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Inverse trigonometric graphs

Transform arcsin, arccos and arctan graphs with exact domains, ranges and endpoints. Find inverses of restricted trig functions and explore reflected areas.

Before you startInverse trig ranges, graph transformations and inverse functions

01 / Transform the graph and its domain

Move the important points before sketching the curve.

An inverse-trig sketch needs more than the right shape. Mark its domain, range, intercepts and included endpoints or asymptotes. Then apply the same coordinate mapping to every key point.

If (u, v) lies on y = F(x), then ((u − c)/b, av + d) lies on y = aF(bx + c) + d, with b ≠ 0.

The model holds the axes fixed while you move from the base graph to the transformed graph. Compare the interval readout as well as the selected point.

Transform points and intervalsExplore

y = arcsin(2x − 1)

Map (u, v) to ((u + 1)/2, v).

Base point (0, 0) becomes (0.500, 0.000).

Domain: [0, 1]. Range: [−π/2, π/2].

Closed dots mark included endpoints. Dashed horizontal lines are unattained arctan asymptotes. The green point is chosen by you; nothing advances on a timer. Point coordinates are rounded to 3 places.

GraphStepBase input

02 / Find the real input interval

The inner expression must stay inside the inverse function’s domain.

Find the domain of y = arcsin(2x − 1)Worked example

Require −1 ≤ 2x − 1 ≤ 1

These are the allowed inputs to arcsin.

Add 1 and divide by 2

0 ≤ x ≤ 1.

The endpoints are included

The inner inputs −1 and 1 both have real inverse-sine values.

Find the domain of y = arccos(1 − 3x)Worked example

Require −1 ≤ 1 − 3x ≤ 1

Solve both inequalities.

0 ≤ x ≤ 2/3

Dividing by a negative coefficient reverses inequality signs.

Endpoint outputs are y = 0 and y = π

At x = 0 the input is 1; at x = 2/3 it is −1.

Arctan accepts every real inner input. An affine expression bx + c therefore does not restrict its real domain, provided b and c are real constants.

03 / Transform the output interval

A negative multiplier reverses the endpoints.

Find the range of y = 2arcsin x + π/2Worked example

Begin with −π/2 ≤ arcsin x ≤ π/2

The base endpoints are closed.

Multiply by 2, then add π/2

−π/2 ≤ y ≤ 3π/2.

Find the range of y = −2arctan(x − 1)Worked example

Begin with −π/2 < arctan(x − 1) < π/2

The endpoints are open.

Multiply by −2 and reorder

−π < y < π.

The horizontal asymptotes are y = ±π

They are never attained by a finite input.

04 / Separate input changes from output changes

Solve the inner expression for x rather than guessing a shift.

Map y = arcsin u to y = arcsin(2x − 1)Worked example

u = 2x − 1, so x = (u + 1)/2

The y-coordinate stays the same.

The base endpoints (−1, −π/2) and (1, π/2)

Become (0, −π/2) and (1, π/2).

The base origin (0, 0)

Becomes (1/2, 0).

Map y = arccos u to y = π − arccos(x/2)Worked example

u = x/2, so x = 2u

Double every horizontal coordinate.

y = π − v

Reflect the output and shift it up by π.

Endpoints (−1, π), (1, 0) become (−2, 0), (2, π)

The transformed curve is increasing on [−2, 2].

05 / Solve for intercepts exactly

A graph can cross an axis away from its centre point.

Find both intercepts of y = 2arcsin x + π/2Worked example

At x = 0, y = π/2

The y-intercept is (0, π/2).

For y = 0, arcsin x = −π/4

This lies in the principal range.

x = sin(−π/4) = −√2/2

The x-intercept is (−√2/2, 0), inside the domain.

For y = arctan(2x) + π/4, setting y = 0 gives x = −1/2. Its horizontal asymptotes are y = −π/4 and y = 3π/4. The vertical shift changes the asymptotes as well as the intercepts.

06 / Invert a restricted transformed sine

Carry the original restriction through the algebra.

Find the inverse of f(t) = 3sin(2t) − 1 for −π/4 ≤ t ≤ π/4Worked example

Let y = 3sin(2t) − 1

Then (y + 1)/3 = sin(2t).

The restriction gives −π/2 ≤ 2t ≤ π/2

So arcsin returns exactly the inner angle 2t.

t = (1/2)arcsin((y + 1)/3)

Replace y by x to write f⁻¹(x).

f⁻¹(x) = (1/2)arcsin((x + 1)/3)

Its domain is [−4, 2] and range [−π/4, π/4].

The inverse domain is the original range. Without the stated restriction, the periodic original would not have a single-valued inverse.

07 / Invert a shifted cosine branch

An interval need not start at zero to be one-to-one.

Find the inverse of g(t) = 2cos(t − π/3) + 1 for π/3 ≤ t ≤ 4π/3Worked example

The inner angle t − π/3 ranges from 0 to π

Cosine is one-to-one on that branch.

(y − 1)/2 = cos(t − π/3)

Apply arccos on its principal branch.

g⁻¹(x) = π/3 + arccos((x − 1)/2)

The inverse domain is [−1, 3].

The inverse range is [π/3, 4π/3]

This is the original restricted domain.

Invert h(t) = 4tan(3t) − 2 for −π/6 < t < π/6Worked example

3t lies in (−π/2, π/2)

Tangent is one-to-one and reaches every real value.

h⁻¹(x) = (1/3)arctan((x + 2)/4)

Its domain is all real numbers and range (−π/6, π/6).

08 / An optional inverse for secant

State the branch convention explicitly.

One common convention restricts sec θ to 0 ≤ θ ≤ π, excluding θ = π/2. The positive and negative branches have disjoint ranges, so the restricted function is one-to-one. Under this convention:

arcsec x = arccos(1/x)
Domain: x ≤ −1 or x ≥ 1
Range: [0, π], excluding π/2

Find arcsec 2 and arcsec(−2) under this conventionWorked example

arcsec 2 = arccos(1/2) = π/3

Use the first-quadrant branch.

arcsec(−2) = arccos(−1/2) = 2π/3

Use the second-quadrant branch.

The input gap (−1, 1) has no real graph. As |x| grows, the output approaches π/2 from the relevant side but never equals it. Other texts may choose a different inverse-secant convention; the stated branch is part of the definition.

09 / Use reflection to compare areas

Reflection preserves area as well as swapping coordinates.

Consider y = sin t on 0 ≤ t ≤ π/2. Its graph lies inside a rectangle of width π/2 and height 1. The region above sine but below the top of that rectangle reflects into the region below y = arcsin x for 0 ≤ x ≤ 1.

Area under arcsin from 0 to 1
= π/2 − area under sine from 0 to π/2

Use the sine area to find the inverse-sine areaWorked example

The rectangle area is (π/2) × 1 = π/2

The complementary regions fill the rectangle.

The sine area is 1

In calculus this follows from [−cos t] from 0 to π/2.

The arcsin area is π/2 − 1

No integration formula for arcsin is needed.

This is an extension connecting inverse graphs with integration. The picture explains the complement; the sine integral supplies its numerical value.

Watch the complementary region reflect

Pause, replay or seek freely. The notes explain the same idea and stay in view.

10 / Check a sketch against its defining equation

Use domains and exact points to catch a plausible-looking error.

A sketch of y = arccos(x + 1) is drawn from x = −1 to x = 1. Correct it.Worked example

Require −1 ≤ x + 1 ≤ 1

The actual domain is −2 ≤ x ≤ 0.

The endpoint at x = −2 has y = π

The endpoint at x = 0 has y = 0.

Why does f(t) = sin t for all real t not have an inverse function, despite a calculator having an arcsin key?

Hint

Consider two different inputs with the same sine.

Worked solution

The unrestricted sine is many-to-one, so reversing it would return several outputs for one input. Arcsin inverts only the chosen restriction to [−π/2, π/2].

12 / Recap

A correct inverse graph includes its intervals and branch choice.

Find the inner-input domain before sketching.
Transform the output range and preserve open or closed endpoints.
Map key points using ((u − c)/b, av + d).
State a one-to-one restriction before inverting a periodic trig function.
Give inverse domain and range as well as its formula.
State the convention when using an inverse secant.
Reflection preserves area; complementary regions can avoid a difficult inverse integral.

Section 1 of 12 · Transform the graph and its domain