Infinitely many angles have a sine of 12. The inverse must pick exactly one.
Find the value
arcsin(−12)
Arcsine must return a value in [−π2, π2], and the only angle in that range with sine −12 is −π6. A and B also have sine −12 but lie outside the principal range, so they are not what arcsin returns. This is not a technicality: without one fixed answer, arcsin would not be a function at all.
Same problem as f(x) = x², solved the same way: cut the domain down until the function is one-to-one, then invert the piece that is left.
| function | domain | range of the inverse |
|---|---|---|
| arcsin x | −1 ≤ x ≤ 1 | −π2 to π2 |
| arccos x | −1 ≤ x ≤ 1 | 0 to π |
| arctan x | all reals | −π2 to π2 |
The domains come from the outputs of the original functions: sine and cosine never leave −1 to 1, so arcsin 2 and arccos 2 do not exist. Tangent takes every real value, so arctan accepts any input at all.
Its range runs from 0 to π, so arccos(−√22) = 3π4, not −π4. Arcsine and arctangent do go negative. Mixing up the two ranges is the commonest error here, and it always produces an angle in the wrong quadrant.
cos θ = −√22
reference angle π4
arccos returns 0 to π
quadrant I or II only
cosine is negative in II
so the answer lies there
π − π4 = 3π4
the principal value
Two questions again: how big is the reference angle, and which quadrant does the principal range allow. Answer them in that order and the placement is forced.
sin(arcsin 0.7) = 0.7
0.7 is in [−1, 1] ✓
arcsin(sin(π5)) = π5
π5 is inside the range ✓
arcsin(sin(2π3)) = ?
2π3 is OUTSIDE the range
= arcsin(√32) = π3
not 2π3
Going out and back always returns you. Going in and out only returns you if you started inside the principal range — otherwise you come back to the equivalent angle that is inside it.
The same two questions — how big, and where the range allows it — in three new settings.
Arccos of a cosine from outside its range: arccos(cos(5π4))
cos(5π4) = −√22
5π4 is in quadrant III
arccos(−√22): reference π4, quadrant II
arccos lives in [0, π]
= π − π4 = 3π4
not 5π4
Arccos can never return 5π4, which is past π. It returns the angle in [0, π] with the same cosine.
A negative arctangent: arctan(−√3)
tan(π3) = √3
reference angle π3
arctan lives in (−π2, π2); negative input
so quadrant IV, the negative side
= −π3
not 2π3
Like arcsine, arctangent answers a negative input with a negative angle.
A triangle in quadrant II: tan(arccos(−35))
θ = arccos(−35), so cos θ = −35
θ is in quadrant II
sin θ = 45
positive in quadrant II
tan θ = (45) ÷ (−35) = −43
sin over cos
The 3-4-5 triangle gives the sizes; the range of arccos puts θ in quadrant II and decides the signs.
let θ = arccos(513)
so cos θ = 513
adjacent 5, hypotenuse 13
opposite = 12
θ is in [0, π] with cos positive
so quadrant I, sine positive
sin θ = 1213
the answer
Naming the inverse expression as an angle turns an intimidating composition into a right triangle you can draw. The principal range then settles the sign without any further thought.
Tests ask for a principal value such as arcsin(−√32) or arccos(0), or for a composition like arcsin(sin(5π6)) or cos(arctan(34)). The fastest route: find the reference angle, then place it inside [−π2, π2] for arcsin and arctan or [0, π] for arccos. A 10-second check: confirm your answer is inside the right range and that its sine, cosine or tangent gives back the input.
Find the value
arccos(−12)
The reference angle is π3, and arccos must return something between 0 and π. Cosine is negative in quadrant II, so the answer is π − π3 = 2π3. A applies the arcsine range, which allows negatives — arccosine never does. D has the right cosine but lies outside the principal range.
Hint: arccos answers live between 0 and π. That rules out every negative option before you calculate anything.
