You already know the values, now see the shapes. The graphs of $\sin^{-1}x$, $\cos^{-1}x$, and $\tan^{-1}x$ each tell a visual story about domains, ranges, and asymptotes. Mastering these graphs is essential for differentiation in HSC Extension 1 and for identifying where inverse trig functions are valid.
The graph of $y = \sin^{-1}x$ is the reflection of $y = \sin x$ in which line? Over what domain is this reflection valid? Write your gut answer before reading on.
Every graph of an inverse trig function question reduces to two operations: identify the key features (domain, range, intercepts, concavity, asymptotes), and use the reflection property in $y = x$ to see how it relates to the original function.
The graph of $y = f^{-1}(x)$ is always the reflection of $y = f(x)$ in the line $y = x$. For inverse trig functions, this reflection is only of the restricted piece of the original function. $y = \sin^{-1}x$ reflects only $y = \sin x$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
The graph of $y = \sin^{-1}x$ is the reflection of $y = \sin x$ restricted to $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, reflected in the line $y = x$.
$y = \sin^{-1}x$ is increasing, passes through the origin, and is an odd function symmetric about $(0,0)$.
The graph of $y = \sin^{-1}x$ is the reflection of $y = \sin x$ restricted to $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, reflected in the line $y = x$.
Pause, copy the key features of $y=\sin^{-1}x$: domain $[-1,1]$, range $[-\pi/2,\pi/2]$, passes through $(-1,-\pi/2),(0,0),(1,\pi/2)$, and sketch the shape into your book.
Quick check: Which point does $y = \sin^{-1}x$ pass through?
We just saw that the graph of $y=\sin^{-1}x$ is the reflection of $y=\sin x$ (restricted to $[-\pi/2,\pi/2]$) in the line $y=x$; it has domain $[-1,1]$, range $[-\pi/2,\pi/2]$, and passes through $(-1,-\pi/2),(0,0),(1,\pi/2)$. That raises a question: how do the graphs of $y=\cos^{-1}x$ and $y=\tan^{-1}x$ differ, specifically, which ones have horizontal asymptotes, and where do they intersect the axes? This card answers it → $\cos^{-1}$ has no asymptote (domain $[-1,1]$); $\tan^{-1}$ has two asymptotes $y=\pm\pi/2$ and passes through the origin.
The graph of $y = \cos^{-1}x$ is the reflection of $y = \cos x$ restricted to $x \in [0, \pi]$, reflected in the line $y = x$.
$y = \cos^{-1}x$ is always decreasing. Unlike $\sin^{-1}x$, it is not an odd function, it has no negative outputs.
The graph of $y = \cos^{-1}x$ is the reflection of $y = \cos x$ restricted to $x \in [0, \pi]$, reflected in the line $y = x$.
Pause, copy the key features of $y=\cos^{-1}x$: domain $[-1,1]$, range $[0,\pi]$, passes through $(-1,\pi),(0,\pi/2),(1,0)$, and sketch the shape into your book.
Did you get this? True or false: the graph of $y = \cos^{-1}x$ is decreasing throughout its domain.
We just saw that $y=\cos^{-1}x$ has domain $[-1,1]$, range $[0,\pi]$, is always decreasing, and has no asymptotes. That raises a question: $\tan^{-1}$ has an unbounded domain unlike $\cos^{-1}$, how does this produce horizontal asymptotes, and what are the exact values they approach? This card answers it → $y=\tan^{-1}x\to\pi/2$ as $x\to+\infty$ and $\to-\pi/2$ as $x\to-\infty$.
The graph of $y = \tan^{-1}x$ is the reflection of $y = \tan x$ restricted to $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, reflected in the line $y = x$. The vertical asymptotes of $\tan x$ become horizontal asymptotes of $\tan^{-1}x$.
$y = \tan^{-1}x$ never reaches $\pm\frac{\pi}{2}$, these are horizontal asymptotes. The curve has an S-shape (inflection at the origin).
The graph of $y = \tan^{-1}x$ is the reflection of $y = \tan x$ restricted to $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, reflected in the line $y = x$. The vertical asymptotes of $\tan x$ become horizontal asymptotes of $\tan^{-1}x$.
Pause, copy the key features of $y=\tan^{-1}x$: domain $\mathbb{R}$, range $(-\pi/2,\pi/2)$, horizontal asymptotes $y=\pm\pi/2$, passes through the origin into your book.
Fill the gap: The horizontal asymptotes of $y = \tan^{-1}x$ are at $y = $ .
Worked examples · 3 in a row, reveal as you go
For the graph of $y = \cos^{-1}x$, state the domain, range, all axis intercepts, and describe the monotonicity.
Show that the graphs of $y = \cos^{-1}x$ and $y = \sin^{-1}x$ intersect at $x = \dfrac{1}{\sqrt{2}}$.
Compare the graphs of $y = \sin^{-1}x$, $y = \cos^{-1}x$, and $y = \tan^{-1}x$: which are odd functions? Which have horizontal asymptotes? Which pass through the origin?
Did you get this? True or false: $y = \cos^{-1}x$ is an odd function.
Odd one out: Which of these is NOT a feature of $y = \tan^{-1}x$?
Activities · practice with the ideas
State the domain and range of $y = \tan^{-1}x$.
Sketch $y = \sin^{-1}x$, showing all key points clearly: endpoints and the origin.
Show that $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$. (Hint: let $\sin^{-1}x = \theta$ and use co-function identities.)
What is the $y$-intercept of $y = \cos^{-1}x$? Why is this different from $y = \sin^{-1}x$?
Explain why $\tan^{-1}x$ has horizontal asymptotes at $y = \pm\frac{\pi}{2}$, but $\sin^{-1}x$ has no asymptotes at all.
Earlier you were asked: The graph of $y = \sin^{-1}x$ is the reflection of $y = \sin x$ in which line? Over what domain is this reflection valid?
The reflection is in the line $y = x$. However, it is not the entire sine curve that reflects, only the portion restricted to $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. This is the domain where $\sin x$ is strictly increasing and one-to-one. Reflecting this restricted piece in $y = x$ gives the curve $y = \sin^{-1}x$ with domain $[-1, 1]$ and range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Why do $\sin^{-1}x$ and $\cos^{-1}x$ have limited domains while $\tan^{-1}x$ has domain $\mathbb{R}$? Because the outputs of $\sin x$ and $\cos x$ are bounded in $[-1,1]$, so their inverses can only accept $x \in [-1,1]$. Tangent, however, takes all real values as outputs on its restricted domain $(-\frac{\pi}{2}, \frac{\pi}{2})$, so $\tan^{-1}x$ can accept any real input.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. State the domain and range of $y = \tan^{-1}x$. (2 marks)
Q2. Sketch $y = \sin^{-1}x$, showing all key points: domain endpoints, $y$-intercept, and any relevant values. (3 marks)
Q3. Prove that $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ for $x \in [-1,1]$. (2 marks)
Activity answers: 1. Domain $\mathbb{R}$, range $(-\frac{\pi}{2}, \frac{\pi}{2})$ 2. Mark $(-1,-\frac{\pi}{2})$, $(0,0)$, $(1,\frac{\pi}{2})$; curve increasing, odd 3. Proof below 4. $\cos^{-1}(0) = \frac{\pi}{2}$ because $\cos\frac{\pi}{2} = 0$; differs from $\sin^{-1}(0)=0$ since $\sin^{-1}$ uses range $[-\frac{\pi}{2},\frac{\pi}{2}]$ 5. $\tan x$ had vertical asymptotes at $x = \pm\frac{\pi}{2}$ (boundaries of the open restricted domain); reflecting in $y=x$ turns these into horizontal asymptotes of $\tan^{-1}x$. $\sin x$ had endpoints (closed interval), not asymptotes.
Q1 (2 marks): Domain: $(-\infty,\infty)$ or $\mathbb{R}$ [1]. Range: $(-\frac{\pi}{2}, \frac{\pi}{2})$, open interval [1].
Q2 (3 marks): Correct domain $[-1,1]$ and range $[-\frac{\pi}{2},\frac{\pi}{2}]$ labelled [1]. Three key points $(-1,-\frac{\pi}{2})$, $(0,0)$, $(1,\frac{\pi}{2})$ marked [1]. Curve drawn increasing, with correct concavity (concave down for $x>0$, concave up for $x<0$) [1].
Q3 (2 marks): Let $\sin^{-1}x = \theta$, so $\sin\theta = x$ with $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ [0.5]. Then $\cos(\frac{\pi}{2}-\theta) = \sin\theta = x$, and since $\frac{\pi}{2}-\theta \in [0,\pi]$ (which is the range of $\cos^{-1}$), we have $\cos^{-1}(x) = \frac{\pi}{2}-\theta$ [1]. Therefore $\sin^{-1}x + \cos^{-1}x = \theta + \frac{\pi}{2} - \theta = \frac{\pi}{2}$ [0.5].
Key takeaways before you fight:
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