi]$.