Inverse Trigonometric Functions
Your calculator has a sin⁻¹ button — but what exactly does it compute? Why does it only return one answer when sine takes infinitely many values? In this lesson you'll discover why domain restriction is the key idea, and you'll master the exact values and identities that appear in every HSC Extension 1 exam.
Why can't we define $\sin^{-1}(x)$ as simply "the angle whose sine is $x$" without any restriction? What goes wrong? Write your gut answer before reading on.
Every inverse trig question in Extension 1 comes down to two operations: identify the restricted range to pick the correct quadrant, and apply the composite identity when functions are nested inside one another.
The restriction $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ for $\sin^{-1}$ and $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$ for $\tan^{-1}$ forces the function to be one-to-one, making the inverse well-defined. For $\cos^{-1}$ the range is $[0, \pi]$ — a different quadrant choice, but the same principle.
Key facts
- $\sin^{-1}x$: domain $[-1,1]$, range $[-\frac{\pi}{2}, \frac{\pi}{2}]$
- $\cos^{-1}x$: domain $[-1,1]$, range $[0, \pi]$
- $\tan^{-1}x$: domain $\mathbb{R}$, range $(-\frac{\pi}{2}, \frac{\pi}{2})$
Concepts
- Why domain restriction is necessary for trig functions to have inverses
- How the restricted ranges determine which quadrant the output lies in
- Why $\sin^{-1}(\sin x) = x$ only holds inside the restricted range
Skills
- Evaluate inverse trig functions using exact values
- Simplify composite expressions like $\sin^{-1}(\sin\frac{5\pi}{6})$
- Apply the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$
Because $\sin$, $\cos$, and $\tan$ are periodic, they are not one-to-one over their natural domains. To define an inverse, we must restrict the domain to an interval on which each function is strictly monotonic and passes the horizontal line test.
Key composite identities:
- $\sin(\sin^{-1}x) = x$ for $x \in [-1, 1]$ — always true
- $\sin^{-1}(\sin x) = x$ only when $x \in \bigl[-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr]$
- $\cos(\cos^{-1}x) = x$ for $x \in [-1, 1]$ — always true
- $\cos^{-1}(\cos x) = x$ only when $x \in [0, \pi]$
- $\tan(\tan^{-1}x) = x$ for all $x \in \mathbb{R}$ — always true
- $\tan^{-1}(\tan x) = x$ only when $x \in \bigl(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\bigr)$
Complement identity: $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ for all $x \in [-1,1]$
Because $\sin$, $\cos$, and $\tan$ are periodic, they are not one-to-one over their natural domains. To define an inverse, we must restrict the domain to an interval on which each function is strictly monotonic and passes the horizontal line test.
Pause — copy all three inverse-trig definitions with their restricted domains and ranges in a three-row table into your book.
Quick check: What is the range of $y = \cos^{-1}x$?
We just saw the three inverse-trig functions defined on restricted domains: $\sin^{-1}:[-1,1]\to[-\pi/2,\pi/2]$; $\cos^{-1}:[-1,1]\to[0,\pi]$; $\tan^{-1}:\mathbb{R}\to(-\pi/2,\pi/2)$. That raises a question: what are the three most common mistakes students make with these functions on HSC exams — and which specific domain or range restriction causes each one? This card answers it → (1) $\sin^{-1}(\sin\theta)\neq\theta$ outside $[-\pi/2,\pi/2]$; (2) $\cos^{-1}$ never gives a negative output; (3) $\tan^{-1}$ outputs are always strictly between $-\pi/2$ and $\pi/2$.
These are the exact mistakes that cost students marks in HSC Extension 1 exams. Read each one carefully.
These are the exact mistakes that cost students marks in HSC Extension 1 exams. Read each one carefully.
Pause — copy the three exam misconceptions with a concrete example showing the error and the correct answer for each into your book.
Did you get this? True or false: $\sin^{-1}(\sin\frac{5\pi}{6}) = \frac{5\pi}{6}$.
Worked examples · 3 in a row, reveal as you go
Find the exact values of: (a) $\sin^{-1}\!\left(\tfrac{1}{2}\right)$, (b) $\tan^{-1}(-1)$, (c) $\cos^{-1}\!\left(-\tfrac{\sqrt{3}}{2}\right)$.
Simplify $\sin^{-1}\!\left(\sin\dfrac{5\pi}{6}\right)$.
If $\sin^{-1}\!\left(\tfrac{3}{5}\right) = \alpha$, find $\cos^{-1}\!\left(\tfrac{3}{5}\right)$ without a calculator.
Fill the gap: $\sin^{-1}\!\left(\sin\dfrac{5\pi}{6}\right) = $ .
Activities · practice with the ideas
Find the exact value of $\cos^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)$.
Evaluate $\sin^{-1}\!\left(-\dfrac{\sqrt{2}}{2}\right)$.
Simplify $\cos^{-1}\!\left(\cos\dfrac{4\pi}{3}\right)$. Show your working.
Given $\cos^{-1}\!\left(\dfrac{5}{13}\right) = \beta$, express $\sin^{-1}\!\left(\dfrac{5}{13}\right)$ in terms of $\beta$.
Explain why $\tan^{-1}x$ has domain $\mathbb{R}$ while $\sin^{-1}x$ has domain $[-1,1]$.
Did you get this? True or false: $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ for all $x \in [-1,1]$.
Odd one out: Which of these is NOT a valid output of $\sin^{-1}x$?
Earlier you were asked: Why can't we define $\sin^{-1}(x)$ as "the angle whose sine is $x$" without restriction?
The issue is that sine is many-to-one: for $x = \frac{1}{2}$, both $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ (and infinitely more) have $\sin\theta = \frac{1}{2}$. A function must assign exactly one output to each input — so without restricting to $[-\frac{\pi}{2}, \frac{\pi}{2}]$, "$\sin^{-1}(x)$" would not be a function at all. The range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is chosen because sine is strictly increasing there and it contains the "principal" angles.
Why do $\sin^{-1}x$ and $\tan^{-1}x$ include negative values in their range, but $\cos^{-1}x$ does not? Because the chosen restricted domain for cosine is $[0, \pi]$ — all non-negative angles — ensuring cosine is strictly decreasing. Sine and tangent are restricted symmetrically around 0, so they produce negative outputs for negative inputs.
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. Find the exact value of $\cos^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)$. (1 mark)
Q2. Evaluate $\sin^{-1}\!\left(-\dfrac{\sqrt{2}}{2}\right)$. (1 mark)
Q3. Simplify $\sin^{-1}\!\left(\sin\dfrac{5\pi}{6}\right)$. Show full working. (2 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $\cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{6}$ 2. $\sin^{-1}(-\frac{\sqrt{2}}{2}) = -\frac{\pi}{4}$ 3. $\cos\frac{4\pi}{3} = -\frac{1}{2}$, so $\cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}$ 4. $\sin^{-1}(\frac{5}{13}) = \frac{\pi}{2} - \beta$ (complement identity) 5. The range of $\sin$ is $[-1,1]$, so $\sin^{-1}$ can only accept inputs in $[-1,1]$; the range of $\tan$ is $\mathbb{R}$, so $\tan^{-1}$ accepts all real $x$.
Q1 (1 mark): Need angle in $[0,\pi]$ with cosine $\frac{\sqrt{3}}{2}$. $\cos^{-1}\!\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$ ✓ [1 mark]
Q2 (1 mark): Need angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ with sine $-\frac{\sqrt{2}}{2}$. The reference angle is $\frac{\pi}{4}$, and since the input is negative, output is negative: $\sin^{-1}\!\left(-\frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}$ [1 mark]
Q3 (2 marks): $\frac{5\pi}{6} \notin [-\frac{\pi}{2}, \frac{\pi}{2}]$ so we cannot apply identity directly [1 mark for recognising this]. $\sin\frac{5\pi}{6} = \sin(\pi - \frac{\pi}{6}) = \sin\frac{\pi}{6} = \frac{1}{2}$. Therefore $\sin^{-1}(\sin\frac{5\pi}{6}) = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ [1 mark]
Key takeaways before you fight:
- Inverse trig functions require restricted domains to be well-defined as functions.
- $\sin^{-1}x$: range $\bigl[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\bigr]$ — outputs in Q4 and Q1
- $\cos^{-1}x$: range $[0, \pi]$ — outputs in Q1 and Q2
- $\tan^{-1}x$: range $\bigl(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\bigr)$ — outputs in Q4 and Q1 (open)
- $\sin^{-1}(\sin x) = x$ only inside $\bigl[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\bigr]$ — adjust otherwise
- $\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2}$ for all $x \in [-1,1]$
Next lesson: Graphs of Inverse Trig Functions.
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