M5.3.6 Chapter 6

Reciprocal and inverse trig functions

What are sec, cosec and cot, and what does arcsin x mean?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A sin⁻¹ x means the angle whose sine is x.
You picked A. Read on. At the end, the chapter comes back to your guess.
B sin⁻¹ x means 1 ÷ sin x.
You picked B. Read on. At the end, the chapter comes back to your guess.
C sin⁻¹ x means the same as sin x, written another way.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Sketch the graphs of sec, cosec and cot
Use 1 + tan²x = sec²x
Give the domain and range of arcsin x

Three more trig functions are the reciprocals of the first three. Reciprocal means 1 divided by it. We define sec x = 1/cos x and cosec x = 1/sin x. We define cot x = cos x/sin x, which is 1/tan x wherever tan x is defined and not 0. Each is defined only where its bottom is not 0.

Where is each defined? The function sec x needs cos x ≠ 0, so x cannot be π/2, 3π/2 and so on. The function cosec x needs sin x ≠ 0, so x cannot be 0, π, 2π and so on. The same holds for cot x. At x = π/2, cot x is defined and equals 0, even though tan x is not defined there.

To sketch the graph of a reciprocal, start with its parent. Where the parent is 1, the reciprocal is 1. Where the parent is −1, so is the reciprocal. Where the parent is 0, the reciprocal has an asymptote. Where the parent is small, the reciprocal is large. The parent and its reciprocal always have the same sign.

y = sec x (solid) from y = cos x (dashed). The asymptotes are where cos x = 0. The curve never enters the strip between −1 and 1.

Two facts follow. Since cos x lies between −1 and 1, the size of sec x = 1/cos x is at least 1. So the range of sec x is y ≤ −1 or y ≥ 1. And the period is 2π, like cos x. The graph has U shaped branches that open up where cos is positive and open down where it is negative.

y = cosec x (solid) from y = sin x (dashed). The asymptotes are at x = 0, π and 2π, where sin x = 0.

y = cot x. It has period π and asymptotes at x = 0, π and 2π. It falls from left to right, and it is 0 at π/2 and 3π/2.

Now two new identities. Start with sin²x + cos²x ≡ 1 and divide every term by cos²x. This gives tan²x + 1 ≡ sec²x. Divide by sin²x instead, and we get 1 + cot²x ≡ cosec²x. The first needs cos x ≠ 0, and the second needs sin x ≠ 0. Learn both: they turn tan into sec, and cot into cosec, in equations.

Worked example

Find the exact values of sec 120°, cosec 150° and cot 135°.

Find the ordinary ratio first, then take the reciprocal.
cos 120° = −1/2, because 120° is in the second quadrant with reference angle 60°.
So sec 120° = 1 ÷ (−1/2) = −2.
sin 150° = 1/2, because 150° has reference angle 30° and sin is positive in the second quadrant. So cosec 150° = 1 ÷ (1/2) = 2.
tan 135° = −1, because the reference angle is 45° and tan is negative in the second quadrant. So cot 135° = 1 ÷ (−1) = −1.

Answer: sec 120° = −2, cosec 150° = 2, cot 135° = −1.

Worked example

Given that tan θ = −3/4 and θ is obtuse, find sec θ and cos θ.

Use 1 + tan²θ ≡ sec²θ. Here tan²θ = (−3/4)² = 9/16.
So sec²θ = 1 + 9/16 = 25/16.
Then sec θ = 5/4 or sec θ = −5/4, because both have a square of 25/16.
θ is obtuse, so it is in the second quadrant, where cos is negative. So sec θ is negative too: sec θ = −5/4.
Then cos θ = 1 ÷ sec θ = 1 ÷ (−5/4) = −4/5.

Answer: sec θ = −5/4 and cos θ = −4/5.

The acute angle φ has tan φ = 3/4 and sec φ = 5/4. The obtuse angle θ = 180° − φ has the same sizes with the signs of the second quadrant.

Exam questions combine the new identities with equations. Replace tan²x by sec²x − 1, or cot²x by cosec²x − 1, so that only one function is left. Then factorise. Remember that |sec x| ≥ 1 and |cosec x| ≥ 1: a factor that gives a value between −1 and 1 has no solution.

Worked example

Solve 3 tan²x + 5 sec x + 1 = 0 for 0 ≤ x ≤ 2π. Give exact answers.

Replace tan²x by sec²x − 1. The equation becomes 3(sec²x − 1) + 5 sec x + 1 = 0.
Expand and collect: 3 sec²x − 3 + 5 sec x + 1 = 0, so 3 sec²x + 5 sec x − 2 = 0.
This is a quadratic in sec x. Factorise: (3 sec x − 1)(sec x + 2) = 0.
So sec x = 1/3 or sec x = −2.
sec x = 1/3 has no solution, because |sec x| ≥ 1. We reject it.
sec x = −2 gives cos x = −1/2. The reference angle is π/3, and cos is negative in the second and third quadrants.
So x = π − π/3 = 2π/3 or x = π + π/3 = 4π/3.

Answer: x = 2π/3 or x = 4π/3

Now the inverse functions. The function sin x takes many angles to the same value. For example sin 30° = sin 150° = 1/2. A function has an inverse only if each output comes from exactly one input. So we cut sin x down to one stretch where it takes each value once: from −π/2 to π/2. On that stretch it rises from −1 to 1 and covers every value once.

On the solid part, from −π/2 to π/2, sin x takes each value from −1 to 1 exactly once. That part has an inverse.

The inverse of this restricted sine is arcsin, also written sin⁻¹. arcsin x is the angle between −π/2 and π/2 whose sine is x. Its domain is the set of x that go in: −1 ≤ x ≤ 1. Its range is the set of values that come out: −π/2 ≤ y ≤ π/2. Its graph is the restricted sine reflected in the line y = x.

y = arcsin x: domain −1 ≤ x ≤ 1, range −π/2 ≤ y ≤ π/2.

The same idea gives the others. cos x is cut to 0 ≤ x ≤ π, where it falls from 1 to −1. Its inverse arccos x has domain −1 ≤ x ≤ 1 and range 0 ≤ y ≤ π. tan x is cut to −π/2 < x < π/2, where it rises through every real value. Its inverse arctan x has domain all real numbers and range −π/2 < y < π/2.

y = arccos x: domain −1 ≤ x ≤ 1, range 0 ≤ y ≤ π. It falls from π to 0.

y = arctan x: domain all real numbers, range −π/2 < y < π/2. The lines y = π/2 and y = −π/2 are asymptotes.

The ranges matter. A calculator, and arcsin, arccos and arctan, give one answer each, taken from the range. So arcsin(sin x) = x only when x is between −π/2 and π/2. For other x, the answer is the angle in the range that has the same sine.

Worked example

Find the exact values of arcsin(−1/2), arccos(−1/2) and arctan 1 in radians.

arcsin(−1/2) is the angle between −π/2 and π/2 whose sine is −1/2. sin(π/6) = 1/2, and the sine is negative below the axis. So the angle is −π/6.
arccos(−1/2) is the angle between 0 and π whose cosine is −1/2. cos(π/3) = 1/2, and cos is negative in the second quadrant. So the angle is π − π/3 = 2π/3.
arctan 1 is the angle between −π/2 and π/2 whose tan is 1. tan(π/4) = 1, so the angle is π/4.

Answer: −π/6, 2π/3 and π/4

Worked example

Find the exact value of arcsin(sin(2π/3)).

First sin(2π/3). The angle is in the second quadrant, with reference angle π/3. So sin(2π/3) = √3/2.
Now arcsin(√3/2) is the angle between −π/2 and π/2 whose sine is √3/2. That angle is π/3.
So arcsin(sin(2π/3)) = π/3, not 2π/3. The angle 2π/3 is outside the range of arcsin.

Answer: π/3, which is 1.047

Worked example

Find the exact value of sin(arccos(3/5)).

Let θ = arccos(3/5). Then cos θ = 3/5, and θ is between 0 and π.
sin θ is not negative between 0 and π, so sin θ = +√(1 − cos²θ).
cos²θ = 9/25, so sin²θ = 1 − 9/25 = 16/25.
So sin θ = 4/5.

Answer: 4/5

Class 12 courses also use the other three inverses. The inverse sec⁻¹ x has domain |x| ≥ 1 and range 0 ≤ y ≤ π, with y ≠ π/2. The inverse cosec⁻¹ x has domain |x| ≥ 1 and range −π/2 ≤ y ≤ π/2, with y ≠ 0. The inverse cot⁻¹ x has domain all real numbers and range 0 < y < π. These are the usual principal ranges, so check the ranges your course uses.

Some results hold wherever both sides are defined. arcsin x + arccos x = π/2 for −1 ≤ x ≤ 1. To see this, let y = arcsin x, so sin y = x. Then cos(π/2 − y) = x, and π/2 − y lies between 0 and π, so arccos x = π/2 − y. Also arcsin(−x) = −arcsin x, and arccos(−x) = π − arccos x.

For the tangent, arctan x + arctan y = arctan((x + y)/(1 − xy)) when xy < 1. It comes from tan(A + B) = (tan A + tan B)/(1 − tan A tan B). When xy > 1 and x and y are positive, the sum is more than π/2, and it equals π + arctan((x + y)/(1 − xy)). When xy > 1 and both are negative, take away π instead. The next example shows this.

Worked example

Find the exact value of arctan 2 + arctan 3.

Let A = arctan 2 and B = arctan 3. Both are acute, so A + B lies between 0 and π.
tan(A + B) = (2 + 3)/(1 − 2 × 3) = 5/(−5) = −1.
Between 0 and π, the angle with tangent −1 is 3π/4. So A + B = 3π/4.
The formula arctan((x + y)/(1 − xy)) would give arctan(−1) = −π/4, which is wrong. It needs xy < 1, and here xy = 6.

Answer: 3π/4

See it three ways

With things

A lift takes you from a floor number to a height. To go back, from a height to a floor number, you reverse it. That reverse is the inverse. It works only if each height belongs to just one floor.

As a picture

In words

arcsin x is the angle between −π/2 and π/2 whose sine is x.

In symbols

y = arcsin x means sin y = x with −π/2 ≤ y ≤ π/2. Domain −1 ≤ x ≤ 1.

Try it

Find an angle with the inverse key

You need: a calculator, paper.

Set the calculator to radian mode. Press sin⁻¹ 0.5 and write the answer. It should be 0.5236, which is π/6.
Press sin⁻¹ of the sine of 2.5 radians. First work out sin 2.5, then press sin⁻¹ of that answer.
Compare the answer with 2.5.
Subtract: π − 2.5 = 0.6416. Compare with your answer.

What you will see: The calculator gives 0.6416, not 2.5. The angle 2.5 is outside the range of arcsin, so the inverse key gives the angle in the range with the same sine, which is π − 2.5.

If you cannot try it, this is what happens.

A wrong idea many people have

sin⁻¹ x is 1/sin x.

Why it feels right: In algebra, a number to the power −1 is its reciprocal. So x⁻¹ = 1/x, and it seems that sin⁻¹ should be the same.

The test that breaks it: Put x = 0.5. Then 1/sin 0.5 = 1/0.4794 = 2.086. But the calculator key sin⁻¹ gives sin⁻¹ 0.5 = 0.5236. These are different numbers, and the key is the inverse function.

What is true: sin⁻¹ x means arcsin x, the inverse function. The reciprocal is written (sin x)⁻¹ or cosec x or 1/sin x. In this book we write arcsin to keep the two apart.

A wrong idea many people have

arcsin(sin x) = x for every x.

Why it feels right: A function and its inverse undo each other, so they should cancel.

The test that breaks it: Take x = 2π/3. sin(2π/3) = √3/2, and arcsin(√3/2) = π/3. The answer is π/3, not 2π/3.

What is true: arcsin(sin x) = x only when −π/2 ≤ x ≤ π/2. For other x, arcsin(sin x) is the angle in that range with the same sine. On the other side, sin(arcsin x) = x for every x from −1 to 1.

A wrong idea many people have

The range of sec x is −1 ≤ y ≤ 1, like cos x.

Why it feels right: sec x is built from cos x, so the values may look alike.

The test that breaks it: Put x = π/3. Then cos x = 1/2, and sec x = 1 ÷ (1/2) = 2, which is outside −1 to 1. Dividing 1 by a number of size at most 1 gives a number of size at least 1.

What is true: The range of sec x is y ≤ −1 or y ≥ 1. The graph has no points in the strip between y = −1 and y = 1. The same holds for cosec x. cot x takes every real value.

Key words

reciprocal: 1 divided by a number. The reciprocal of 2 is 1/2.
inverse function: the function that undoes another. arcsin is the inverse of sin on −π/2 ≤ x ≤ π/2.
domain: the set of values that can go into a function.
range: the set of values that can come out of a function.

Back to your guess

The answer is A: sin⁻¹ x means the angle whose sine is x.

sin⁻¹ x means the angle whose sine is x. It is the inverse function of sin, also written arcsin x. The reciprocal 1 ÷ sin x has its own name: cosec x. The same small −1 is used for both ideas in different places, which is a common cause of mistakes.

Keep it

Copy this into your exercise book.

sec = 1/cos, cosec = 1/sin, cot = 1/tan. 1 + tan²x = sec²x, 1 + cot²x = cosec²x. arcsin: domain −1 to 1, range −π/2 to π/2. arccos range 0 to π. arctan range −π/2 to π/2.

Draw: The graph of sec x with its asymptotes and the cos x curve dashed beneath it.