iGCSE (2021 Edition)

Lesson

In this lesson, we will be taking the skills studied previously and using radians as a measure of an angle instead of degrees.

### Radian mode on the calculator

To evaluate trigonometric functions that use radians on the calculator, we need to set the calculator to be in radian mode. This is very important for getting the correct answer when evaluating expressions such as $\sin3.2^c$sin3.2c where the angle is in radians.

Different scientific calculators will have their own instructions to do this. For the Casio fx-82AU Plus calculator, you can change the mode by pressing: shift - setup - 4.

Alternatively, we could convert the radian angle to degrees first to avoid having to change the calculator to radian mode.

Careful!

Always check to see that your calculator is in the right mode, either degree or radians, when doing trigonometric calculations.

On the top of your calculator screen, it may say "D" for degree mode or "R" for radian mode.

#### Practice questions

##### Question 1

Evaluate $\sin\frac{35\pi}{16}$sin35π16 correct to two decimal places.

##### Question 2

Evaluate $\cos6.87$cos6.87, where the angle given is measured in radians. Give your answer correct to two decimal places.

### Exact trigonometric ratios in radians

In the last chapter we found the exact trigonometric ratios for angles $30^\circ$30°$60^\circ$60° and $45^\circ.$45°.

We can now express these trigonometric ratios in radians rather than degrees:

 $30^\circ=\frac{\pi}{6}$30°=π6​ $45^\circ=\frac{\pi}{4}$45°=π4​ $60^\circ=\frac{\pi}{3}$60°=π3​

The two triangles that give exact trigonometric ratios can be drawn using radians instead of degrees.

From the diagram we can read off the following function values:

 $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$sinπ3​=√32​ $\sin\frac{\pi}{6}=\frac{1}{2}$sinπ6​=12​ $\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}$sinπ4​=1√2​ $\cos\frac{\pi}{3}=\frac{1}{2}$cosπ3​=12​ $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$cosπ6​=√32​ $\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}$cosπ4​=1√2​ $\tan\frac{\pi}{3}=\sqrt{3}$tanπ3​=√3 $\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}$tanπ6​=1√3​ $\tan\frac{\pi}{4}=1$tanπ4​=1

#### Practice questions

##### Question 3

Find the exact value of $\sin\frac{\pi}{3}$sinπ3.

##### Question 4

Evaluate $\sin\frac{\pi}{6}\cos\frac{\pi}{4}$sinπ6cosπ4, leaving your answer in exact form with a rational denominator.

### Angles of any magnitude using radians

The unit circle can be expressed in radians as shown in this diagram:

The steps in finding the value of a trigonometric function for any angle remain the same, i.e. identifying the quadrant, finding the related acute angle, and working out the trigonometric function for the related acute angle. It is just a matter of getting used to thinking in radians instead of degrees!

#### Practice questions

##### Question 5

Find the reference angle for $\frac{2\pi}{3}$2π3.

##### Question 6

Find the exact value of $\cos\left(-\frac{5\pi}{3}\right)$cos(5π3)

##### Question 7

By rewriting each ratio in terms of the related acute angle, evaluate the expression:

$\frac{\left(\sin\frac{2\pi}{3}\right)\left(\cos\frac{2\pi}{3}\right)\left(\tan\frac{3\pi}{4}\right)}{\tan\left(-\frac{\pi}{4}\right)}$(sin2π3)(cos2π3)(tan3π4)tan(π4)

When sketching trigonometric functions using radians, we can simply change the values on the horizontal axis from degrees to radians. The domain also changes accordingly.

Now that we are working in radians it is easier to graph other functions together with trigonometric functions in order to solve equations graphically. This is because we are no longer using degrees but radians on the $x$x-axis, meaning that other functions of $x$x are possible.

#### Practice questions

##### Question 8

Consider the graph of $y=\sin x$y=sinx given below.

1. Using the graph, what is the sign of $\sin\frac{13\pi}{12}$sin13π12?

Positive

A

Negative

B

Positive

A

Negative

B
2. Which quadrant does an angle with measure $\frac{13\pi}{12}$13π12 lie in?

Quadrant $I$I

A

Quadrant $II$II

B

Quadrant $IV$IV

C

Quadrant $III$III

D

Quadrant $I$I

A

Quadrant $II$II

B

Quadrant $IV$IV

C

Quadrant $III$III

D

##### Question 9

Consider the identity $\sec x=\frac{1}{\cos x}$secx=1cosx and the table of values below.

 $x$x $\cos x$cosx $0$0 $\frac{\pi}{4}$π4​ $\frac{\pi}{2}$π2​ $\frac{3\pi}{4}$3π4​ $\pi$π $\frac{5\pi}{4}$5π4​ $\frac{3\pi}{2}$3π2​ $\frac{7\pi}{4}$7π4​ $2\pi$2π $1$1 $\frac{1}{\sqrt{2}}$1√2​ $0$0 $-\frac{1}{\sqrt{2}}$−1√2​ $-1$−1 $-\frac{1}{\sqrt{2}}$−1√2​ $0$0 $\frac{1}{\sqrt{2}}$1√2​ $1$1
1. For which values of $x$x in the interval $\left[0,2\pi\right]$[0,2π] is $\sec x$secx not defined?

Write all $x$x-values on the same line separated by commas.

2. Complete the table of values:

 $x$x $\sec x$secx $0$0 $\frac{\pi}{4}$π4​ $\frac{\pi}{2}$π2​ $\frac{3\pi}{4}$3π4​ $\pi$π $\frac{5\pi}{4}$5π4​ $\frac{3\pi}{2}$3π2​ $\frac{7\pi}{4}$7π4​ $2\pi$2π $\editable{}$ $\editable{}$ undefined $\editable{}$ $\editable{}$ $\editable{}$ undefined $\editable{}$ $\editable{}$
3. What is the minimum positive value of $\sec x$secx?

4. What is the maximum negative value of $\sec x$secx?

5. Plot the graph of $y=\sec x$y=secx on the same set of axes as $y=\cos x$y=cosx.