Inverse Trigonometric Functions

Sine repeats forever, so its inverse has to be told which answer to give back.

Ask for the angle whose sine is 0.5.

30° works. So does 150°. So does 390°, and 510°, and on it goes.

one height, and the wave keeps returning to it 0.5 only here which one should arcsin return? so the domain is cut down until exactly one answer is left

The wave crosses that height again and again, so the question has infinitely many right answers.

A function is only allowed to return one. So mathematicians made a choice and wrote it down.

Cut the domain until each height is reached exactly once, and the inverse becomes a function. That restriction is not a technicality — it is the whole reason sin-1 exists.

What each inverse function is allowed to return each inverse hands back exactly one angle, from a fixed stretch of the circle arcsin −90° to 90° arccos 0° to 180° arctan −90° to 90°

Why a restriction is needed at all

A function has an inverse only if it is one-to-one, and sine fails that badly — it hits every height in its range infinitely often.

So before inverting, keep just enough of the curve: one rise from bottom to top, and throw the rest away.

For sine that stretch is -90° to 90°, where the curve climbs steadily from -1 to 1 without repeating.

One rise is enough

Any stretch covering the full range exactly once would do. These particular choices are conventions, agreed so that everybody’s calculator gives the same answer.

The three ranges, and why they differ

arcsin returns -90° to 90° — quadrants IV and I, where sine sweeps its whole range once.

arccos returns 0° to 180° — quadrants I and II. Cosine is falling there, again covering everything once.

arctan returns strictly between -90° and 90°, since tangent runs from -∞ to ∞ across that stretch.

Cosine is the odd one out

Sine and tangent both return angles around zero; cosine returns angles in the upper half. Forgetting that is the most common source of a wrong quadrant.

Reading a calculator answer honestly

sin-1(0.5) gives 30° and stops. It is not wrong — it is answering the restricted question.

If the problem wants an obtuse angle, you must find it yourself: 180° - 30° = 150°.

So treat the calculator output as the reference angle, then place it in the quadrant the problem demands.

The calculator cannot know your triangle

In an obtuse triangle the sine rule can give sin A = 0.5 with A = 150°. The machine will still say 30°, and only you know which one the picture needs.

Composing a function with its inverse

sin(arcsin x) = x for any x in [-1, 1] — the inner one picks an angle, the outer one undoes it.

The other order is not always so tidy: arcsin(sin 150°) = 30°, not 150°.

That is the restriction showing its hand. The answer comes back in the allowed range, whatever went in.

One direction always works

Going out and back through the inverse is safe. Going in and back out lands you in the principal range, which may not be where you started.

Exact values worth knowing

arcsin12 = 30°, arcsin√22 = 45°, arcsin√32 = 60° — the same special angles, read backwards.

arccos(-12) = 120°, because cosine is negative in the upper-left part of its range.

Sketching the unit circle and marking the allowed arc settles every one of these without memorising a table.

Draw the arc first

Shade the range the inverse is allowed to return, then find the point at the right height. The answer has to lie in the shaded part.

Worked examples

A principal value

Find arccos(-12) in degrees.

  1. The reference angle has cos = 12, which is 60°. Size first.
  2. Arccosine must return something between 0° and 180°, and cosine is negative in quadrant II. The restriction decides the quadrant.
  3. So the answer is 180° - 60° = 120°. Inside the allowed range, as it must be.

120°

When the calculator is not enough

In a triangle, sin A = 0.6 and A is known to be obtuse. Find A to the nearest degree.

  1. sin-1(0.6) ≈ 37°. The calculator returns the principal value.
  2. That is acute, but the problem says A is obtuse. Use 37° as the reference angle.
  3. A = 180° - 37° = 143°. Check: sin 143° ≈ 0.6, so both fit the equation.

143°

A composition

Evaluate arcsin(sin 200°).

  1. 200° is in quadrant III, with reference angle 20°, so sin 200° = -sin 20°. Sine is negative there.
  2. arcsin must return an angle between -90° and 90°. The restriction applies to the output.
  3. The angle in that range with that sine is -20°. Not 200° — the function cannot return what it is not allowed to.

-20°

Words to know

Arcsine
The inverse of sine, restricted to outputs from -90° to 90°.
Arccosine
The inverse of cosine, restricted to outputs from 0° to 180°.
Arctangent
The inverse of tangent, restricted to outputs strictly between -90° and 90°.
Principal value
The single answer the restricted inverse returns.
Restricted domain
The stretch of the original function kept so that it is one-to-one.

Mistakes to watch for

Reading sin-1x as 1sin x. The first is an angle, the second is a number: sin-1(0.5) = 30° while 1sin 0.5 is about 2.09. The -1 marks an inverse function, not a reciprocal.

Accepting the calculator’s answer when the problem needs the other one. sin-1(0.6) gives 37°, but an obtuse angle with the same sine is 143°. The machine returns the principal value and knows nothing about your triangle.

Using the arcsine range for arccosine. Arcsine returns -90° to 90°; arccosine returns 0° to 180°. A negative arccosine answer is always wrong.

Assuming arcsin(sinθ) = θ. It holds only when θ is already inside the restricted range — arcsin(sin 150°) is 30°, not 150°.

Taking an arcsine of something outside [-1, 1]. Sine never exceeds 1, so arcsin(2) has no answer, and a calculator returning an error there is correct.

The notation sin-1 is unfortunate and everyone knows it. Many textbooks now prefer arcsin, which cannot be mistaken for a reciprocal.

"Arc" is literal: on the unit circle the answer is the arc length, because radius one makes arc and angle the same number.

Your calculator has already made the restriction for you. That is why sin-1 never returns an obtuse angle, however much the problem wants one.

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