↩️Inverse Trigonometric Functions

Astronomer Vera keeps a brass sighting ring at the Unit Circle Observatory. Turn the ring to an angle and a pointer reads off the height of the star: that is sine, angle in, ratio out. The question this skill answers is the other one. The height is known. What was the angle?

That question has too many answers. sin(30°) = 1/2, and so does sin(150°), and so does sin(390°), and so does every one of the infinitely many angles that land on the same height. Draw the wave y = sin(x) and lay a ruler across it at height 1/2: the ruler crosses the curve over and over, for ever. That is the horizontal line test, and sine does not pass it.

A function is only allowed one output per input, so the reverse of sine cannot be a function until the wave is cut down to a piece that no horizontal ruler crosses twice. Mathematics makes that cut once, and everybody uses the same one:

• ◆sine is kept on -pi/2 <= x <= pi/2, where it climbs steadily from -1 up to 1;
• ◆cosine is kept on 0 <= x <= pi, where it falls steadily from 1 down to -1;
• ◆tangent is kept on -pi/2 < x < pi/2, where it climbs from far below zero to far above it.

Each of those pieces hits every possible value exactly once, so each can be reversed. Reflect the piece in the line y = x and you have the graph of the inverse. That is all arcsin, arccos and arctan are: three restricted pieces, turned round.

You will also see them written sin^-1, cos^-1 and tan^-1. That little -1 is not a power. sin^-1(x) means "the angle whose sine is x"; it does not mean 1/sin(x), which is something else entirely. This lesson writes arcsin, arccos and arctan throughout, because that spelling cannot be misread.

The restricted piece becomes the range of the inverse: the only angles it is allowed to give back. Learn these three lines and most of the skill is done.

arcsin(x) input -1 <= x <= 1 output -pi/2 <= y <= pi/2 (-90° to 90°)

arccos(x) input -1 <= x <= 1 output 0 <= y <= pi (0° to 180°)

arctan(x) input any real number output -pi/2 < y < pi/2 (strictly between)

Three things in that table are worth saying out loud.

The inputs of arcsin and arccos stop at 1. A sine and a cosine are heights and widths on a circle of radius 1, so they never get bigger than 1. That is why arcsin(2) has no value at all — no angle anywhere has sine 2. A tangent has no such limit, which is why arctan will accept any number you give it.

arccos never returns a negative angle. Its whole range sits in the upper half of the circle: quadrant 1 for a positive input, quadrant 2 for a negative one. So arccos(-1/2) is 2pi/3, an obtuse angle, and never -pi/3.

arcsin and arctan never return an obtuse angle. Their ranges are quadrant 1 for a positive input and quadrant 4 — that is, a negative angle — for a negative one. So arcsin(-1/2) is -pi/6. It is not 7pi/6, and it is not 11pi/6, even though both of those angles do have sine -1/2. They are simply not in the range, and the range is what makes arcsin a function at all.

The ends of the arctan range are open, because at pi/2 and -pi/2 the cosine underneath the tangent is 0 and the tangent has no value there. arctan can get as close to pi/2 as you like by feeding it a huge number, but it never arrives.

To evaluate one of these exactly, read it as a question and answer it with the unit circle.

Step 1. Say what is being asked. arccos(-sqrt(3)/2) asks: which angle has cosine -sqrt(3)/2?

Step 2. Find the reference angle from the size of the number, ignoring its sign. cos(30°) = sqrt(3)/2, so the reference angle here is 30°, or pi/6.

Step 3. Let the range place it. The input is negative and this is arccos, so the answer sits in quadrant 2: 180° - 30° = 150°, which is 5pi/6.

Notice the division of labour, because it is the whole method: the number gives the size, the range gives the position. Nothing else is involved.

The nine values worth knowing by heart are the ones the unit circle already gave you, read backwards:

arcsin: 0 -> 0, 1/2 -> pi/6, sqrt(2)/2 -> pi/4, sqrt(3)/2 -> pi/3, 1 -> pi/2

arccos: 1 -> 0, sqrt(3)/2 -> pi/6, sqrt(2)/2 -> pi/4, 1/2 -> pi/3, 0 -> pi/2

arctan: 0 -> 0, sqrt(3)/3 -> pi/6, 1 -> pi/4, sqrt(3) -> pi/3

A negative input reflects the arcsin and arctan answers straight down into negative angles, and reflects the arccos answers across into obtuse ones.

Answers come in two units and the question always says which. In degrees an exact answer is a whole number: 150. In radians it is an exact expression built from pi: pi/6, -pi/4, 5pi/6, 2pi/3. A decimal such as 2.618 is not an exact answer to a question that asked for one, however many places you carry it to.

Putting a function next to its inverse is where most of the interesting work sits, and there are exactly two cases.

Inverse first, function second — nothing to do. sin(arcsin(0.3)) = 0.3, and cos(arccos(-3/5)) = -3/5. The inner function hands over an angle, the outer one reads its ratio straight back, and you are where you started. This works for every input the inverse accepts.

Function first, inverse second — check the range. arcsin(sin(t)) equals t only when t is already inside the range of arcsin. Otherwise the answer is the one angle in the range that shares the same sine. For example:

arcsin(sin(2pi/3)) = arcsin(sqrt(3)/2) = pi/3 not 2pi/3

arccos(cos(5pi/4)) = arccos(-sqrt(2)/2) = 3pi/4 not 5pi/4

arctan(tan(3pi/4)) = arctan(-1) = -pi/4 not 3pi/4

Work inside out and the trap disappears: evaluate the inner ratio first, then ask the inverse for its own angle.

A composition of two different functions is a triangle. To evaluate sin(arccos(3/5)), name the inside: let t = arccos(3/5), so cos t = 3/5. Draw a right triangle with adjacent 3 and hypotenuse 5; Pythagoras gives the opposite side 4. Then sin t = 4/5, and that is the answer.

The sign is the only part that needs care, and the range decides it once again. In cos(arctan(-4/3)) the angle t is in quadrant 4, because arctan of a negative number returns a negative angle — and cosine is positive in quadrant 4, so cos t = 3/5, positive. In sin(arccos(-8/17)) the angle is in quadrant 2, because arccos of a negative number returns an obtuse angle — and sine is positive in quadrant 2, so sin t = 15/17, positive.

The same triangle handles a letter as easily as a number. For sin(arctan(x)) with x positive, tan t = x/1, so the legs are x and 1 and the hypotenuse is sqrt(x^2 + 1), giving

sin(arctan(x)) = x / sqrt(x^2 + 1)

On a calculator these live above the sin, cos and tan keys, behind shift or 2nd. Two habits stop nearly every wrong answer.

Set the mode first. arcsin(0.3) is 17.5 in degree mode and 0.305 in radian mode, and the calculator will not warn you which one you asked for. Decide from the question, set the mode, then press the keys.

Check the answer against the range. arccos of a negative number must come out obtuse; arcsin of a negative number must come out negative. Anything else means a wrong key or a wrong mode.

Most real questions are angle questions in disguise. A ladder 6 m long reaching 5 m up a wall makes an angle t with the ground where sin t = 5/6, so t = arcsin(0.8333) = 56.4°. A tower 12 m tall casting a 20 m shadow gives an angle of elevation where tan t = 12/20, so t = arctan(0.6) = 31.0°. A road with a grade of 8% climbs 8 m for every 100 m measured horizontally, so its angle is arctan(0.08) = 4.6°. Rise and run are two legs, so they are always a tangent; a slanted length is a hypotenuse, so it is a sine or a cosine.

One last thing the inverse cannot do for you. Asked to solve sin t = 0.6 on 0° <= t < 360°, the calculator returns one angle, 36.9°, because that is all its range allows. The circle has a second one at 180° - 36.9° = 143.1°. For a cosine equation the second solution is 360° minus the first instead, since cosine mirrors across the horizontal axis rather than the vertical one. The inverse finds the reference angle; the unit circle finds the rest.