Inverse trigonometric functions
Master the inverse trigonometric functions — arcsine, arccosine and arctangent — for Year 12 Specialist Mathematics in Queensland (QCAA). Each one undoes a restricted sine, cosine or tangent and returns a single principal value.
You will learn the domain and range of each function, sketch and read their graphs (including the asymptotes of arctangent), evaluate exact values, and work out compositions such as \(\cos(\arcsin x)\) — the groundwork for differentiating and integrating inverse trig functions in the integration techniques topic.
Theory
The inverse trigonometric functions — arcsine, arccosine and arctangent — undo the restricted sine, cosine and tangent in Year 12 Specialist Mathematics (QCAA, Queensland). Each returns the single principal value whose sine, cosine or tangent is \(x\). This page sets out their domains, ranges and graphs, and evaluates exact values and compositions.
Sine, cosine and tangent are many-to-one, so they are not invertible until each is restricted to an interval on which it is one-to-one. The inverse function then returns the single angle — the principal value — in that interval.
Arcsine. \(y=\arcsin x\) means \(\sin y=x\) with \(-\dfrac{\pi}{2}\le y\le\dfrac{\pi}{2}\). Its domain is \(-1\le x\le 1\) and its range is \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\). The alternative notation is \(\sin^{-1}x\).
Arccosine. \(y=\arccos x\) means \(\cos y=x\) with \(0\le y\le\pi\). Its domain is \(-1\le x\le 1\) and its range is \([0,\pi]\). A negative input gives a second-quadrant angle, never a negative one.
Arctangent. \(y=\arctan x\) means \(\tan y=x\) with \(-\dfrac{\pi}{2}
Because they are inverses, \(\sin(\arcsin x)=x\) for every \(x\) in \([-1,1]\); but \(\arcsin(\sin\theta)=\theta\) only when \(\theta\) already lies in the range. A transformation such as \(y=\arcsin\!\left(\dfrac{x}{2}\right)\) or \(y=a\arctan x\) shifts the domain or stretches the range, and the new domain and range can be read straight off the graph.
Each inverse is defined by the equation it solves, together with the principal-value range:
The domains and ranges (the principal values) are:
| Function | Domain | Range |
|---|---|---|
| \(\arcsin x\) | \([-1,1]\) | \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\) |
| \(\arccos x\) | \([-1,1]\) | \([0,\pi]\) |
| \(\arctan x\) | \(\mathbb{R}\) | \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\) |
How to evaluate an inverse trigonometric function
- Name the angle. Set \(y\) equal to the inverse expression, e.g. \(y=\arccos x\), and write down the range \(y\) must lie in.
- Apply the trig function. Take \(\sin\), \(\cos\) or \(\tan\) of both sides to get an ordinary equation such as \(\cos y=x\).
- Find the special angle in the correct range whose sine, cosine or tangent is \(x\) — using a known exact value or a right-triangle triad for a composition.
- Check the interval. Confirm the answer lies in the range; for \(\arccos\) of a negative number it is a second-quadrant angle, and for domains solve the inner inequality \(-1\le(\text{argument})\le 1\).
Ask for the angle in \([0,\pi]\) whose cosine is \(\dfrac{1}{2}\):
| \(y\) | \(=\) | \(\arccos\dfrac{1}{2}\) |
| \(\cos y\) | \(=\) | \(\dfrac{1}{2},\quad 0\le y\le\pi\) |
| \(\cos\dfrac{\pi}{3}\) | \(=\) | \(\dfrac{1}{2}\quad(\text{known value})\) |
| \(y\) | \(=\) | \(\dfrac{\pi}{3}\) |
\(\arccos\!\left(\dfrac{1}{2}\right)=\dfrac{\pi}{3}\).
Tangent is odd, so the angle lies in \(\left(-\dfrac{\pi}{2},0\right)\):
| \(y\) | \(=\) | \(\arctan(-\sqrt{3})\) |
| \(\tan y\) | \(=\) | \(-\sqrt{3},\quad -\dfrac{\pi}{2} |
| \(\tan\dfrac{\pi}{3}\) | \(=\) | \(\sqrt{3}\) |
| \(y\) | \(=\) | \(-\dfrac{\pi}{3}\) |
\(\arctan(-\sqrt{3})=-\dfrac{\pi}{3}\).
Name the inner angle; on the arcsine range cosine is positive, so use \(\sin^2\theta+\cos^2\theta=1\):
| \(\theta\) | \(=\) | \(\arcsin\dfrac{4}{5}\) |
| \(\sin\theta\) | \(=\) | \(\dfrac{4}{5},\quad -\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}\) |
| \(\cos\theta\) | \(=\) | \(\sqrt{1-\left(\dfrac{4}{5}\right)^2}\) |
| \(=\) | \(\sqrt{\dfrac{25-16}{25}}\) | |
| \(=\) | \(\dfrac{3}{5}\) |
\(\cos\!\left(\arcsin\dfrac{4}{5}\right)=\dfrac{3}{5}\).
Arctangent accepts every real input, so the domain is unchanged:
| \(\arctan x\) | \(\text{ is defined for all }\) | \(x\in\mathbb{R}\) |
| \(\text{domain}\) | \(=\) | \(\text{all real numbers}\) |
The factor \(3\) stretches the arctan range \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\) by \(3\):
| \(-\dfrac{\pi}{2}<\arctan x\) | \(<\) | \(\dfrac{\pi}{2}\) |
| \(-\dfrac{3\pi}{2}<3\arctan x\) | \(<\) | \(\dfrac{3\pi}{2}\) |
| \(\text{asymptotes: } y\) | \(=\) | \(\pm\dfrac{3\pi}{2}\) |
Domain all real \(x\); range \(-\dfrac{3\pi}{2}
Common pitfalls
Frequently asked questions
What are the domains and ranges of arcsin, arccos and arctan?
\(\arcsin x\) has domain \([-1,1]\) and range \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\); \(\arccos x\) has domain \([-1,1]\) and range \([0,\pi]\); \(\arctan x\) has domain all real numbers and range \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\).
Why is arccos of a negative number not negative?
Because the range of \(\arccos\) is \([0,\pi]\), which contains no negative angles. A negative input gives a second-quadrant angle, for example \(\arccos\!\left(-\dfrac{\sqrt{3}}{2}\right)=\dfrac{5\pi}{6}\).
What is the difference between \(\sin^{-1}x\) and \(\dfrac{1}{\sin x}\)?
\(\sin^{-1}x\) is another name for \(\arcsin x\), the inverse function that returns an angle. The reciprocal \(\dfrac{1}{\sin x}=(\sin x)^{-1}=\operatorname{cosec} x\) is a completely different thing.
Does the graph of \(y=\arctan x\) have asymptotes?
Yes. Because its range is the open interval \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\), the graph has horizontal asymptotes \(y=\dfrac{\pi}{2}\) and \(y=-\dfrac{\pi}{2}\), which it approaches but never reaches. \(\arcsin\) and \(\arccos\) have no asymptotes.
How do you evaluate something like \(\cos(\arcsin x)\)?
Let \(\theta=\arcsin x\) so \(\sin\theta=x\); then use \(\cos\theta=\sqrt{1-x^2}\) (positive on the arcsine range). A Pythagorean triad such as 3-4-5 or 5-12-13 often gives a neat fraction.
How do you find the domain of \(y=\arcsin(\tfrac{x}{2})\) or a similar transformed function?
The argument of \(\arcsin\) (or \(\arccos\)) must lie in \([-1,1]\), so solve \(-1\le\dfrac{x}{2}\le 1\) to get \(-2\le x\le 2\). A vertical factor changes the range instead, not the domain.