Inverse Trigonometric Functions

Why the inverse of sine needs a rule first

In Class 10 you found angles from ratios. A pole is 6 m tall, and its shadow is 6 m long. Find the angle of the sun. The tan of that angle is 6 over 6, which is 1. You knew the angle whose tan is 1, and wrote 45°. There was one answer, and you never doubted it.

Your calculator does the same thing. Type 0.5 and press the $\sin^{-1}$ key. In degree mode it shows 30. One ratio goes in, and one angle comes out.

But many angles share a sine. Give it an angle and you get one height. Now ask the question backwards. Which angle has this height? Many do. $\sin 30^\circ = 1/2$, and $\sin 150^\circ = 1/2$ as well. Go once more round the circle and $390^\circ$ gives $1/2$ again. The list never ends.

Why sine cannot simply be reversed

In Chapter 1 you saw when a function can be run backwards. It must be one-one and onto. Sine is not one-one, and neither are cosine and tangent. Each one repeats its values for ever. In radians, $\sin 0 = \sin \pi = \sin 2\pi = 0$.

So which one does the calculator show? It shows 30, never 150 or 390. Everyone has agreed on one branch of angles to answer from.

The principal branch

Here is that agreement, written as a definition. $\sin^{-1} x$ is the angle whose sine is $x$, taken from the agreed branch $-\pi/2$ to $\pi/2$. This is the PRINCIPAL BRANCH. On it, each height belongs to one angle, so the inverse has one answer. Each trigonometric function gets its own branch. On it the function must reach every value it can take. It must reach each value only once.

Here is the question for the whole chapter. The sine of an angle is always one number. The angles behind a sine are many. Which one do you read? By the last page you will know the principal branch of all six functions. You will also know which rules still hold on those branches.

0.5

→

sin⁻¹ key

→

π/6

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Before you start

You met every one of these facts once already, in Class 11. Try each check below.

• Radian measure of an angle (Class 11, Mathematics (NCERT Class 11), page 45).
  Here: the branch that defines sine inverse is itself an interval of radians, never degrees.
  Check: $2\pi$ radians make one full turn, the same $360^\circ$ you already know.

• Standard values of sine, cosine, tangent (Class 11, Mathematics (NCERT Class 11), page 51).
  Here: each worked example below undoes one of these standard values, the opposite way round.
  Check: $\sin(\pi/6) = 1/2$.

• Sign of sine by quadrant (Class 11, Mathematics (NCERT Class 11), page 52).
  Here: the branch chosen for sine inverse, $[-\pi/2, \pi/2]$, takes the fourth-quadrant angles, where sine is negative, and the first-quadrant ones, where it is positive.
  Check: $\sin(\pi/4) > 0$ and $\sin(-\pi/4) < 0$, and both angles lie in the branch.

• Domain and range of sine and cosine (Class 11, Mathematics (NCERT Class 11), page 52).
  Here: cosine inverse accepts exactly cosine’s own range as its input, the interval from negative one to one.
  Check: cosine’s range is $[-1, 1]$, so every value cosine inverse accepts lies there.

• Periodicity of sine and cosine (Class 11, Mathematics (NCERT Class 11), page 50).
  Here: sine’s repeating every $2\pi$ is exactly why a single sine value can trace back to infinitely many angles, not one.
  Check: $\sin(0) = 0$ and $\sin(2\pi) = 0$: one output, already two of infinitely many inputs.

• Domain and range of tangent (Class 11, Mathematics (NCERT Class 11), page 53).
  Here: tangent inverse is defined for every real number because tangent’s own range already covers every real number.
  Check: $\tan(0) = 0$, a real number, and tangent inverse accepts any real number back.

If any of these felt new, read the page named before going on.

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The principal branches

Picture a ladder leaning on a wall. The ladder is 4 metres long. Its top touches the wall 2 metres up. How steep is it? Height over length gives 2 over 4, which is a half. A half is the sine of 30°. So the ladder makes 30° with the ground.

But 150° also has a sine of one half. Could the ladder lean at 150°? Not against this wall. A ladder leans somewhere between lying flat and standing upright. You never even think about the other angles. You read the one that makes sense.

Writing the habit down

Mathematics writes that habit down. Look at the sine curve from $-\pi/2$ to $\pi/2$. Over that branch it climbs once, from $-1$ at the left end to $1$ at the right end. It never turns back. So every height from $-1$ to $1$ is reached, and reached only once. On that branch sine is one-one and onto $[-1, 1]$.

domain

range

This interval, $[-\pi/2, \pi/2]$, is the PRINCIPAL BRANCH of sine. The INVERSE SINE undoes sine on this branch. It takes a height and gives the one angle on the branch with that height. In symbols, $\sin^{-1}: [-1, 1] \to [-\pi/2, \pi/2]$. So $\sin^{-1} y$ is the one $x$ in $[-\pi/2, \pi/2]$ with $\sin x = y$. *Read $\sin^{-1} x$ aloud* as “the angle whose sine is $x$”. It is an angle, never a ratio.

*Whatever sits inside $\sin^{-1}$* must land in $[-1, 1]$. Take $\sin^{-1}(2x - 1)$. 1. The condition: $-1 \leq 2x - 1 \leq 1$, since that is what $\sin^{-1}$ needs. 2. Add $1$ to all three parts: $0 \leq 2x \leq 2$. 3. Halve all three parts: $0 \leq x \leq 1$. So the domain of $\sin^{-1}(2x - 1)$ is $[0, 1]$.

π/2

→

sin

→

sin⁻¹

→

π/2

0

→

sin

→

sin⁻¹

→

0

−π/2

→

sin

→

sin⁻¹

→

−π/2

Let us check two values. $\sin(\pi/2) = 1$, and $\pi/2$ is the right end of the branch. So $\sin^{-1}(1) = \pi/2$. Next, $\sin 0 = 0$, and $0$ sits in the middle. So $\sin^{-1}(0) = 0$.

range

-1

Why both ends are allowed

Both ends belong to the branch, and the square brackets say so. An answer of exactly $\pi/2$ or $-\pi/2$ is allowed. Every other angle with the same sine lies outside the branch. It is simply not a candidate. The ladder’s 30° is $\pi/6$, and it is on the branch. So $\sin^{-1}(1/2) = \pi/6$.

The branch is what makes the inverse exist at all. With it, your calculator’s inverse-sine key gives one answer, and you know which one.

Your turn: what is $\sin^{-1}(1)$? Check that your angle is on the branch before you write it down.

Cosine needs its own branch

*$\cos^{-1} x$* is the angle whose cosine is $x$. Many angles share a cosine, just as many share a sine. So cosine needs an agreed branch too. Sine’s branch, $-\pi/2$ to $\pi/2$, will not do.

On sine’s branch, cosine repeats itself. $\cos(-x) = \cos x$, so $-x$ and $x$ always give the same height. For example, $\cos(-\pi/3)$ and $\cos(\pi/3)$ are both $1/2$. Cosine also never goes below $0$ there. So no angle on that branch gives $-1/2$.

Cosine falls once from 0 to π

Now take the angles from $0$ to $\pi$. $\cos 0 = 1$, $\cos 90^\circ = 0$ and $\cos 180^\circ = -1$. On $[0, \pi]$ cosine falls the whole way, from $1$ down to $-1$. It reaches every value in $[-1, 1]$, and each one only once.

So cosine gets its own PRINCIPAL BRANCH, $[0, \pi]$. On sine’s branch, cosine is neither one-one nor onto $[-1, 1]$. On $[0, \pi]$ it is both.

domain

range

Defining the inverse cosine

The INVERSE COSINE is $\cos^{-1}: [-1, 1] \to [0, \pi]$. It undoes cosine on that branch: $\cos^{-1} y$ is the one angle $x$ in $[0, \pi]$ with $\cos x = y$. The branch starts at $0$, so no answer is ever negative.

Check the two ends. $\cos 0 = 1$, and $0$ is in $[0, \pi]$. So $\cos^{-1}(1) = 0$. $\cos \pi = -1$, and $\pi$ is the far end. So $\cos^{-1}(-1) = \pi$. Both ends are allowed.

sin⁻¹

cos⁻¹

0

0

Each inverse gets a branch chosen for its own function. When you meet $\cos^{-1}$, think “0 to $\pi$” before anything else.

Your turn: what is $\cos^{-1}(1)$? Before you answer, say which branch your angle must come from.

The slope is a tangent

The slope of a ramp is the tangent of its angle. A slope of 1 means $\tan 45^\circ = 1$. As the angle grows from $-\pi/2$ towards $\pi/2$, the tangent climbs from far below zero to far above it. It reaches every real number, and each one only once.

A ramp rises 1 metre for every 1 metre it runs along the ground. Its slope is 1. A wheelchair ramp is much gentler than that. A slide in a park is steeper. Slope is rise over run, and a slope can be any number at all. Flat ground has slope 0. A road going downhill has a negative slope. A very steep climb might have slope 10 or 100.

Yet the angle of a ramp never reaches upright. A wall’s slope cannot be written as a number, because its run is zero.

The two ends are left out. At $-\pi/2$ and at $\pi/2$ the tangent has no value, just as a wall’s slope has none. So tangent’s PRINCIPAL BRANCH is open at both ends: $(-\pi/2, \pi/2)$. On it, $\tan$ is one-one and onto all of $\mathbb{R}$.

domain

range

The INVERSE TANGENT is $\tan^{-1}: \mathbb{R} \to (-\pi/2, \pi/2)$. It undoes tangent on that branch. $\tan^{-1} y$ is the one angle $x$ in $(-\pi/2, \pi/2)$ with $\tan x = y$.

0

→

tan⁻¹

→

0

1

→

tan⁻¹

→

π/4

Two quick checks. $\tan 0 = 0$, so $\tan^{-1}(0) = 0$. $\tan(\pi/4) = 1$, and $\pi/4$ is inside the branch. So $\tan^{-1}(1) = \pi/4$, the ramp with slope 1.

Every real number is a legal input to $\tan^{-1}$, which sine and cosine do not have. You never need a domain check on $\tan^{-1}$ itself. Only what you put inside it must be defined. But an answer of exactly $\pi/2$ is always wrong.

Your turn: a ramp has slope $1000$. Is its angle $\pi/2$? Say why, using the branch.

Three more functions, built as reciprocals

Cotangent, secant and cosecant are reciprocals. $\cot x = 1/(\tan x)$, $\sec x = 1/(\cos x)$ and $\operatorname{cosec} x = 1/(\sin x)$. Cosine and sine never go beyond $1$ in size. So secant and cosecant never come below $1$ in size. No value strictly between $-1$ and $1$ is ever reached. Cotangent is different: $\cot x = \cos x/\sin x$, a ratio that passes through every real number, including values between $-1$ and $1$. So $\cot^{-1}$ takes every real number as its input, unlike $\sec^{-1}$ and $\operatorname{cosec}^{-1}$.

Four friends share one pizza. Each friend gets a quarter of a pizza. Turn the fact around: four quarters make one pizza, so the reciprocal of a quarter is four.

Take the reciprocal of a small number and you get a big one. Take the reciprocal of a number between minus one and one, other than zero, and you land outside that range. So the reciprocal of a number that started small can never end up small.

That gives the inverses their domains. $\sec^{-1}$ and $\operatorname{cosec}^{-1}$ take only inputs with $|x| \geq 1$. Their domain is $\mathbb{R} - (-1, 1)$. *Check $|x| \geq 1$ first,* every time.

domain

range

Three branches, three ranges

Each one also gets a PRINCIPAL BRANCH where it is one-one. $\sec^{-1}$ has range $[0, \pi] - \{\pi/2\}$, because $\sec$ has no value at $\pi/2$. $\operatorname{cosec}^{-1}$ has range $[-\pi/2, \pi/2] - \{0\}$, because $\operatorname{cosec}$ has no value at $0$. $\cot^{-1}: \mathbb{R} \to (0, \pi)$, and it never touches $0$ or $\pi$.

2

→

sec⁻¹

→

π/3

2

→

1/x

→

cos⁻¹

→

π/3

Let us try one. $\cos(\pi/3) = 1/2$, so $\sec(\pi/3) = 2$. And $\pi/3$ is on secant’s branch. So $\sec^{-1}(2) = \pi/3$.

−√3

→

cot⁻¹

→

5π/6

−√3

→

1/x

→

tan⁻¹

→

−π/6

Board papers ask about these three much less than the first three. You can turn them back into the first three by taking the reciprocal of the input. $\sec^{-1} x = \cos^{-1}(1/x)$ for $|x| \geq 1$. In the same way, $\operatorname{cosec}^{-1} x = \sin^{-1}(1/x)$ for $|x| \geq 1$. For $\cot^{-1}$ the reciprocal needs a sign check. $\cot^{-1} x = \tan^{-1}(1/x)$ only for $x > 0$. For $x < 0$ it is $\pi + \tan^{-1}(1/x)$. For example, $\cot^{-1}(-1) = \pi + (-\pi/4) = 3\pi/4$, which is inside $(0, \pi)$.

So you carry three branches in your head, not six. The other three come from a reciprocal.

Your turn: find $\cot^{-1}(-1/\sqrt{3})$. Which of the two rules for $\cot^{-1}$ do you need?

−1/√2

→

drop the sign

→

cos⁻¹

→

π minus

→

3π/4

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Graphs of the inverse functions

Make a table with two columns, angle and sine. One row reads 30° and one half. Another reads 90° and 1. Now swap the two columns. The new table starts from a height and gives an angle. That swapped table is the inverse.

Swapping is a mirror

Plot both tables on graph paper. Every point moves to the place with its two numbers swapped. The point (3, 5) would move to (5, 3). Swapping like this is a mirror. The mirror is the diagonal line through the origin.

*That line is $y = x$. A function and its inverse are always mirror images in it. You saw this in Chapter 1. Here it works* on three curves you already know.

Let us test one point. $(\pi/6, 1/2)$ is on sine’s branch. Swap it and you get $(1/2, \pi/6)$, which is on the graph of $\sin^{-1} x$.

Three shapes, mirrored

The graph of $\sin^{-1} x$ rises from $(-1, -\pi/2)$ through the origin to $(1, \pi/2)$. It climbs the whole way, like sine’s own branch on $[-\pi/2, \pi/2]$. The graph of $\cos^{-1} x$ falls from $(-1, \pi)$ to $(1, 0)$. It falls the whole way, like cosine’s branch on $[0, \pi]$. A mirror keeps rising curves rising and falling curves falling.

*$\tan^{-1} x$ is different.* It climbs across all of $\mathbb{R}$, but it never reaches $\pi/2$ or $-\pi/2$. The lines $y = \pi/2$ and $y = -\pi/2$ are horizontal asymptotes. The curve flattens towards them as $x \to \infty$ and $x \to -\infty$, and never touches either.

The other three follow their own branches. $\cot^{-1} x$ falls from $\pi$ towards $0$ across all of $\mathbb{R}$. $\sec^{-1} x$ and $\operatorname{cosec}^{-1} x$ each break into two pieces. Nothing is drawn between $-1$ and $1$.

So if a branch slips your memory, sketch the function’s branch and reflect it in $y = x$.

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When sine inverse of sine is not x

In Chapter 1, an inverse undoes its function: $f^{-1}(f(x)) = x$. With sine, the undo works only on the branch. Start from $3\pi/4$, which is off it. Its sine is $1/\sqrt{2}$, and $\sin^{-1}(1/\sqrt{2}) = \pi/4$. You come back with $\pi/4$, not $3\pi/4$.

Check one that works. Take $x = \pi/6$. Its sine is $1/2$, and $\sin^{-1}(1/2) = \pi/6$. *You get $x$ again,* because $\pi/6$ was on the branch.

π/6

→

sin

→

sin⁻¹

→

π/6

3π/4

→

sin

→

sin⁻¹

→

π/4

−π/3

→

cos

→

cos⁻¹

→

π/3

3π/4

→

tan

→

tan⁻¹

→

−π/4

sin⁻¹

cos⁻¹

tan⁻¹

0

The same check applies to the other two. $\cos^{-1}(\cos x) = x$ only for $x$ in $[0, \pi]$. And $\tan^{-1}(\tan x) = x$ only for $x$ in $(-\pi/2, \pi/2)$. Each one carries its own branch.

*When $x$ is off the branch,* bring it back first. 1. Sine: $\sin^{-1}(\sin x) = \pi - x$ for $x$ in $[\pi/2, 3\pi/2]$, and $x - 2\pi$ for $x$ in $[3\pi/2, 5\pi/2]$. For example, $4\pi/5$ is in $[\pi/2, 3\pi/2]$, so $\sin^{-1}(\sin(4\pi/5)) = \pi - 4\pi/5 = \pi/5$. 2. Cosine: $\cos^{-1}(\cos x) = 2\pi - x$ for $x$ in $[\pi, 2\pi]$. For example, $3\pi/2$ is in $[\pi, 2\pi]$, so $\cos^{-1}(\cos(3\pi/2)) = 2\pi - 3\pi/2 = \pi/2$. 3. Tangent: $\tan^{-1}(\tan x) = x - k \pi$, with the integer $k$ chosen so the answer lands in $(-\pi/2, \pi/2)$. For example, $2\pi/3$ needs $k = 1$, so $\tan^{-1}(\tan(2\pi/3)) = 2\pi/3 - \pi = -\pi/3$.

3π/4

→

π minus x

→

sin⁻¹ of sin

→

π/4

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Two different meanings for the same exponent

A car has a gear marked R. It does not make the car go slower. It does not turn the speed into one over the speed. It runs the car backwards, along the road it came by. The R names a reverse.

The undo mark, in symbols

Sine is a function too. An angle goes in and a height comes out. Run it backwards and a height goes in and an angle comes out. Mathematics marks that undo with a small raised minus one.

2

→

f: 3x−2

→

f⁻¹

→

2

π/6

→

sin

→

sin⁻¹

→

π/6

So $\sin^{-1}$ is the INVERSE SINE, sine run backwards. It is the same $-1$ that names $f^{-1}$ in Chapter 1.

The trouble is that a raised $-1$ on a number does mean “one over”. $2^{-1} = 1/2$. So $\sin^{-1} x$ looks as if it means $1/(\sin x)$. It does not. The number $1/(\sin x)$ already has its own name. It is $\operatorname{cosec} x$.

1/2

→

sin⁻¹

→

≈0.52

1/2

→

sin

→

1/x

→

≈2.09

Putting numbers on both

Put numbers on both. $\sin^{-1}(1/2) = \pi/6 \approx 0.52$. That is an angle in radians. Now work out $1/(\sin(1/2))$. Here $1/2$ is an angle of half a radian. Its sine is about $0.48$, and one over that is about $2.09$. That is a plain ratio. Different kinds of answer, and far apart.

Try it yourself. Is $\cos^{-1}(1/2)$ an angle or a number? And $1/(\cos(1/2))$?

This gives you a quick test. An inverse trigonometric answer is always an angle. A reciprocal is always a ratio. Ask which one the question wants.

Your turn: is $\sin^{-1}(1)$ equal to $1/(\sin 1)$? Work out both, in radians, and compare.

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Complementary angle identities

Go back to the ladder against the wall. The ladder, the wall and the ground make a right-angled triangle. Its other two angles add up to 90°. One sits at the foot of the ladder. The other sits at the top, against the wall.

Reading one number two ways

The ladder was 4 metres long, with its top 2 metres up. At the foot, height over length is 2 over 4. That is the sine of the foot angle, 30°. At the top, the same 2 over 4 is the cosine of the top angle, 60°. One number, read as a sine at one corner and as a cosine at the other. The two angles add to 90°.

That is a COMPLEMENTARY pair: two angles that fill a right angle. In radians, $\sin^{-1} x + \cos^{-1} x = \pi/2$ for every $x$ in $[-1, 1]$.

Let us see why it holds for every such $x$, negatives too. 1. Call $\theta = \sin^{-1} x$. Then $\theta$ is in $[-\pi/2, \pi/2]$. 2. So $\pi/2 - \theta$ is in $[0, \pi]$, which is cosine’s branch. 3. And $\cos(\pi/2 - \theta) = \sin \theta = x$, the Class 11 co-function rule. 4. So $\cos^{-1} x = \pi/2 - \theta$, the definition of $\cos^{-1}$ on its branch. 5. Add $\theta$ to both sides: $\sin^{-1} x + \cos^{-1} x = \pi/2$.

1/2

→

sin⁻¹

→

+ cos⁻¹(1/2)

→

π/2

0

→

sin⁻¹ + cos⁻¹

→

π/2

−1

→

sin⁻¹ + cos⁻¹

→

π/2

A quick check at $x = 0$. $\sin^{-1}(0) = 0$ and $\cos^{-1}(0) = \pi/2$. They add to $\pi/2$.

sin⁻¹, cos⁻¹

sin⁻¹cos⁻¹

tan⁻¹, cot⁻¹

tan⁻¹cot⁻¹

sec⁻¹, cosec⁻¹

sec⁻¹cosec⁻¹

sin⁻¹ + cos⁻¹

tan⁻¹ + cot⁻¹

sec⁻¹ + cosec⁻¹

Three pairs, same rule

Two more pairs work the same way, each on its own domain.

• $\tan^{-1} x + \cot^{-1} x = \pi/2$, for every real $x$.
• $\sec^{-1} x + \operatorname{cosec}^{-1} x = \pi/2$, only for $|x| \geq 1$.

Each is proved by the same step as the first pair. 1. Tangent and cotangent: call $\theta = \tan^{-1} x$, in $(-\pi/2, \pi/2)$. Then $\pi/2 - \theta$ is in $(0, \pi)$, cotangent’s branch, and $\cot(\pi/2 - \theta) = \tan \theta = x$. So $\cot^{-1} x = \pi/2 - \theta$. 2. Secant and cosecant: for $|x| \geq 1$, call $\theta = \sec^{-1} x$. Then $\pi/2 - \theta$ is in cosecant’s branch, and $\operatorname{cosec}(\pi/2 - \theta) = \sec \theta = x$. So $\operatorname{cosec}^{-1} x = \pi/2 - \theta$.

The secant and cosecant pair needs $|x| \geq 1$, because neither inverse exists between $-1$ and $1$. The branches do not break any of these rules. They only fix which angle each side names.

This saves you work. Given one member of a pair, the other is $\pi/2$ minus it. You never compute the harder one.

Your turn: check $\tan^{-1} x + \cot^{-1} x = \pi/2$ at $x = -1$. Does it still hold for a negative input?

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Negating the input, two different ways

A thermometer scale runs from minus 10 to 10, with 0 in the middle. Change a reading’s sign to put it on the other side of zero. 3 becomes minus 3, and it is still on the scale.

Two kinds of scale

Now take a test marked out of 100. Change the sign of 30 to get minus 30 and you fall off the scale. On this scale, taking it from 100 is what stays on: 30 becomes 100 minus 30, which is 70.

sin⁻¹

0

0

cos⁻¹

0

0

A scale centred on zero changes sign. A scale that starts at zero takes the reading away from the top. The six inverse functions split the same way, three and three.

−1/2

→

sin⁻¹

→

−π/6

1/2

→

sin⁻¹

→

minus

→

−π/6

Three branches centred on zero

Three branches are centred on $0$. Negating the input just negates the answer.

• $\sin^{-1}(-x) = -\sin^{-1} x$, for $x$ in $[-1, 1]$.
• $\tan^{-1}(-x) = -\tan^{-1} x$, for every real $x$.
• $\operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1} x$, for $|x| \geq 1$.

−1/2

→

cos⁻¹

→

2π/3

1/2

→

cos⁻¹

→

π minus

→

2π/3

Three branches that start at zero

Three branches start at $0$. Negating the input takes the answer away from $\pi$.

• $\cos^{-1}(-x) = \pi - \cos^{-1} x$, for $x$ in $[-1, 1]$.
• $\cot^{-1}(-x) = \pi - \cot^{-1} x$, for every real $x$.
• $\sec^{-1}(-x) = \pi - \sec^{-1} x$, for $|x| \geq 1$.

1/2

→

cos⁻¹

→

minus

→

−π/3

−1/2

→

cos⁻¹

→

2π/3

Why can cosine, secant and cotangent not use a minus sign? A minus sign would give a negative angle, and their branches hold no negative angles at all. Here is the step that works instead. 1. Cosine: call $\theta = \cos^{-1} x$, in $[0, \pi]$. Then $\cos(\pi - \theta) = -\cos \theta = -x$, and $\pi - \theta$ is in $[0, \pi]$ too, so it is the branch angle with cosine $-x$: $\cos^{-1}(-x) = \pi - \theta = \pi - \cos^{-1} x$. 2. Secant: the same step gives $\sec^{-1}(-x) = \pi - \sec^{-1} x$, since $\sec(\pi - \theta) = -\sec \theta$ and $\pi - \theta$ stays off $\pi/2$ whenever $\theta$ does. 3. Cotangent: the same step gives $\cot^{-1}(-x) = \pi - \cot^{-1} x$, since $\cot(\pi - \theta) = -\cot \theta$ and $\pi - \theta$ is in $(0, \pi)$ whenever $\theta$ is.

Let us check one. $\cos^{-1}(1/2) = \pi/3$. So $\cos^{-1}(-1/2) = \pi - \pi/3 = 2\pi/3$. That is inside $[0, \pi]$, as it must be.

centred on 0: minus sign

sin⁻¹tan⁻¹cosec⁻¹

starts at 0: π minus

cos⁻¹cot⁻¹sec⁻¹

Sort the six by shape once, and you have two rules to hold instead of six: a minus sign for the three centred on 0, and $\pi$ minus for the three that start at 0.

Your turn: find $\cos^{-1}(-1)$ with the rule. What would a plain minus sign have given?

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The addition formula for tangent inverse

Turn a camera on its stand through 30°. Then turn it a further 15°. In all it has turned 45°. Angles add. Slopes do not. A 30° slope is about 0.58, and a 15° slope is about 0.27. They add to about 0.85. But a 45° slope is exactly 1.

So if you know two slopes and want the slope of the two angles added together, you need a rule. You met it in Class 11 as the tangent addition formula.

['1/2', '1/3']

→

add the two tan⁻¹

→

π/4

['1/2', '1/3']

→

(x + y)/(1 − xy)

→

tan⁻¹

→

π/4

The addition formula, for inverses

Written for inverses, it is the ADDITION FORMULA for tangent inverse. For $x y < 1$, $\tan^{-1} x + \tan^{-1} y = \tan^{-1}((x + y)/(1 - x y))$.

The condition $x y < 1$ is part of the formula. Call $\alpha = \tan^{-1} x$ and $\beta = \tan^{-1} y$. The reason is in the sign of $\tan(\alpha + \beta) = (x + y)/(1 - x y)$, which splits into three cases. 1. $x, y$ both positive: $x y < 1$ keeps $1 - x y$ positive, so $\tan(\alpha + \beta)$ is positive; since $\alpha + \beta$ is between $0$ and $\pi$, that puts it under $\pi/2$, inside tangent’s branch, where $\tan^{-1}$ undoes $\tan$. 2. $x, y$ opposite signs: one angle is above $0$ and the other below it, so the sum stays inside the branch on its own. 3. $x, y$ both negative: the same argument as case 1 runs in mirror image, below $0$. So the formula holds for every pair with $x y < 1$.

A pair that breaks the formula

Let us try a pair that breaks it. Take $x = y = 2$. Then $x y = 4$. Each $\tan^{-1}(2)$ is about $1.107$, so the true sum is about $2.214$. That is more than $\pi/2$. The fraction gives $(2 + 2)/(1 - 4) = -4/3$, and $\tan^{-1}(-4/3)$ is about $-0.927$. The formula’s answer is off by $\pi$.

['2', '2']

→

add the two tan⁻¹

→

≈2.21

['2', '2']

→

(x + y)/(1 − xy)

→

tan⁻¹

→

≈−0.93

*So when $x y > 1$* the plain formula lands in the wrong place. A $+ \pi$ brings it back when $x$ and $y$ are both positive, and a $- \pi$ when both are negative. At $x y = 1$ the bottom of the fraction is $0$, and the formula gives nothing at all.

['1/2', '1/3']

→

xy

→

1/6

*So work out $x y$* on its own line first. Then decide whether the plain formula applies.

Your turn: you want $\tan^{-1}(3) + \tan^{-1}(1/2)$. Work out $x y$ first. Can you use the formula as it stands?

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Inverse trig functions in integrals

You learn a new word in an English lesson. A year later it turns up in a science paper. You do not have to learn it again. You only have to recognise it. The inverse functions in these pages will turn up again in the same way. Their new role seems to have nothing to do with angles.

sin⁻¹ x

→

derivative, Chapter 5

→

integral, Chapter 7

→

sin⁻¹ x + C

They come back as answers to integrals. In Chapter 5 you will find that the derivative of $\sin^{-1} x$ is $1/\sqrt{1 - x^2}$. In Chapter 7 you will run that backwards. You will integrate $1/\sqrt{1 - x^2}$ and get $\sin^{-1} x + C$.

You need no integration here.

√(1 − x²)

sin⁻¹cos⁻¹

1 + x²

tan⁻¹cot⁻¹

x√(x² − 1)

sec⁻¹cosec⁻¹

A square root of $1 - x^2$ in the bottom of a fraction is a strong hint. An inverse trigonometric function is hiding behind it.

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Restriction, and checking the branch first

The question at the start was this. The sine of an angle is one number, but many angles share it. Which one do you read? You read the one on the principal branch. $\sin^{-1} x$ is the angle whose sine is $x$, taken from that branch. Each function has its own branch, and on it the inverse has one answer. Each rule works only where its condition holds.

• PRINCIPAL BRANCHES: $\sin^{-1}$ answers from $[-\pi/2, \pi/2]$, $\cos^{-1}$ from $[0, \pi]$, $\tan^{-1}$ from $(-\pi/2, \pi/2)$. So $\cos^{-1}(-1) = \pi$, never $-\pi$.
• UNDO: $\sin^{-1}(\sin x) = x$ only for $x$ in $[-\pi/2, \pi/2]$. So $\sin^{-1}(\sin(3\pi/4)) = \pi/4$.
• NOT A RECIPROCAL: $\sin^{-1}(1/2) = \pi/6 \approx 0.52$, while $1/(\sin(1/2)) \approx 2.09$.
• COMPLEMENTARY PAIRS: $\sin^{-1} x + \cos^{-1} x = \pi/2$ on $[-1, 1]$. At $x = 1/2$, $\pi/6 + \pi/3 = \pi/2$.
• NEGATIVE INPUTS: $\cos^{-1}(-x) = \pi - \cos^{-1} x$. So $\cos^{-1}(-1/2) = 2\pi/3$.
• ADDITION: $\tan^{-1} x + \tan^{-1} y = \tan^{-1}((x + y)/(1 - x y))$ for $x y < 1$. So $\tan^{-1}(1/2) + \tan^{-1}(1/3) = \pi/4$.

1 m, 12 m

→

rise ÷ run

→

tan⁻¹

→

≈4.8°

Both halves of the answer are the Chapter 1 idea, used on real curves. A function can be run backwards only where it is one-one and onto. The branch is where that is true. And every identity here carries a condition. Skip the check and you get a wrong answer that looks right. So fix the branch first, then check the condition, every time.

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Where you will meet this

You have already guessed at angles from ratios without knowing the name for the step.

• A ramp’s angle from its rise and run. A wheelchair ramp’s rise and its horizontal run give a ratio of two measured lengths.
  The maths: the inverse tangent turns that rise-over-run ratio back into the ramp’s own angle.
  A ramp rising $1$ m over a $12$ m run has angle $\tan^{-1}(1/12) \approx 4.8^\circ$.

• A ladder’s lean against a wall. A ladder’s length and the distance of its foot from the wall give another such ratio.
  The maths: the inverse cosine recovers the lean angle from the ratio of foot-distance to length.
  A $5$ m ladder with its foot $3$ m from the wall leans at $\cos^{-1}(3/5) \approx 53^\circ$.

• The sun’s elevation from a shadow. A flagpole’s height and the length of its shadow give the sun’s elevation angle.
  The maths: the same ratio-to-angle step as the ramp finds the sun’s elevation, through the inverse tangent.
  A $6$ m pole with a $6$ m shadow gives elevation $\tan^{-1}(6/6) = 45^\circ$.

• A ship’s bearing from two distances. A navigator plots how far north and how far east a ship has sailed, as two separate distances.
  The maths: for a leg that runs north and east, the inverse tangent of the east-distance over the north-distance gives the bearing angle from north.
  Sailing $4$ km north and $3$ km east gives a bearing $\tan^{-1}(3/4) \approx 36.9^\circ$ east of north.

• A fielder’s throw angle. A fielder’s throw speed and the range it covers fix two possible release angles, a low one and a high one.
  The maths: halving the inverse sine of a ratio built from range and speed recovers the lower of the two release angles.
  Range $20$ m at speed $15$ m/s needs angle $\theta = 1/2 \sin^{-1}(R g/v^2)$, with $g = 9.8$ m/s²: $1/2 \sin^{-1}(20 \times 9.8/15^2) \approx 30.3^\circ$.

• A camera’s field of view. A camera’s frame width and its distance from the subject give two measured lengths.
  The maths: half the field of view is an inverse tangent of half the frame width over that distance.
  A $36$ mm frame at $50$ mm distance gives a half-angle $\tan^{-1}(18/50) \approx 19.8^\circ$.

Your turn. A kite string is $50$ m long and the kite sits $30$ m above the ground. Which inverse trig function gives the string’s angle with the ground, and from what ratio?

Answer: The inverse sine of $30/50$, since sine is the opposite side over the hypotenuse, the string itself.

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Practice set: Exercise 2.1

Exercise 2.1 is about principal values. You evaluate all six inverse functions at standard values like $1/2$, $1/\sqrt{2}$ and $\sqrt{3}/2$, and at their negatives. Every answer must come from the principal branch.

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Practice set: Exercise 2.2

Exercise 2.2 practises the properties. You simplify $\sin^{-1}(\sin x)$ when $x$ is outside the branch. You also use the complementary pairs, and you add two or three inverse tangents with the formula.

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Practice set: miscellaneous

The Miscellaneous Exercise mixes the whole chapter. A typical problem reduces an angle into its branch, then uses an identity, then simplifies. Nobody tells you which identity to reach for first.