Class 12 Mathematics · Chapter 2

Inverse
Trigonometry.

Ordinary trigonometry turns an angle into a ratio. Inverse trigonometry runs it backwards: give it a ratio, and it hands back the angle. The whole chapter is about making that "the" possible.

sin takes the angle π/6 to the ratio ½. sin⁻¹ takes ½ back to π/6. But sin(5π/6) is also ½, so we have to agree on which angle to give back.

Why sin has no inverse (yet)Branches Mirror graphs Principal values sin⁻¹x ≠ 1/sin x sin⁻¹(sin x)Triangle trick Substitutions Self-check

New to this? Start with the basics →Walk through every example →

Part 1 · The problem

Trig functions repeat, so they can't be undone.

A function can be undone only when it is one-one and onto. Sine is neither on its natural domain: its wave repeats forever.

What an inverse needs

one-one and onto

In Chapter 1 you saw that a function has an inverse only when it is one-one and onto.

Why sine fails

sin 0 = sin π = sin 2π = 0

The wave repeats forever, so every height between −1 and 1 is reached infinitely many times. The table below shows the same problem for all six trig functions.

Basics for thisdomain and range of sin why tan and cot have holes what an inverse needs

Function	Natural domain	Range	Why no inverse as it stands
sin	ℝ	[−1, 1]	sin 0 = sin π = sin 2π = 0 : many-one
cos	ℝ	[−1, 1]	cos 0 = cos 2π = 1 : many-one
tan	ℝ − {(2n + 1)π/2}	ℝ	tan 0 = tan π = 0 : many-one
cot	ℝ − {nπ}	ℝ	repeats every π
sec	ℝ − {(2n + 1)π/2}	ℝ − (−1, 1)	repeats every 2π
cosec	ℝ − {nπ}	ℝ − (−1, 1)	repeats every 2π

In plain words

Imagine a Ferris wheel. If I tell you "your seat is 10 m above the ground", you can't tell whether you're going up or down, or which lap you're on. Many moments give the same height. To answer "when?", we first have to agree to look at one stretch of the ride only. That agreement is called restricting the domain.

Part 2 · Branches

Cut out one good piece of the wave.

Choose a stretch where the wave climbs (or falls) from −1 to 1 exactly once. On that stretch, the function can be undone.

One good piece

sin : [−π2, π2] → [−1, 1]

Pick an interval where the graph climbs (or falls) all the way from −1 to 1 exactly once. On that piece the function is one-one and onto [−1, 1], so it has an inverse. Each such piece gives a branch.

The standard one

cos : [0, π] → [−1, 1]

The standard branch is called the principal value branch: [−π/2, π/2] for sine and [0, π] for cosine. The cards below explain why.

Basics for thisthe sine wave from the circle reverse the arrows

Window testChoose a window and slide the level y = c. A good window is hit exactly once at every level between −1 and 1.

Window for x

Level c

Hits inside the window

One-one on the window?

Range on the window

sine

Principal branch [−π/2, π/2]

Sine rises steadily from −1 to 1 here. Other valid branches: [π/2, 3π/2], [−3π/2, −π/2], … but [0, π] fails, because it goes up and then back down.

cosine

Principal branch [0, π]

Cosine falls steadily from 1 to −1 here. [−π/2, π/2] fails for cosine: it is symmetric about 0, so cos(−θ) = cos θ.

Why these choices?

They contain the familiar angles

Both principal branches include 0 to π/2, the acute angles from right-angle triangles. That keeps answers like sin⁻¹(½) = π/6 natural.

Part 3 · Graphs of the six inverses

Every inverse is a mirror image.

Swap the two coordinates of every point and you get the graph of the inverse. That swap is a reflection in the line y = x.

Swap the pair

(a, b) ⟷ (b, a)

If (a, b) is on the graph of y = sin x, then (b, a) is on the graph of y = sin⁻¹ x. Swapping the coordinates is the same as reflecting in the line y = x.

Bold and faint

y = sin⁻¹ x

In the lab, the bold piece is the principal branch; the faint copies are the other branches.

Basics for thisradians on the axes graphs of tan, cot, sec, cosec

Mirror labChoose a function. Slide x to read the inverse and see its twin point on the original graph.

Input x

Output

Blue: sin x. Orange: the inverse. Grey dashes: the mirror y = x. Axis ticks are in radians (π ≈ 3.14).

Inverse	Domain (what you can feed in)	Principal range (what comes out)	Shape of the graph
sin⁻¹	[−1, 1]	[−π/2, π/2]	Rising S from (−1, −π/2) to (1, π/2)
cos⁻¹	[−1, 1]	[0, π]	Falling from (−1, π) to (1, 0)
cosec⁻¹	ℝ − (−1, 1)	[−π/2, π/2] − {0}	Two pieces, flattening toward 0
sec⁻¹	ℝ − (−1, 1)	[0, π] − {π/2}	Two pieces, approaching π/2
tan⁻¹	ℝ	(−π/2, π/2)	Rising, trapped between two horizontal asymptotes
cot⁻¹	ℝ	(0, π)	Falling, from near π down toward 0

Swap and restrict: the domain of each inverse is the range of the original, and its range is the restricted domain of the original.

Part 4 · Principal values

Where the answers live on the circle.

Each inverse gives back one answer, called its principal value. On the unit circle, all those answers live in just two halves.

Principal value

sin⁻¹(12) = π6

The value of an inverse trig function that lies in its principal range is its principal value.

Two halves

right half: [−π2, π2]

top half: [0, π]

On the unit circle, the principal ranges are just these two halves. Remember this picture and you'll never mix them up.

Basics for thisthe special angles signs in each quadrant

Memory trick

sin, tan and cosec answer from the right half (−π/2 to π/2): negative inputs give negative angles.

cos, sec and cot answer from the top half (0 to π): the answer is never negative, so negative inputs land in the second quadrant.

Principal value finderChoose a function and a standard value. The circle shows the one angle in the principal range with that ratio, and a faint "impostor" angle with the same ratio that is not allowed.

Input value

Example 1

Find the principal value of sin⁻¹(1/√2).

Let sin⁻¹(1/√2) = y, so sin y = 1/√2.
y must lie in [−π/2, π/2].
sin(π/4) = 1/√2, and π/4 is in that range.

Principal value = π/4.

Example 2

Find the principal value of cot⁻¹(−1/√3).

cot y = −1/√3 = −cot(π/3).
Use cot(π − θ) = −cot θ: −cot(π/3) = cot(π − π/3) = cot(2π/3).
2π/3 lies in (0, π), the principal range of cot⁻¹.

Principal value = 2π/3, not −π/3. Negative cot inputs land in the second quadrant.

Note · don't confuse these

sin⁻¹ x means "the angle whose sine is x". It is not this:

(sin x)⁻¹ = 1sin x = cosec x

For example, these two are completely different numbers. An angle:

sin⁻¹(0.5) = π6 ≈ 0.524

A ratio:

(sin 0.5)⁻¹ = 1sin(0.5 rad) ≈ 2.086

The "−1" here is function-inverse notation, like f⁻¹, not a power.

When no branch is mentioned, always use the principal value branch.

Try it · Exercise 2.1

Find the principal values.

Say which half of the circle the answer must come from before you work out the angle. Most can also be checked in the finder above. Watch every one solved on the circle →

Q1sin⁻¹(−½)

−π/6. sin(π/6) = ½, and sin⁻¹ takes negatives to negative angles (right half).

Q2cos⁻¹(√3/2)

π/6. cos(π/6) = √3/2, and π/6 is in [0, π].

Q3cosec⁻¹(2)

π/6. cosec y = 2 means sin y = ½.

Q4tan⁻¹(−√3)

−π/3. tan(π/3) = √3; tan⁻¹ answers from (−π/2, π/2), so the negative input gives −π/3.

Q5cos⁻¹(−½)

2π/3. Use cos(π − θ) = −cos θ: π − π/3 = 2π/3. The answer can't be negative for cos⁻¹.

Q6tan⁻¹(−1)

−π/4.

Q7sec⁻¹(2/√3)

π/6. sec y = 2/√3 means cos y = √3/2.

Q8cot⁻¹(√3)

π/6.

Q9cos⁻¹(−1/√2)

3π/4. π − π/4 = 3π/4.

Q10cosec⁻¹(−√2)

−π/4. sin y = −1/√2, answered from the right half.

Q11tan⁻¹(1) + cos⁻¹(−½) + sin⁻¹(−½)

π/4 + 2π/3 − π/6 = (3π + 8π − 2π)/12 = 3π/4.

Q12cos⁻¹(½) + 2 sin⁻¹(½)

π/3 + 2(π/6) = 2π/3.

Part 5 · Properties

Undo, then redo, but mind the window.

Because they are inverses, each pair cancels. But the two orders are not equally safe.

Basics for thisfull turns sine as height

Always safe (inside the domain)

sin(sin⁻¹ x) = x

x ∈ [−1, 1]

Find the angle whose sine is x, then take its sine: you get x back. Every time.

Only inside the principal range

sin⁻¹(sin x) = x

x ∈ [−π/2, π/2]

For other angles, sin⁻¹ can only return an angle from [−π/2, π/2]. It returns the one with the same sine, which is not x.

The folding graphy = sin⁻¹(sin x) is not the line y = x. It follows the line only inside the principal window, then folds back like a zigzag.

Angle x

Textbook cases

Working

Example 6 · The classic trap

Find sin⁻¹(sin 3π/5).

Tempting answer: 3π/5. But 3π/5 = 108° is outside [−π/2, π/2].
Use sin(π − θ) = sin θ: sin(3π/5) = sin(π − 3π/5) = sin(2π/5).
2π/5 = 72° is inside [−π/2, π/2].

sin⁻¹(sin 3π/5) = 2π/5.

The folding rules

sin⁻¹(sin x): first remove whole turns of 2π. If the angle is past π/2, use π − x; if it is below −π/2, use −π − x.
cos⁻¹(cos x): remove turns of 2π to land in [−π, π]. If negative, use −x (cosine is even).
tan⁻¹(tan x): tan repeats every π, so subtract a multiple of π to land in (−π/2, π/2).

Part 6 · The triangle trick

Turn an inverse into a right triangle.

Any inverse trig value is an angle θ. Draw that angle in a right triangle, and every other ratio of θ can be read off the picture.

The idea

θ = tan⁻¹ x ⇒ tan θ = x

The input tells you one ratio of θ. That ratio gives you two sides of a right triangle.

How to use it

Call the inverse value θ, for example θ = tan⁻¹ x.
Draw a right triangle with that angle and label two sides from the input (for tan⁻¹ x: opposite x, adjacent 1).
Find the third side by Pythagoras (here √(1 + x²)).
Read every other ratio of θ straight off the picture.

Basics for thisright-triangle ratios Pythagoras

Triangle builderPick θ = tan⁻¹ x, sin⁻¹ x, cos⁻¹ x or sec⁻¹ x. Drag x and watch the triangle and all six ratios change.

x

So the sides are

Angle θ

Misc. Exercise Q13

sin(tan⁻¹ x), |x| < 1, equals?

Let θ = tan⁻¹ x: opposite = x, adjacent = 1.
Hypotenuse = √(1 + x²).
sin θ = opposite/hypotenuse.

sin(tan⁻¹ x) = x/√(1 + x²). Option (D). The highlighted sin card in the builder shows it.

Example 5

Write cot⁻¹(1/√(x² − 1)), x > 1, in simplest form.

Put x = sec θ. Then √(x² − 1) = √(sec²θ − 1) = tan θ.
So 1/√(x² − 1) = 1/tan θ = cot θ.
cot⁻¹(cot θ) = θ.

= sec⁻¹ x. In the builder, choose sec⁻¹ x: its cot card reads 1/√(x² − 1).

Exercise 2.2 Q12

Evaluate tan(sin⁻¹(3/5) + cot⁻¹(3/2)).

sin⁻¹(3/5): a 3-4-5 triangle, so tan = 3/4.
cot⁻¹(3/2): adjacent 3, opposite 2, so tan = 2/3.
tan(A + B) = (3/4 + 2/3)/(1 − (3/4)(2/3)) = (17/12)/(1/2).

= 17/6.

Misc. Exercise Q3

Prove 2 sin⁻¹(3/5) = tan⁻¹(24/7).

Let θ = sin⁻¹(3/5). The 3-4-5 triangle gives tan θ = 3/4.
tan 2θ = 2 tan θ / (1 − tan²θ) = (3/2)/(7/16) = 24/7.
θ is acute and 2θ < π/2 (since tan θ < 1), so 2θ = tan⁻¹(24/7).

Part 7 · Simplest form

Spot the pattern, substitute, simplify.

Long expressions inside an inverse often hide a trig identity. The right substitution makes it appear.

How to use it

Spot the pattern inside the inverse, using the table below.
Substitute x = sin θ, cos θ or tan θ so the identity appears.
Simplify with that identity.
Cancel the inverse with the function, for example sin⁻¹(sin 2θ) = 2θ. Check that the angle stays in the principal range (Part 5).

If you see	Substitute	Because
√(1 − x²)	x = sin θ or cos θ	1 − sin²θ = cos²θ
√(1 + x²)	x = tan θ	1 + tan²θ = sec²θ
√(x² − 1)	x = sec θ	sec²θ − 1 = tan²θ
√(a² − x²)	x = a sin θ	a² − a² sin²θ = a² cos²θ
2x/(1 + x²), (1 − x²)/(1 + x²)	x = tan θ	sin 2θ, cos 2θ in terms of tan θ
3x − 4x³	x = sin θ	sin 3θ = 3 sin θ − 4 sin³θ
4x³ − 3x	x = cos θ	cos 3θ = 4 cos³θ − 3 cos θ
(1 − cos x)/(1 + cos x)	half angle	= tan²(x/2)

Example 3

Show sin⁻¹(2x√(1 − x²)) = 2 sin⁻¹ x, for −1/√2 ≤ x ≤ 1/√2.

Put x = sin θ, so θ = sin⁻¹ x and √(1 − x²) = cos θ.
2x√(1 − x²) = 2 sin θ cos θ = sin 2θ.
sin⁻¹(sin 2θ) = 2θ, valid because |x| ≤ 1/√2 keeps 2θ inside [−π/2, π/2].

= 2 sin⁻¹ x. (With x = cos θ and 1/√2 ≤ x ≤ 1, the same expression equals 2 cos⁻¹ x.)

Example 4

Simplify tan⁻¹(cos x/(1 − sin x)), −3π/2 < x < π/2.

Write cos x = cos²(x/2) − sin²(x/2) and 1 − sin x = (cos(x/2) − sin(x/2))².
Cancel one factor: (cos + sin)/(cos − sin) of x/2.
Divide by cos(x/2): (1 + tan(x/2))/(1 − tan(x/2)) = tan(π/4 + x/2).

= π/4 + x/2.

Exercise 2.2 answers

See each proof animated and checked by graph →

Q13 sin⁻¹ x = sin⁻¹(3x − 4x³), x ∈ [−½, ½]

Put x = sin θ: 3x − 4x³ = sin 3θ. |x| ≤ ½ means |θ| ≤ π/6, so 3θ ∈ [−π/2, π/2] and sin⁻¹(sin 3θ) = 3θ.

Q23 cos⁻¹ x = cos⁻¹(4x³ − 3x), x ∈ [½, 1]

Put x = cos θ with θ ∈ [0, π/3]: 4x³ − 3x = cos 3θ, and 3θ ∈ [0, π].

Q3tan⁻¹((√(1 + x²) − 1)/x), x ≠ 0

Put x = tan θ: (sec θ − 1)/tan θ = (1 − cos θ)/sin θ = tan(θ/2). Answer: ½ tan⁻¹ x.

Q4tan⁻¹√((1 − cos x)/(1 + cos x)), 0 < x < π

The root is tan(x/2), and x/2 ∈ (0, π/2). Answer: x/2.

Q5tan⁻¹((cos x − sin x)/(cos x + sin x)), −π/4 < x < 3π/4

Divide by cos x: (1 − tan x)/(1 + tan x) = tan(π/4 − x), and π/4 − x ∈ (−π/2, π/2). Answer: π/4 − x.

Q6tan⁻¹(x/√(a² − x²)), |x| < a

Put x = a sin θ: the fraction becomes tan θ. Answer: sin⁻¹(x/a).

Q7tan⁻¹((3a²x − x³)/(a³ − 3ax²)), a > 0, −a/√3 < x < a/√3

Put x = a tan θ: the fraction is (3 tan θ − tan³θ)/(1 − 3 tan²θ) = tan 3θ. Answer: 3 tan⁻¹(x/a).

Q8tan⁻¹[2 cos(2 sin⁻¹ ½)]

sin⁻¹ ½ = π/6, so 2 cos(π/3) = 1, and tan⁻¹ 1 = π/4.

Q9tan ½[sin⁻¹(2x/(1 + x²)) + cos⁻¹((1 − y²)/(1 + y²))], |x| < 1, y > 0, xy < 1

With x = tan α, y = tan β the bracket is 2α + 2β. So the expression is tan(α + β) = (x + y)/(1 − xy).

Q10–11sin⁻¹(sin 2π/3) and tan⁻¹(tan 3π/4)

π/3 and −π/4. Both are in the folding graph's textbook cases.

Try it · Miscellaneous exercise

Mixed practice.

Q1cos⁻¹(cos 13π/6)

13π/6 − 2π = π/6, which is in [0, π]. Answer: π/6.

Q2tan⁻¹(tan 7π/6)

7π/6 − π = π/6. Answer: π/6.

Q4sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36)

Triangles 8-15-17 and 3-4-5 give tangents 8/15 and 3/4. tan(A + B) = (8/15 + 3/4)/(1 − (8/15)(3/4)) = (77/60)/(36/60) = 77/36.

Q11Solve 2 tan⁻¹(cos x) = tan⁻¹(2 cosec x)

Left side = tan⁻¹(2 cos x/(1 − cos²x)) = tan⁻¹(2 cos x/sin²x). Equate: 2 cos x/sin²x = 2/sin x, so cos x = sin x. Answer: x = π/4.

Q12Solve tan⁻¹((1 − x)/(1 + x)) = ½ tan⁻¹ x, x > 0

Left side = tan⁻¹ 1 − tan⁻¹ x = π/4 − tan⁻¹ x. So (3/2) tan⁻¹ x = π/4, tan⁻¹ x = π/6. Answer: x = 1/√3.

Self-check

Seven exam-style questions.

The multiple-choice questions from the chapter. Tap an option to check it and see the reasoning.

Score0 / 77 left to answer

Summary

Remember this.

Trig functions repeat, so we restrict their domains to make them one-one and onto. Each valid restriction is a branch; the standard one is the principal value branch.

Right half [−π/2, π/2]: sin⁻¹, tan⁻¹ (open ends), cosec⁻¹ (without 0).

Top half [0, π]: cos⁻¹, cot⁻¹ (open ends), sec⁻¹ (without π/2).

The graph of each inverse is the mirror image of the restricted graph in the line y = x.

sin⁻¹ x ≠ (sin x)⁻¹. The first is an angle, the second is cosec x.

sin(sin⁻¹ x) = x for x ∈ [−1, 1], but sin⁻¹(sin x) = x only for x ∈ [−π/2, π/2]. Otherwise fold the angle back.

To simplify, draw the triangle or substitute x = sin θ, tan θ, sec θ so an identity appears.

Historical note

A shadow, a staff and a pyramid.

Thales (about 600 BC) is said to have found the height of a great pyramid from shadows.

Same angle

HS = hs = tan(sun's altitude)

The sun's rays hit the pyramid and a small staff at the same angle. Inverse trigonometry turns the measured shadow straight back into that angle.

Sun's altitude α

Pyramid height 146 m (the Great Pyramid as built). Staff height 2 m. The staff is drawn larger than scale so you can see it.

Measure the staff's shadow s

Recover the angle: α = tan⁻¹(h/s)

Pyramid height H = S · tan α

476Aryabhata

Early Indian results in trigonometry, including tables of sines.

598Brahmagupta

Further results in the siddhantas, Sanskrit astronomical works.

≈600Bhaskara I

Formulae for sines of angles greater than 90°.

1114Bhaskara II

Exact values of sines and cosines of 18°, 36°, 54°, 72°.

1500sYuktibhasa

A Malayalam work containing a proof for the expansion of sin(A + B).

1813John F. W. Herschel

Suggests the symbols sin⁻¹ x, cos⁻¹ x for arc sin x, arc cos x.

The sine function itself is one of the main contributions of Indian mathematics. The knowledge travelled from India to Arabia and then to Europe.

Worked examples · all 6 from the textbook

Every example,
step by step.

Each textbook example is a small animated story on the unit circle, a triangle or a graph. Press Next step (or Play) and watch the picture build the solution. Every example also has a Try it panel for testing your own numbers. Under the controls, each step is explained in detail, with links to the exact basics it uses. After reading a basic, press ← Back to return to the same step.

Section 2.2 · Basic concepts

Principal values: Examples 1–2

Draw the ratio on the unit circle, find every angle that fits, and keep the one inside the principal half.

Section 2.3 · Properties

Simplest form: Examples 3–5

A clever substitution turns the messy expression into a single trig ratio, and a graph shows exactly where the answer is valid.

Miscellaneous examples

The classic trap: Example 6

sin⁻¹(sin x) is not always x. Watch the angle fold back into the principal range.

Real world · Chapter 2 in action

Turning measurements
back into angles.

Builders, sailors, engineers and your phone can easily measure lengths and ratios. What they often need is the angle. Here are six scenes where an inverse trig function gets it, and one where forgetting the principal range sends you the wrong way.

Ramps and roads Ladder safety Where is the boat?Best view of a statue Optical fibres Phone tilt Quick check

Scene 1 · tan⁻¹

How steep is a wheelchair ramp?

Builders measure a ramp by its rise and run. Accessibility guidelines allow at most 1 unit of rise for every 12 along. Since tan θ = rise/run, the angle is θ = tan⁻¹(rise/run).

Rise

Run

Angle θ = tan⁻¹(rise/run)

tan⁻¹ turns rise ÷ run into an angletan⁻¹(1/12) ≈ 4.76° the ramp limitprincipal range (−π/2, π/2) covers every slope

Scene 2 · cos⁻¹

Is the ladder at a safe angle?

A ladder is safest at about 75° to the ground. Too shallow and the foot slides out; too steep and it can tip backwards. Painters use the 1-in-4 rule: put the foot one quarter of the ladder's length from the wall. You know the length L and the distance d, so cos θ = d/L.

Ladder length L

Foot distance from wall d

Angle θ = cos⁻¹(d/L)

cos⁻¹(1/4) ≈ 75.5° the 1-in-4 ruledomain d/L is always in [0, 1]range [0, π] gives a sensible ground angle

Scene 3 · tan⁻¹ and the principal range

Where is the boat? Don't trust tan⁻¹ alone.

A lighthouse radar measures a boat at x km east and y km north. Its direction is an angle whose tangent is y/x. But tan⁻¹ only answers between −90° and 90°, the right half of the circle. Tap or drag on the sea to move the boat.

Boat position

tan⁻¹(y/x) says

True direction (anticlockwise from east)

(4, 3) and (−4, −3) share the ratio 3/4principal range (−π/2, π/2) = right half onlyatan2(y, x) adds 180° when x < 0

Scene 4 · tan⁻¹ − tan⁻¹

Where do you get the best view of a statue?

A 5 m statue stands on a pedestal, so its feet are 4 m and its head 9 m above your eyes. Too close and you stare up at the pedestal; too far and the statue looks tiny. Each elevation angle is a tan⁻¹, and the statue fills the difference.

Your distance x

Viewing angle θ = tan⁻¹(9/x) − tan⁻¹(4/x)

θ(x) = tan⁻¹(b/x) − tan⁻¹(a/x)best at x = √(ab) = √(4 × 9) = 6 msame idea best cinema seat, rugby conversion kicks

Scene 5 · sin⁻¹ and its domain

Why light stays trapped inside an optical fibre.

When light passes from glass into air it bends. Snell's law gives the new angle: sin θ₂ = 1.5 sin θ₁, so θ₂ = sin⁻¹(1.5 sin θ₁). That only works while 1.5 sin θ₁ ≤ 1. Beyond that, sin⁻¹ has no answer, and the light cannot escape.

Angle in the glass θ₁

1.5 × sin θ₁

θ₂ = sin⁻¹(1.5 sin θ₁)

domain of sin⁻¹ [−1, 1]critical angle sin⁻¹(1/1.5) ≈ 41.8°total internal reflection carries internet signals across oceans

Scene 6 · sin⁻¹

How your phone knows it's tilted.

A phone's accelerometer can't measure angles directly. It measures how much of gravity (9.8 m/s²) pulls along the screen's width. Tilt the phone by θ and that pull is 9.8 sin θ, so the spirit-level app works backwards with sin⁻¹.

Tilt the phone

Sensor reading a = 9.8 sin θ

App computes θ = sin⁻¹(a/9.8)

a/9.8 always lies in [−1, 1]principal range [−90°, 90°] covers every tiltface up or down? needs a second axis, like atan2

Scene 7 · Quick check

Which inverse do you need? Tap to flip.

For each situation, decide which inverse trig function turns the measurement into the angle, then tap the card to check.

Exercises 2.1, 2.2 and Miscellaneous · animated solutions

Every exercise,
solved visually.

Every question in Exercises 2.1, 2.2 and the miscellaneous exercise has a walkthrough here or in the Lesson's quiz (the multiple-choice ones). Pick a question from the chips above each player. Press Play to watch it solved, or step through with Next. Yellow boxes ask you to predict before the answer appears.

Exercise 2.1: principal values Folding: sin⁻¹(sin x)Triangle trick Exercise 2.2: identities Misc: proofs and equations

Exercise 2.1 · Q1–Q12

Find the principal values.

Every question is solved the same visual way: draw the ratio, find every angle that fits, and keep the one in the principal half. Q11 and Q12 add angles by turning around the circle.

Basics for thisspecial angles sin and cos on the circle tan, cot, sec, cosec principal ranges

Exercise 2.2 Q10, Q11, Q13 · Miscellaneous Q1, Q2

Fold the angle back into range.

Find the angle inside the principal range with the same ratio, and slide to it along the dashed line.

Basics for thisradians and full turns sine as height the folding graph

Exercise 2.2 Q12 · Miscellaneous Q3–Q7, Q13

Stack triangles to add angles.

Stack one triangle on the hypotenuse of the other. The angles add at the corner, and the slope of the outer ray is the tangent of the sum.

Basics for thisright-triangle ratios Pythagoras and special triangles the triangle trick

Exercise 2.2 · Q1–Q9

Prove it, then see it.

Each identity is proved step by step, then checked by graph: the two sides trace the same curve exactly inside the condition in the question, and split apart outside it.

Basics for thisidentities on the circle domain and range substitution table

Miscellaneous exercise · Q8–Q12, Q14

Prove by picture, solve by looking.

Q8–Q10 are identities: prove them, then watch both sides trace the same curve. Q11, Q12 and Q14 are equations: draw both sides and look for where they cross.

Basics for thistan, cot and their holes domain and range

Basics · start here

Build it up
from the roots.

Inverse trigonometry rests on a few older ideas. Here each one is explained from its roots: not just what it is, but why it exists and where it comes from. Every example and exercise links back here exactly where you need it, and a button takes you straight back to the problem.

1Angles and radianswhy 360°, and why mathematicians prefer radians 2sin and coswhy ratios, why a circle, and the mirror rules 3The values you needspecial triangles and reference angles 4Domain and rangewhat they mean, and why sin gives ℝ → [−1, 1] 5tan, cot, sec, cosecdivision by zero and the holes it makes 6What "inverse" meanswhy a window is needed, and the name "arcsin"

Step 1 · Angles and radians

An angle is an amount of turning, and there are two ways to count it.

Stand at the centre of a circle facing right (east) and turn. The angle is how far you turned. Anticlockwise counts as positive, clockwise as negative, and you may keep turning past a full circle. The only question is what unit to count the turning in.

1a

Degrees: a choice made by people

Degrees split one full turn into 360 equal parts. Nothing in nature forces 360. It comes from the ancient Babylonians, who counted in sixties, and it may echo the roughly 360 days of a year. It stuck because 360 divides evenly in so many ways: by 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120 and 180. Halves, thirds, quarters, sixths and eighths of a turn all come out as whole numbers.

Angle turnerDrag the slider or tap a preset. Watch the arc you sweep out on a circle of radius 1.

Turn by

In degrees

In radians (arc length on a radius-1 circle)

1b

The radian: let the circle measure itself

Here is a way to measure turning that does not depend on any human choice. Take the radius of the circle, bend it like a piece of string, and lay it along the edge. The angle at the centre that this one radius-length of arc makes is 1 radian. So a radian answers the question: "how many radius-lengths of arc did I sweep?"

Wrap the radius around the circlePress Play. Copies of the radius are laid end to end around the edge. Count how many fit.

Arc laid so far (in radius-lengths)

One radian

≈ 57.3°

One radius-length of arc. Slightly less than 60°: the equilateral triangle's 60° corner has a straight side one radius long, but the curved arc needs a bit less turning to reach that length.

Half a turn

π radians = 180°

Half the circumference is πr, which is π ≈ 3.14 radius-lengths. That is where π enters every angle in this chapter.

A full turn

2π radians = 360°

The whole circumference is 2πr, so a full turn is 2π ≈ 6.28 radius-lengths: six whole radii and a little bit more.

1c

Why the size of the circle doesn't matter

A radian is arc ÷ radius. Make the circle twice as big and the arc for the same angle also doubles, so the ratio stays the same. (The sectors are similar shapes, just zoomed.) That is why the radian is a property of the angle alone.

Same angle, any circleChange the radius. The arc length changes, but arc ÷ radius does not.

Angle

Radius r

Arc length s

Angle in radians = s ÷ r

θ (in radians) = arc length sradius rradians = degrees × π180

no unit length ÷ length, so a radian is a pure number30° = 30 × π/180 = π/63π/5 = 3/5 × 180° = 108°

1d

Why mathematicians prefer radians

Reason 1

An angle becomes a real number

Because a radian is length ÷ length, "π/6" is just the number 0.5236. So sin can take any real number as input, and sin⁻¹ can return a real number. That is what lets us write sin : ℝ → [−1, 1].

Reason 2

Formulas lose their clutter

Arc length is s = rθ and sector area is ½r²θ. In degrees they carry an awkward factor: s = πrθ/180. The 180 is left over from the Babylonians' choice.

Reason 3

For small angles, sin θ ≈ θ

In radians, a tiny arc is almost straight, so its height almost equals its length: sin 0.1 = 0.0998. In degrees that nice fact breaks: sin(0.1°) = 0.0017. Calculus (Chapter 5) depends on this.

Radians versus degrees near zeroBoth panels plot y = x and y = sin x. On the left x is in radians, on the right in degrees.

Probe x

In plain words

Degrees are like measuring a road in "laps of a stadium that someone once built": fine, but arbitrary. Radians measure the road with the circle's own ruler, its radius. Using the circle's own ruler makes every formula simpler, which is why this textbook writes π/6 and not 30°.

Watch out

Calculators have a DEG/RAD switch. sin(30) in DEG mode is 0.5; in RAD mode it is the sine of 30 radians, about −0.988. When you see π in an angle, you are in radians.

Next step: sin and cos →

Step 2 · sin and cos

sin and cos are the coordinates of a point on a circle.

To see why that definition is the right one, start where trigonometry started: with triangles.

2a

Why ratios? Because the shape fixes them

Take any right triangle with a 30° corner, big or small. All such triangles are similar: the same shape, scaled. So the ratio opposite ÷ hypotenuse is the same for all of them. That fixed number belongs to the angle, and we call it sin 30°. Likewise cos = adjacent ÷ hypotenuse.

Scale the triangleChange the size: the sides change, the ratios don't. Change the angle: the ratios change.

Angle θ

Size (hypotenuse)

opposite ÷ hypotenuse = sin θ

adjacent ÷ hypotenuse = cos θ

2b

Why a circle? Because triangles run out

A right triangle can only have corners between 0° and 90°, so the triangle definition cannot handle 120°, 300° or −45°. The fix is to put the triangle inside a circle of radius 1. Then the hypotenuse is 1, so opposite = sin θ and adjacent = cos θ: the point at angle θ is (cos θ, sin θ). That definition works for every angle, and agrees with the triangle for acute ones.

Class 10 view

A triangle with hypotenuse 1

sin θ = opp/1 = opp and cos θ = adj/1 = adj. Only works for 0 < θ < 90°.

Class 12 view

A point that goes round

For θ = 150° the point is up and to the left, so cos θ is negative and sin θ positive. The signs come out automatically.

From circle to wavePress Play. The point runs round the circle; its height is sin θ. Plot that height against the angle and the sine wave draws itself.

Angle θ

sin θ (height, orange)

cos θ (across, blue)

2c

Two facts that come free with the circle

Pythagoras on the circle

sin²θ + cos²θ = 1

The point (cos θ, sin θ) is on the circle x² + y² = 1, which is just Pythagoras for the radius. Substitute: cos²θ + sin²θ = 1, for every angle.

Signs by quadrant

Right is +cos, up is +sin

Quadrant I: all positive. II: only sin positive. III: only tan positive. IV: only cos positive. (A common memory aid: "All Students Take Coffee".)

2d

The mirror rules: why many angles share a value

The circle is symmetric, and each symmetry is an identity you will use all chapter long. Reflecting in the vertical axis keeps the height, so sin(π − θ) = sin θ. Reflecting in the horizontal axis keeps the across-distance, so cos(−θ) = cos θ. A half turn through the centre flips both, and a full turn changes nothing.

Mirror rulesPick a symmetry and an angle. Compare the two points' heights and across-distances.

Angle θ

The rule

In plain words

Imagine a seat on a Ferris wheel of radius 1 m. sin θ is how high the seat is above the centre and cos θ is how far it is to the right. The seat at 30° and the seat at 150° are at exactly the same height, one on each side. That is why sin 30° = sin 150°, and it is the whole reason sin needs a restricted window before it can be inverted.

Next step: the values you need →

Step 3 · Special angles

Where ½, √3/2 and 1/√2 come from.

Almost every answer in this chapter is built from five angles: 0, π/6, π/4, π/3 and π/2. Their values come from two triangles you can draw yourself, and a "reference angle" carries them round to every other quadrant. There is nothing to memorise blindly.

3a

Two triangles you can build

30°-60°-90°

Cut an equilateral triangle in half

An equilateral triangle of side 2 has all angles 60°. Its height cuts the base in half (1 and 1) and the top angle in half (30°). Pythagoras gives the height:

height = √(2² − 1²) = √3

So:

sin 30° = 12, cos 30° = √32

sin 60° = √32, cos 60° = 12

45°-45°-90°

Cut a square along its diagonal

A square of side 1 cut corner to corner gives two equal angles of 45°. Pythagoras gives the diagonal:

diagonal = √(1² + 1²) = √2

So:

sin 45° = cos 45° = 1√2

This is also written √22 (multiply top and bottom by √2) and is about 0.707.

θ	0	π/6 (30°)	π/4 (45°)	π/3 (60°)	π/2 (90°)
sin θ	0	1/2	1/√2	√3/2	1
cos θ	1	√3/2	1/√2	1/2	0
tan θ = sin ÷ cos	0	1/√3	1	√3	not defined (÷ 0)

Memory trick

Write sin for 0°, 30°, 45°, 60°, 90° as √0/2, √1/2, √2/2, √3/2, √4/2. That gives 0, ½, 1/√2, √3/2, 1. cos is the same list backwards, because cos θ = sin(90° − θ): the two acute angles of a right triangle swap opposite and adjacent.

3b

Reference angles: carrying the values into every quadrant

Drop a line from the point to the horizontal axis and you get a right triangle again. Its angle at the centre, measured from the nearest part of the horizontal axis, is the reference angle (always between 0° and 90°). The values of the angle are the values of the reference angle, with the signs of its quadrant. This is exactly how exercise answers like 2π/3 and −π/6 are found.

Find the reference angleTap an angle. The triangle to the horizontal axis appears, with its reference angle and the signs.

Reference angle

So the values are

Next step: domain and range →

Step 4 · Domain and range

sin takes any real number, but only gives values from −1 to 1.

To explain that sentence we need three words from Chapter 1, and a clear picture of each.

4a

Domain, codomain and range

Domain

What you may put in

For sin: every angle for which sin makes sense. It is written first in f : A → B.

Codomain

Where outputs are allowed to land

The set written after the arrow, f : A → B. It may be bigger than what actually comes out.

Range

What actually comes out

The set of outputs really produced. When range = codomain, the function is onto (Chapter 1).

4b

Why the domain of sin is all of ℝ

In radians, an angle is just a real number (Step 1). And every real number is an amount of turning you can actually do: 2.5 means turn 2.5 radius-lengths, −100 means turn 100 radius-lengths clockwise. Whatever you turn by, you land on some point of the circle, and that point has a height. So sin θ exists for every real θ: the domain is ℝ. The same is true of cos.

4c

Why the range of sin is exactly [−1, 1]

Two separate reasons, and you need both. Never outside: the point lies on a circle of radius 1, so its height can't be above the top (1) or below the bottom (−1). Every value inside: the point moves smoothly, without jumping, from the bottom to the top, so it passes through every height in between, like a lift that stops at every floor. Together, the range is exactly [−1, 1].

Can sin reach this value?Choose a target value c. The dashed level shows every point with that height, on the circle and on the wave.

Target value c

4d

Why the textbook writes sin : ℝ → [−1, 1]

You could write sin : ℝ → ℝ, but then the codomain ℝ would be bigger than the range, and sin would not be onto. Choosing the codomain [−1, 1] makes sin onto, which is one of the two conditions for an inverse. The other condition, one-one, fails: sin repeats every 2π, so each value is produced by infinitely many angles. Step 6 fixes that.

Domain of sin, cos

ℝ

You can turn by any real amount.

Range of sin, cos

[−1, 1]

The circle has radius 1, and the point passes every height.

Period 2π

Values repeat

sin(θ + 2π) = sin θ, so sin is many-one on ℝ.

Watch out

"Domain" is about what you put in (angles); "range" is about what comes out (ratios). For sin⁻¹ they swap: its inputs are ratios from [−1, 1] and its outputs are angles. That swap is the whole reason sin⁻¹ only accepts numbers between −1 and 1.

Next step: tan, cot, sec, cosec →

Step 5 · The other four ratios

tan, cot, sec and cosec are fractions of sin and cos, so they inherit holes.

Each of the other four is a fraction made from sin and cos. Wherever the bottom of the fraction is 0, the function does not exist. Those angles are removed from its domain.

5a

Why you can't divide by zero

Dividing means "undo multiplying": 6 ÷ 2 = 3 because 3 × 2 = 6. So 1 ÷ 0 would have to be a number that gives 1 when multiplied by 0. But anything times 0 is 0, never 1. No such number exists, so 1 ÷ 0 is undefined. It isn't a rule someone invented; there is simply no answer.

tan θ = sin θcos θcot θ = cos θsin θsec θ = 1cos θcosec θ = 1sin θ

cos θ = 0 at the top and bottom of the circle: θ = π/2, 3π/2, … = (2n + 1)π/2sin θ = 0 at the left and right: θ = 0, π, 2π, … = nπ

5b

What happens near a hole

Close to 90°, cos θ is tiny, and dividing by a tiny number gives a huge one. tan does not settle on any value: it grows without limit, then flips to huge negative numbers just past 90°. On a graph this shows up as a vertical asymptote, a line the curve races along but never touches.

θ	80°	89°	89.9°	89.99°	90°	90.1°
cos θ	0.174	0.0175	0.00175	0.000175	0	−0.00175
tan θ	5.67	57.3	573	5 730	not defined	−573

Where the holes arePick a function and slide the angle. Near a hole the value shoots off to ±∞; exactly at the hole it doesn't exist.

Angle θ

5c

Why sec and cosec skip the values between −1 and 1

cos θ always lies in [−1, 1]. Flipping a number that small, 1 ÷ (something between −1 and 1), always gives something at least 1 in size:

1 ÷ 0.5 = 2

1 ÷ 0.25 = 4

1 ÷ (−0.5) = −2

The only way to get exactly 1 or −1 is cos θ = ±1. So sec θ is never strictly between −1 and 1.

The range of sec

ℝ − (−1, 1)

The same holds for cosec.

Function	Built as	Undefined when	Domain	Range, and why
sin	height of the point	never	ℝ	[−1, 1]: the circle has radius 1
cos	across-distance of the point	never	ℝ	[−1, 1]: the circle has radius 1
tan	sin ÷ cos	cos θ = 0, θ = (2n + 1)π/2	ℝ − {(2n + 1)π/2}	ℝ: the ray can meet the line x = 1 at any height
cot	cos ÷ sin	sin θ = 0, θ = nπ	ℝ − {nπ}	ℝ: the ray can meet the line y = 1 anywhere
sec	1 ÷ cos	cos θ = 0	ℝ − {(2n + 1)π/2}	ℝ − (−1, 1): \|cos\| ≤ 1, so \|1/cos\| ≥ 1
cosec	1 ÷ sin	sin θ = 0	ℝ − {nπ}	ℝ − (−1, 1): \|sin\| ≤ 1, so \|1/sin\| ≥ 1

This is exactly the table the textbook lists in Section 2.2. Now you know where every entry comes from. One more fact: tan and cot repeat every π (not 2π), because the opposite point through the centre flips the signs of both sin and cos, and their ratio stays the same.

Next step: what "inverse" means →

Step 6 · Inverses

An inverse runs the machine backwards, but only if nothing collides.

A function is a machine: angle in, ratio out. Its inverse would be the reverse machine: ratio in, angle out.

6a

What an inverse must do

From Chapter 1: g is the inverse of f when g undoes f and f undoes g. In symbols, g(f(x)) = x for every input and f(g(y)) = y for every output. Such a g exists exactly when f is one-one (no two inputs share an output, so the way back is never ambiguous) and onto (every output is used, so the way back always has an answer).

Reverse the arrowsStep through: first sin sends angles to values, then we try to send the values back.

6b

Why this particular window?

Many windows would make sin one-one and onto [−1, 1]; [π/2, 3π/2] works too. The textbook's choice, [−π/2, π/2], is used worldwide for three reasons.

It is one piece

No gaps, no repeats

From −π/2 to π/2 the height climbs steadily from −1 to 1, hitting every value exactly once.

It contains the acute angles

0 to π/2 is inside

So the answers agree with right-triangle trigonometry: sin⁻¹(½) = π/6, the familiar 30°.

It is balanced

Symmetric about 0

Negative inputs give the matching negative angles: sin⁻¹(−x) = −sin⁻¹ x.

6c

Where the name "arcsin" comes from

On a circle of radius 1, an angle in radians equals the length of its arc (Step 1). So sin⁻¹ x literally answers: "how long is the arc whose sine is x?". That is why it is also written arcsin x, and why calculators label the key "asin". The "−1" is inverse-function notation, like f⁻¹. It does not mean 1 ÷ sin x.

Rule 1

Domain of the inverse = range of the original

sin produces values in [−1, 1], so sin⁻¹ can only accept inputs from [−1, 1].

Rule 2

Range of the inverse = the window we kept

Because we kept [−π/2, π/2], sin⁻¹ only gives answers from [−π/2, π/2].

Rule 3

The other five work the same way

Each function gets its own window, chosen for the same three reasons. The Lesson shows all six.

In plain words

If three students all have the same roll number, the teacher can't call "number 5" and know who stands up. sin gives the same value to infinitely many angles, so we agree to look at only one stretch of angles, where every value appears once. That agreement is the principal value branch.

Now start the chapter →

Trig functions repeat, so they can't be undone.

Cut out one good piece of the wave.

Principal branch [−π/2, π/2]

Principal branch [0, π]

They contain the familiar angles

Every inverse is a mirror image.

Where the answers live on the circle.

Find the principal values.

Undo, then redo, but mind the window.

sin(sin−1 x) = x

sin−1(sin x) = x

Turn an inverse into a right triangle.

Spot the pattern, substitute, simplify.

Exercise 2.2 answers

Mixed practice.

Seven exam-style questions.

Remember this.

A shadow, a staff and a pyramid.

Principal values: Examples 1–2

Simplest form: Examples 3–5

The classic trap: Example 6

How steep is a wheelchair ramp?

Is the ladder at a safe angle?

Where is the boat? Don't trust tan⁻¹ alone.

Where do you get the best view of a statue?

Why light stays trapped inside an optical fibre.

How your phone knows it's tilted.

Which inverse do you need? Tap to flip.

Find the principal values.

Fold the angle back into range.

Stack triangles to add angles.

Prove it, then see it.

Prove by picture, solve by looking.

An angle is an amount of turning, and there are two ways to count it.

Degrees: a choice made by people

The radian: let the circle measure itself

≈ 57.3°

π radians = 180°

2π radians = 360°

Why the size of the circle doesn't matter

Why mathematicians prefer radians

An angle becomes a real number

Formulas lose their clutter

For small angles, sin θ ≈ θ

sin and cos are the coordinates of a point on a circle.

Why ratios? Because the shape fixes them

Why a circle? Because triangles run out

A triangle with hypotenuse 1

A point that goes round

Two facts that come free with the circle

sin²θ + cos²θ = 1

Right is +cos, up is +sin

The mirror rules: why many angles share a value

Where ½, √3/2 and 1/√2 come from.

Two triangles you can build

Cut an equilateral triangle in half

Cut a square along its diagonal

Reference angles: carrying the values into every quadrant

sin takes any real number, but only gives values from −1 to 1.

Domain, codomain and range

What you may put in

Where outputs are allowed to land

What actually comes out

Why the domain of sin is all of ℝ

Why the range of sin is exactly [−1, 1]

Why the textbook writes sin : ℝ → [−1, 1]

ℝ

[−1, 1]

Values repeat

tan, cot, sec and cosec are fractions of sin and cos, so they inherit holes.

Why you can't divide by zero

What happens near a hole

Why sec and cosec skip the values between −1 and 1

An inverse runs the machine backwards, but only if nothing collides.

What an inverse must do

Why this particular window?

No gaps, no repeats

0 to π/2 is inside

Symmetric about 0

Where the name "arcsin" comes from

sin(sin⁻¹ x) = x

sin⁻¹(sin x) = x