iGCSE (2021 Edition)

# 7.04 Reciprocal relationships

Lesson

### Reciprocal relationships

The unit circle definitions of the trigonometric functions began with sine and cosine: other functions are derived from these, starting with the tangent function, which we know is $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ. Here are three other functions that we will utilise, particularly to simplify expressions involving trigonometric identities.

Reciprocal trigonometric functions

The secant function is the reciprocal of the cosine function:

$\sec\theta=\frac{1}{\cos\theta}$secθ=1cosθ

The cosecant function is the reciprocal of the sine function:

$\operatorname{cosec}\theta=\frac{1}{\sin\theta}$cosecθ=1sinθ

The cotangent function is the reciprocal of the tangent function:

$\cot\theta=\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}$cotθ=1tanθ=cosθsinθ

To remember the pairs look at the third letter of the new functions cosec, sec and cot. It will match to the first letter of its trigonometric pair - sin, cos, tan.

It is possible to derive the values of all six of the trigonometric functions if we are given the value of any one and information regarding the quadrant within which the angle lies. The reciprocal ratios have the same signs as their related ratios across the four quadrants. Our knowledge of exact values will now be extended to the three new functions.

### Exact values sec, cosec, cot

Consider the diagram below showing a point on the unit circle.

From the coordinates, and our knowledge of exact values in the second quadrant, we have:

 $\cos120^\circ$cos120° $=$= $-\frac{1}{2}$−12​

And:

 $\sin120^\circ$sin120° $=$= $\frac{\sqrt{3}}{2}$√32​

From these we can find values for the other four trigonometric functions:

 $\tan120^\circ$tan120° $=$= $-\sqrt{3}$−√3 $\sec120^\circ$sec120° $=$= $-2$−2 $\operatorname{cosec}120^\circ$cosec120° $=$= $\frac{2}{\sqrt{3}}$2√3​ $\cot120^\circ$cot120° $=$= $-\frac{1}{\sqrt{3}}$−1√3​

#### Practice questions

##### Question 1

Consider the graph of the unit circle shown below.

1. State the value of $\sin30^\circ$sin30°.

2. Hence, state the value of $\operatorname{cosec}30^\circ$cosec30°.

##### Question 2

If $\cos\theta=\frac{7}{18}$cosθ=718, what is the value of $\sec\theta$secθ?

### Complementary relationships

There are connections between the trigonometric functions that can make simplifications possible.

In a right-angled triangle, the two acute angles together make a right-angle. We say the acute angles are complementary (to one another). If the two acute angles have measures $\alpha$α and $\beta$β, then $\alpha+\beta=90^\circ$α+β=90° and so, $\beta=90^\circ-\alpha$β=90°α.

The diagram below illustrates the following relationships.

$\cos\alpha=\frac{b}{h}=\sin\beta=\sin\left(90^\circ-\alpha\right)$cosα=bh=sinβ=sin(90°α)

$\sin\alpha=\frac{a}{h}=\cos\beta=\cos\left(90^\circ-\alpha\right)$sinα=ah=cosβ=cos(90°α)

$\cot\alpha=\frac{b}{a}=\tan\beta=\tan\left(90^\circ-\alpha\right)$cotα=ba=tanβ=tan(90°α)

Thus, the 'co' in complementary explains the meaning of cosine in relation to sine, and to cotangent in relation to tangent.

If we take reciprocals of both sides of the complementary relationships linking sine and cosine, we can also see that:

$\operatorname{cosec}\theta=\sec\left(90^\circ-\theta\right)$cosecθ=sec(90°θ)

And:

$\sec\theta=\operatorname{cosec}\left(90^\circ-\theta\right)$secθ=cosec(90°θ)

In summary:

Complementary trigonometric functions

$\cos\alpha\equiv\sin\left(90^\circ-\alpha\right)$cosαsin(90°α) and $\sin\alpha\equiv\cos\left(90^\circ-\alpha\right)$sinαcos(90°α)

$\cot\alpha\equiv\tan\left(90^\circ-\alpha\right)$cotαtan(90°α) and $\tan\alpha\equiv\cot\left(90^\circ-\alpha\right)$tanαcot(90°α)

$\operatorname{cosec}\alpha\equiv\sec\left(90^\circ-\alpha\right)$cosecαsec(90°α) and $\sec\alpha\equiv\operatorname{cosec}\left(90^\circ-\alpha\right)$secαcosec(90°α)

These statements are called identities because they are true whatever the value of the angle $\alpha$α. This can be confirmed by thinking about the geometry in the unit circle diagram that is used for defining the trigonometric functions of angles of any magnitude.

#### Practice questions

##### Question 3

Complete the blank in the following expression:

1. $\cos$cos$\editable{}$$^\circ$°= $\sin59^\circ$sin59°

##### QUESTION 4

Simplify $\cos\left(90^\circ-\theta\right)\operatorname{cosec}\left(180^\circ+\theta\right)$cos(90°θ)cosec(180°+θ).

### Graphs of reciprocal trigonometrical functions

Since $\sec x=\frac{1}{\cos x}$secx=1cosx, it follows that the graph of $y=\sec x$y=secx is created by taking the reciprocal of every function value of the graph of $y=\cos x$y=cosx. The same applies for the graphs of $y=\operatorname{cosec}x$y=cosecx and $y=\cot x$y=cotx.

The graph of each function is drawn below.

 Graph of $y=\operatorname{cosec}x$y=cosecx

 Graph of $y=\sec x$y=secx

 Graph of $y=\cot x$y=cotx

Some of the features of these graphs are listed below:

• All three of these reciprocal trigonometric functions have asymptotes. These occur at points where the relevant parent function $(\sin x$(sinx ,$\cos x$cosx or $\tan x)$tanx) has a $y$y-value of zero. For example, $\sec x=\frac{1}{\cos x}$secx=1cosx is undefined at $x=90^\circ$x=90° or at $x=270^\circ$x=270°, and so on, because at these points $\cos x=0$cosx=0.
• Because the parent functions are periodic, the reciprocal functions are also all periodic, with the same periods as their respective originals.
• Since $\operatorname{cosec}x$cosecx and $\sec x$secx are reciprocals of the functions $\sin x$sinx and $\cos x$cosx, the reciprocal functions can never take $y$y-values between $y=-1$y=1 and $y=1$y=1. Also, these reciprocal functions do not have any upper or lower bounds. This is because as one of the original function values approaches zero, the value of its reciprocal function becomes arbitrarily large.
• You should notice that $\cot x$cotx can attain any value, since $\tan(x)$tan(x) can do likewise.
• Perhaps the most interesting feature of the $\cot(x)$cot(x) graph is that it passes through the $x$x-axis at the values where $\tan(x)$tan(x) had asymptotes. This can be simply demonstrated by considering how $\tan x=\frac{\sin x}{\cos x}$tanx=sinxcosx has its asymptotes when $\cos x=0$cosx=0. When this is substituted into the numerator of $\cot\theta=\frac{\cos\theta}{\sin\theta}$cotθ=cosθsinθ, the function will always be equal to zero. And so, $\cot(x)$cot(x) is defined at these points which will specify the locations of the $x$x-intercepts.

#### Practice questions

##### Question 5

Consider the graph of $y=\sec x$y=secx below.

1. When $x=45^\circ$x=45°, $y=\sqrt{2}$y=2.

What is the next positive $x$x-value for which $y=\sqrt{2}$y=2?

2. What is the period of the graph?

3. What is the smallest value of $x$x greater than $360^\circ$360° for which $y=\sqrt{2}$y=2?

4. What is the first $x$x-value less than $0^\circ$0° for which $y=\sqrt{2}$y=2?

##### Question 6

Consider the angle $270^\circ$270°.

1. What are the coordinates of the point on the unit circle that corresponds to $270^\circ$270°?

2. Find the value of $\cos270^\circ$cos270°.

3. Find the value of $\sin270^\circ$sin270°.

4. Which of the following are undefined? Choose all correct answers.

$\tan270^\circ$tan270°

A

$\cot270^\circ$cot270°

B

$\csc270^\circ$csc270°

C

$\sec270^\circ$sec270°

D

$\tan270^\circ$tan270°

A

$\cot270^\circ$cot270°

B

$\csc270^\circ$csc270°

C

$\sec270^\circ$sec270°

D

### Outcomes

#### 0606C10.1

Know the six trigonometric functions of angles of any magnitude (sine, cosine, tangent, secant, cosecant, cotangent).