Year 12 Maths Extension 1 (2027) Trigonometric functions

Definitions of inverse trig functions (domain/range)

Q: What is arcsin (sin inverse)?

It is the inverse of sine restricted to minus pi over 2 to pi over 2. It takes a value in minus one to one and returns the angle in that interval whose sine is that value.

Q: What are the domain and range of arcsin, arccos and arctan?

Arcsin: domain minus one to one, range minus pi over 2 to pi over 2. Arccos: domain minus one to one, range zero to pi. Arctan: domain all real numbers, range minus pi over 2 to pi over 2.

Q: Why do we restrict the domain of sine to define arcsin?

Because sine repeats and fails the horizontal line test, it is not one-to-one. Restricting it to minus pi over 2 to pi over 2 makes it one-to-one so a single-valued inverse exists.

Q: How do you evaluate cos(arcsin x)?

Let theta be arcsin x, use the range to fix the sign of cos theta, then use cos theta equals the square root of one minus x squared.

Q: Why is arccos of a negative number an obtuse angle?

Because the range of arccos is zero to pi. A negative cosine corresponds to an angle between pi over 2 and pi, that is, obtuse.

Q: Is arctan defined for all real numbers?

Yes. Arctan accepts every real input and returns an angle strictly between minus pi over 2 and pi over 2.

20 practice questions 2 video lessons Theory + worked examples

Meet the inverse trigonometric functions in NSW Year 12 Mathematics Extension 1. Because sine, cosine and tangent repeat, their domains must be restricted before each can have a genuine inverse.

You will learn how the restricted domains define arcsin, arccos and arctan, state the domain and range of each, and evaluate exact values — the groundwork for graphing and differentiating inverse trig functions later in Extension 1.

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Practice questions

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Inverse Trig Domain & Range Watch
Inverse Trigonometric Functions: Domain, Range, and Graphs Watch

Theory

The inverse trigonometric functions $\sin^{- 1}$ , $\cos^{- 1}$ and $\tan^{- 1}$ recover an angle from a ratio. This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-03) defines each as the inverse of a restricted trig function and fixes its domain and range.

The inverse trigonometric functions $\sin^{- 1}$ , $\cos^{- 1}$ and $\tan^{- 1}$ (also written $\arcsin$ , $\arccos$ , $\arctan$ ) reverse sine, cosine and tangent: they recover an angle from a ratio.

Because $\sin$ , $\cos$ and $\tan$ repeat, they fail the horizontal line test and are not one-to-one. Each must first be restricted to a one-to-one piece before an inverse can exist. The chosen restriction becomes the range of the inverse.

$\sin^{- 1} x$ is the inverse of $y = \sin x$ restricted to $- \frac{π}{2} \leq x \leq \frac{π}{2}$ ; $\cos^{- 1} x$ is the inverse of $y = \cos x$ restricted to $0 \leq x \leq π$ ; and $\tan^{- 1} x$ is the inverse of $y = \tan x$ restricted to \(-\dfrac{\pi}{2}

NESA link. The Year 12 Inverse trigonometric functions focus area, outcome ME1-12-03 ("solves problems involving inverse trigonometric functions") with MAO-WM-01. The syllabus asks students to examine domain restrictions of $\sin$ , $\cos$ , $\tan$ , define each inverse, and evaluate and simplify expressions such as $\cos (\sin^{- 1} (- \frac{1}{4}))$ .

y = \sin x

meets a horizontal line repeatedly, so it is restricted to the gold piece before inverting.

Let

θ

be the inverse angle, then read the other ratios off a right triangle.

Each inverse is defined on the restriction that makes the trig function one-to-one:

\sin^{- 1} x : inverse of \sin on [- \frac{π}{2}, \frac{π}{2}], \cos^{- 1} x : on [0, π], \tan^{- 1} x : on (- \frac{π}{2}, \frac{π}{2}) .

arcsin inverse of sin on -pi/2..pi/2; arccos on 0..pi; arctan on -pi/2..pi/2

Domain and range

	Domain	Range
$y = \sin^{- 1} x$	$- 1 \leq x \leq 1$	$- \frac{π}{2} \leq y \leq \frac{π}{2}$
$y = \cos^{- 1} x$	$- 1 \leq x \leq 1$	$0 \leq y \leq π$
$y = \tan^{- 1} x$	all real $x$	\(-\dfrac{\pi}{2}

Evaluating a composite such as $\cos (\sin^{- 1} a)$ : let $θ$ be the inverse angle, use the range to fix the sign, then read the missing ratio from a right triangle or from $\sin^{2} θ + \cos^{2} θ = 1$ .

How to evaluate a composite like $\cos (\sin^{- 1} a)$

Name the angle. Let $θ = \sin^{- 1} a$ , so $\sin θ = a$ .
Fix the quadrant. Use the inverse's range to decide the sign of the other ratio (e.g. $\cos^{- 1}$ gives $θ \in [0, π]$ , so $\sin θ \geq 0$ ).
Find the missing ratio. Draw a right triangle with the known ratio, or use $\sin^{2} θ + \cos^{2} θ = 1$ .
Read off the answer in radians, keeping the sign from step 2.

Example 1 — Exact values

Evaluate

\tan^{- 1} (- 1)

and

\cos^{- 1} (- \frac{\sqrt{3}}{2})

Solution

$\tan^{- 1} (- 1)$ : the angle in $(- \frac{π}{2}, \frac{π}{2})$ with $\tan = - 1$ is $- \frac{π}{4}$ .

$\cos^{- 1} (- \frac{\sqrt{3}}{2})$ : the angle in $[0, π]$ with $\cos = - \frac{\sqrt{3}}{2}$ is $\frac{5 π}{6}$ .

So $- \frac{π}{4}$ and $\frac{5 π}{6}$ .

Example 2 — A composite value

Find the exact value of

\sin (\cos^{- 1} \frac{4}{5})

Solution

Let $θ = \cos^{- 1} \frac{4}{5}$ , so $\cos θ = \frac{4}{5}$ with $θ \in [0, π]$ , giving $\sin θ \geq 0$ .

$\sin θ$	$=$	$\sqrt{1 - {(\frac{4}{5})}^{2}}$
	$=$	$\sqrt{1 - \frac{16}{25}} = \frac{3}{5}$

sin(arccos(4/5)) = 3/5

So $\sin (\cos^{- 1} \frac{4}{5}) = \frac{3}{5}$ .

Example 3 — Handling the sign

Find the exact value of

\tan (\sin^{- 1} (- \frac{5}{13}))

Solution

Let $θ = \sin^{- 1} (- \frac{5}{13})$ , so $\sin θ = - \frac{5}{13}$ with $θ \in [- \frac{π}{2}, \frac{π}{2}]$ , giving $\cos θ \geq 0$ .

$\cos θ$	$=$	$\frac{12}{13}$
$\tan θ$	$=$	$\frac{- 5 / 13}{12 / 13} = - \frac{5}{12}$

tan(arcsin(-5/13)) = -5/12

So the value is $- \frac{5}{12}$ .

Example 4 — An algebraic form

For

- 1 \leq x \leq 1

, write

\sin (\cos^{- 1} x)

without inverse trig functions.

Solution

Let $θ = \cos^{- 1} x$ , so $\cos θ = x$ with $θ \in [0, π]$ , giving $\sin θ \geq 0$ .

\sin (\cos^{- 1} x)

=

\sin θ = \sqrt{1 - x^{2}}

sin(arccos x) = sqrt(1 - x^2)

So $\sin (\cos^{- 1} x) = \sqrt{1 - x^{2}}$ .

Common pitfalls

Forgetting the range fixes the sign.

\cos^{- 1}

outputs lie in

[0, π]

, so a negative input gives an obtuse angle, e.g.

\cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}

Sign of the output.

\sin^{- 1}

and

\tan^{- 1}

return negative values for negative inputs;

\cos^{- 1}

never does.

Degrees. Always give inverse-trig values in radians unless told otherwise.

Domain of the input.

\sin^{- 1}

and

\cos^{- 1}

only accept inputs in

[- 1, 1]

;

\tan^{- 1}

accepts all reals.

Frequently asked questions

What is arcsin (sin inverse)?

It is the inverse of $\sin x$ restricted to $[- \frac{π}{2}, \frac{π}{2}]$ . It takes a value in $[- 1, 1]$ and returns the angle in that interval whose sine is that value.

What are the domain and range of arcsin, arccos and arctan?

$\sin^{- 1}$ : domain $[- 1, 1]$ , range $[- \frac{π}{2}, \frac{π}{2}]$ . $\cos^{- 1}$ : domain $[- 1, 1]$ , range $[0, π]$ . $\tan^{- 1}$ : domain all reals, range $(- \frac{π}{2}, \frac{π}{2})$ .

Why do we restrict the domain of sine to define arcsin?

Because $\sin$ repeats and fails the horizontal line test, it is not one-to-one. Restricting it to $[- \frac{π}{2}, \frac{π}{2}]$ makes it one-to-one so a single-valued inverse exists.