01 · Exact arcsin
Find arcsin(√2/2).
Hint
Choose an angle in [−π/2, π/2].
Worked solution
π/4.
Understand · explore · practise
Arcsin, arccos and arctan
Understand arcsin, arccos and arctan, their principal values, domains and ranges. Reflect restricted trig graphs, calculate exact angles and avoid inverse-function mistakes.
Before you startTrig graphs, radians, exact values and inverse functions
01 / Why a restriction is needed
An inverse function must return just one output.
Sine takes the same value at many angles. For example, sin(π/6) = sin(5π/6) = 1/2. Reversing the unrestricted sine function would give several possible outputs for one input, so it would not be a function.
Restrict the original domain → reverse the one-to-one function
Arcsin, arccos and arctan each return a chosen principal angle. Move the model through each restricted domain, and compare a point with its reflection in y = x.
Swap input and outputExplore
sin(π/4) = 0.707 (rounded).
arcsin(sin(π/4)) = π/4.
Restrict sine to −π/2 ≤ θ ≤ π/2.
Arcsin domain: [−1, 1]. Range: [−π/2, π/2].
Blue: the restricted original. Gold: its inverse. The green segment joins swapped coordinates. Both axes use the same scale; decimals are rounded to 3 places.
Watch restricted cosine reflect into arccos
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Inverse is different from reciprocal
The superscript −1 can carry two different meanings.
Compare arcsin(1/2) with cosec(π/6)Worked example
arcsin(1/2) = π/6
Input a sine value; output its principal angle.
cosec(π/6) = 1/sin(π/6) = 2
Input an angle; output the reciprocal of its sine.
sin⁻¹(1/2) usually means arcsin(1/2)
In inverse-function notation the −1 means inverse, not reciprocal.
Use arcsin, arccos and arctan when the notation might be ambiguous. The expression (sin x)⁻¹ is a reciprocal, whereas the inverse-function expression sin⁻¹ x is normally arcsin x.
03 / Arcsin chooses the central sine branch
Its principal angles can be negative.
Original: sin θ for −π/2 ≤ θ ≤ π/2
Arcsin domain: −1 ≤ x ≤ 1
Arcsin range: −π/2 ≤ y ≤ π/2
Find arcsin(−√3/2)Worked example
An angle with sine −√3/2 is −π/3
This lies inside the chosen principal range.
Therefore arcsin(−√3/2) = −π/3
Although 4π/3 also has that sine, it is not a principal arcsin output.
On this branch sine is increasing and takes every value from −1 to 1 exactly once. Its inverse is therefore increasing too. The endpoints are included: arcsin(−1) = −π/2 and arcsin(1) = π/2.
04 / Arccos chooses the top half-turn
Its principal outputs are never negative.
Original: cos θ for 0 ≤ θ ≤ π
Arccos domain: −1 ≤ x ≤ 1
Arccos range: 0 ≤ y ≤ π
Find arccos(−√2/2)Worked example
Cosine is negative in the second quadrant of the principal interval
The required angle is 3π/4.
arccos(−√2/2) = 3π/4
The angle −3π/4 has the same cosine but lies outside [0, π].
Cosine decreases from 1 to −1 on the selected branch, so arccos is decreasing. Keep the reversed endpoints straight: arccos(1) = 0, arccos(0) = π/2 and arccos(−1) = π.
05 / Arctan has an open principal range
A finite input cannot produce ±π/2.
Original: tan θ for −π/2 < θ < π/2
Arctan domain: all real x
Arctan range: −π/2 < y < π/2
Find arctan(−√3)Worked example
tan(−π/3) = −√3
The angle lies strictly inside the principal interval.
arctan(−√3) = −π/3
The alternative angle 2π/3 is outside that interval.
The vertical asymptotes of the restricted tangent graph become horizontal asymptotes of arctan, y = ±π/2. The curve approaches them but never reaches them. For example, arctan(1000) is close to π/2, not equal to it.
06 / Read coordinates after reflection
The inverse swaps the two coordinates of every point.
Use cos(π/3) = 1/2 to locate a point on arccosWorked example
Restricted cosine contains (π/3, 1/2)
The first coordinate is its angle input.
Arccos contains (1/2, π/3)
Swap input and output.
The line y = x lies halfway between the two points
The graphs are reflections because both axes use the same scale.
The original range becomes the inverse domain, and the restricted original domain becomes the inverse range. This is why checking the original restriction is enough to determine both inverse intervals.
07 / Calculate without losing the unit
Calculator mode determines how the angle is displayed.
Evaluate arcsin(0.37) to 3 decimal places in radiansWorked example
Use the inverse sine key in RAD mode
The unrounded value is about 0.3790090207.
arcsin(0.37) ≈ 0.379 radians
Round once, at the end.
In degrees the same angle is about 21.716°
These are two units for the same angle, not two different branches.
Check the result against the principal range. A radian answer for arcsin must lie between about −1.571 and 1.571. A calculator domain error for arccos(1.2) is expected: 1.2 is outside the real domain.
08 / A principal value is only a starting angle
Solving a trig equation can require other branches.
Compare arcsin(1/2) with solving sin x = 1/2 for 0 ≤ x < 2πWorked example
arcsin(1/2) = π/6
This function returns one principal value.
The equation also has x = 5π/6
Sine is positive in both Quadrants I and II.
The equation answers are π/6 and 5π/6
Use symmetry and the requested interval after finding a reference angle.
Solve cos x = −0.4 for 0 ≤ x < 2πWorked example
Let α = arccos(−0.4)
It lies in Quadrant II.
The two roots are α and 2π − α
Cosine has the same value below the x-axis.
Do not use π − α as the second root
That would change the sign of the cosine.
09 / Check the order of the two operations
Undoing a function works only on the matching domain.
sin(arcsin x) = x for −1 ≤ x ≤ 1
arcsin(sin θ) = θ only when −π/2 ≤ θ ≤ π/2
Evaluate arcsin(sin(5π/6))Worked example
sin(5π/6) = 1/2
The inner function is defined for this angle.
arcsin(1/2) = π/6
Return the principal angle, not the original nonprincipal one.
Evaluate arccos(cos(7π/6)) and arctan(tan(3π/4))Worked example
cos(7π/6) = −√3/2, so arccos returns 5π/6
The output must be in [0, π].
tan(3π/4) = −1, so arctan returns −π/4
The output must be in (−π/2, π/2).
The next lesson develops these compositions and their branch restrictions in detail. For now, always write down the outer inverse function’s range before selecting an angle.
10 / Your turn
Give exact principal angles unless decimals are requested.
02 · Negative input
Find arcsin(−1/2).
Hint
The principal arcsin angle is negative.
Worked solution
−π/6. Although 7π/6 also has sine −1/2, it is outside the principal range.
03 · Arccos
Find arccos(−1/2).
Hint
Choose the second-quadrant angle in [0, π].
Worked solution
2π/3.
04 · Arctan
Find arctan(−1).
Hint
The principal tangent interval contains −π/4.
Worked solution
−π/4.
05 · Endpoints
Evaluate arcsin(1), arccos(1) and arccos(−1).
Hint
Use the closed principal intervals.
Worked solution
Respectively π/2, 0 and π.
06 · Real domains
Which exist as real numbers: arcsin(1.1), arccos(−1.1), arctan(1.1)?
Hint
Arcsin and arccos need an input in [−1, 1].
Worked solution
Only arctan(1.1) exists as a real number. Arctan accepts every finite real input.
07 · Reflect a point
Sin(−π/4) = −√2/2. State the corresponding point on y = arcsin x.
Hint
Swap the original input and output.
Worked solution
(−√2/2, −π/4).
08 · Reciprocal or inverse?
Evaluate arcsin(1/2) and 1/sin(π/6). Explain why their answers differ.
Hint
One returns an angle; the other is a reciprocal value.
Worked solution
Arcsin(1/2) = π/6; 1/sin(π/6) = 2. These are different operations with different inputs.
09 · Reverse composition
Find arcsin(sin(3π/4)).
Hint
First evaluate the inner sine, then choose the principal output.
Worked solution
Sin(3π/4) = √2/2, so the result is π/4.
10 · Another principal branch
Find arccos(cos(5π/3)).
Hint
Arccos must return an angle in [0, π].
Worked solution
Cos(5π/3) = 1/2, so the result is π/3.
11 · Tangent periodicity
Find arctan(tan(5π/4)).
Hint
Tangent has period π.
Worked solution
Tan(5π/4) = 1, so the principal output is π/4.
12 · An unreachable output
Can arctan x = π/2 hold for a finite real x?
Hint
The range endpoints are open.
Worked solution
No. Arctan approaches π/2 as x grows without bound but never reaches it at a finite input.
13 · Every equation root
Solve sin x = −1/2 for 0 ≤ x < 2π. Why is −π/6 not one of the listed answers?
Hint
The principal arcsin value is useful but outside the requested interval.
Worked solution
x = 7π/6 or 11π/6. The angle −π/6 is outside [0, 2π), so use its coterminal angle 11π/6 and the other sine branch.
14 · Calculator and mode
Evaluate arccos(0.6) to 3 decimal places in radians, then explain what changes in degree mode.
Hint
Check the answer lies between 0 and π.
Worked solution
Arccos(0.6) ≈ 0.927 radians. Degree mode gives about 53.130° for the same angle. Use the unit requested by the question.
11 / Recap
The principal range decides which angle the inverse returns.
- Restrict a trig function to a one-to-one branch before taking its inverse.
- Arcsin: domain [−1, 1], range [−π/2, π/2].
- Arccos: domain [−1, 1], range [0, π].
- Arctan: real domain, open range (−π/2, π/2).
- Inverse graphs reflect in y = x; coordinates swap.
- A principal value alone may not solve an entire trig equation.
- Check the outer function’s range in a reversed composition.
Section 1 of 11 · Why a restriction is needed