What is arcsin x plus arccos x?

It equals pi over 2 for every x in minus one to one.

What is arcsin x plus arccos x?

It equals pi over 2 for every x in minus one to one.

Year 12 Maths Extension 1 (2027) Trigonometric functions

Properties, composite functions & transformations

Q: Does sin(arcsin x) = x always?

On the domain minus one to one, yes. The same holds for cos(arccos x) on minus one to one and tan(arctan x) for all real numbers.

Q: When does arcsin(sin theta) = theta?

Only when theta is in minus pi over 2 to pi over 2. For other theta, reduce it to the angle in that range with the same sine first.

Q: What are the inverse trig properties?

Arcsin of minus x equals minus arcsin x; arctan of minus x equals minus arctan x; arccos of minus x equals pi minus arccos x; and arcsin x plus arccos x equals pi over 2.

Q: How do you simplify arccos(cos theta) outside 0 to pi?

Find the angle in zero to pi with the same cosine as theta; that angle is the value of arccos(cos theta).

Q: How do transformations affect inverse trig graphs?

A factor inside f of b x shrinks the domain; a factor outside a f of x stretches the range; a minus sign reflects; and plus c or x minus h shift the graph.

20 practice questions 1 video lesson Theory + worked examples

Go further with inverse trigonometric properties and transformations in NSW Year 12 Mathematics Extension 1. Useful identities, such as the sum of arcsin and arccos, simplify expressions and support proofs.

You will learn to apply inverse trig identities, analyse composite functions using domain and range, and graph translations, reflections and dilations of inverse trig curves — extending your command of these functions in the Extension 1 course.

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Compositions of Inverse Trig Functions Watch

Theory

Because inverse trig functions undo the trig functions, composites simplify — within the right domain and range. This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-03) covers the composite identities, the four properties, and transformations of $\sin^{- 1}$ , $\cos^{- 1}$ and $\tan^{- 1}$ .

Because inverse trig functions undo the trig functions, composites simplify — but only within the correct domain and range.

Applying a function after its inverse always returns the input: $\sin (\sin^{- 1} x) = x$ , $\cos (\cos^{- 1} x) = x$ , $\tan (\tan^{- 1} x) = x$ . The reverse, $\sin^{- 1} (\sin θ)$ , only returns $θ$ when $θ$ already lies in the inverse's range; otherwise you must first reduce $θ$ into that range.

Four symmetry properties help simplify, solve and prove: $\sin^{- 1} (- x) = - \sin^{- 1} x$ , $\tan^{- 1} (- x) = - \tan^{- 1} x$ , $\cos^{- 1} (- x) = π - \cos^{- 1} x$ , and $\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$ .

NESA link. Outcome ME1-12-03 (with MAO-WM-01). The syllabus asks students to graph the composites $y = \sin (\sin^{- 1} x)$ and $y = \sin^{- 1} (\sin x)$ (and the $\cos$ , $\tan$ versions), establish where each equals $x$ , prove the four properties, and apply transformations — e.g. graph $y = - 2 \cos^{- 1} (3 x)$ (Example 4).

y = \sin^{- 1} (\sin x)

equals

x

only on

[- \frac{π}{2}, \frac{π}{2}]

, then folds back.

At every

x

\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}

Function after inverse (always true on the domain):

\sin (\sin^{- 1} x) = x, \cos (\cos^{- 1} x) = x, \tan (\tan^{- 1} x) = x .

sin(arcsin x)=x, cos(arccos x)=x, tan(arctan x)=x

Inverse after function (only on the inverse's range):

\sin^{- 1} (\sin θ) = θ on [- \frac{π}{2}, \frac{π}{2}], \cos^{- 1} (\cos θ) = θ on [0, π], \tan^{- 1} (\tan θ) = θ on (- \frac{π}{2}, \frac{π}{2}) .

arcsin(sin t)=t only for t in -pi/2..pi/2; similarly arccos on 0..pi; arctan on -pi/2..pi/2

The four properties

Property	Statement
Odd (sin)	$\sin^{- 1} (- x) = - \sin^{- 1} x$
Odd (tan)	$\tan^{- 1} (- x) = - \tan^{- 1} x$
Reflection (cos)	$\cos^{- 1} (- x) = π - \cos^{- 1} x$
Complementary	$\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$

Transformations. Inverse trig graphs transform like any function: $y = a f (x)$ stretches the range by $a$ ; $y = f (b x)$ shrinks the domain (so $\sin^{- 1} (b x)$ needs $- 1 \leq b x \leq 1$ ); $y = - f (x)$ reflects in the $x$ -axis; $y = f (x) + c$ and $y = f (x - h)$ shift.

How to simplify $\sin^{- 1} (\sin θ)$ when $θ$ is outside the range

Check the range. If $θ \in [- \frac{π}{2}, \frac{π}{2}]$ the answer is just $θ$ .
Reduce $θ$ . Use an identity (e.g. $\sin θ = \sin (π - θ)$ ) to find the angle in the range with the same sine (or cosine/tangent).
Read off the value once the reduced angle lies in the inverse's range.

How to apply a transformation (e.g. $y = a f (b x)$ )

Domain. Solve the inside constraint (for $\sin^{- 1}$ / $\cos^{- 1}$ , $- 1 \leq b x \leq 1$ ).
Range. Start from the base range and apply the vertical factor $a$ (and any reflection/shift).
Describe the dilations, reflections and translations, and find intercepts or asymptotes.

Example 1 — Angle outside the range

Find the exact value of

\sin^{- 1} (\sin \frac{4 π}{5})

Solution

$\frac{4 π}{5}$ is outside $[- \frac{π}{2}, \frac{π}{2}]$ , so reduce it.

\sin \frac{4 π}{5}

=

\sin (π - \frac{4 π}{5}) = \sin \frac{π}{5}

arcsin(sin(4pi/5)) = pi/5

Since $\frac{π}{5} \in [- \frac{π}{2}, \frac{π}{2}]$ : $\sin^{- 1} (\sin \frac{4 π}{5}) = \frac{π}{5}$ .

Example 2 — Reduce to the range

Find the exact value of

\cos^{- 1} (\cos \frac{5 π}{4})

Solution

Find the angle in $[0, π]$ with the same cosine.

$\cos \frac{5 π}{4}$	$=$	$- \frac{\sqrt{2}}{2}$
$\cos^{- 1} (- \frac{\sqrt{2}}{2})$	$=$	$\frac{3 π}{4}$

arccos(cos(5pi/4)) = 3pi/4

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

Example 3 — Using a property

Use

\cos^{- 1} (- x) = π - \cos^{- 1} x

to evaluate

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

Solution

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

arccos(-1/2) = pi - pi/3 = 2pi/3

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

Example 4 — NESA: a transformation

Describe

y = - 2 \cos^{- 1} (3 x)

as a transformation of

y = \cos^{- 1} x

, and state its domain and range. (A NESA syllabus example.)

Solution

Domain: need $- 1 \leq 3 x \leq 1$ , so $- \frac{1}{3} \leq x \leq \frac{1}{3}$ .

Range: $\cos^{- 1}$ gives $[0, π]$ ; $\times (- 2)$ gives $[- 2 π, 0]$ .

y = \cos^{- 1} x

\to

y = - 2 \cos^{- 1} (3 x)

y = -2 arccos(3x): domain -1/3..1/3, range -2pi..0

Horizontal dilation factor $\frac{1}{3}$ , vertical dilation factor $2$ , then reflection in the $x$ -axis. Domain $[- \frac{1}{3}, \frac{1}{3}]$ , range $[- 2 π, 0]$ .

Common pitfalls

Cancelling too fast.

\sin^{- 1} (\sin θ) = θ

only when

θ \in [- \frac{π}{2}, \frac{π}{2}]

; otherwise reduce

θ

into that range first.

The wrong reflection.

\cos^{- 1} (- x) = π - \cos^{- 1} x

, not

- \cos^{- 1} x

Domain of $f (b x)$ . For

y = \sin^{- 1} (b x)

\cos^{- 1} (b x)

the domain shrinks to

- \frac{1}{| b |} \leq x \leq \frac{1}{| b |}

Order of transformations. Apply the horizontal factor to the domain and the vertical factor (and reflection) to the range separately.

Frequently asked questions

Does sin(arcsin x) = x always?

On the domain $[- 1, 1]$ , yes — $\sin (\sin^{- 1} x) = x$ . The same holds for $\cos (\cos^{- 1} x)$ on $[- 1, 1]$ and $\tan (\tan^{- 1} x)$ for all reals.

When does arcsin(sin theta) = theta?

Only when $θ \in [- \frac{π}{2}, \frac{π}{2}]$ . For other $θ$ , reduce it to the angle in that range with the same sine first.

What are the inverse trig properties?

$\sin^{- 1} (- x) = - \sin^{- 1} x$ , $\tan^{- 1} (- x) = - \tan^{- 1} x$ , $\cos^{- 1} (- x) = π - \cos^{- 1} x$ , and $\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$ .

How do you simplify arccos(cos theta) outside 0 to pi?

Find the angle in $[0, π]$ with the same cosine as $θ$ ; that angle is the value of $\cos^{- 1} (\cos θ)$ .

How do transformations affect inverse trig graphs?

A factor inside, $f (b x)$ , shrinks the domain; a factor outside, $a f (x)$ , stretches the range; a minus sign reflects; and $+ c$ or $x - h$ shift the graph.