# Sin Graph

Here we will learn about sine graphs, including how to recognise the graph of the sine function, sketch the sine curve and label important values, and interpret the sine graph.

There are also sine graphs worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What is the sin graph?

The sin graph is a visual representation of the sine function for a given range of angles.

The horizontal axis of a trigonometric graph represents the angle, usually written as \theta , and the y -axis is the sine function of that angle.

• The graph does not start at the origin but it does pass through it.
• The graph is continuous and  repeats every 360^o
• The maximum value is 1 and the minimum value is -1 .

We will only look at the graph of the sine function in this lesson for angles in degrees although this can also be represented in radians.

## How to plot the sin graph

Remember that \sin(\theta) is a relationship between the opposite side and the hypotenuse of a right angle triangle:

Let’s look at 3 triangles where we would use the sine ratio to calculate the size of the angle \theta . For each triangle, the hypotenuse is the same but the length of the opposite side and the associated angle change.

Here we can see that as \sin(\theta)=\frac{opp}{hyp} , as the angle \theta increases, the length of the side opposite to the angle also increases. So for each triangle we have:

• Triangle 1: \sin(\theta)=\frac{2}{10}=0.2
• Triangle 2: \sin(\theta)=\frac{6}{10}=0.6
• Triangle 3: \sin(\theta)=\frac{9}{10}=0.9

So what would happen if the opposite side to the angle is equal to 10

\sin(\theta)=\frac{10}{10}=1

So when the opposite side is equal to the hypotenuse, we get \sin(\theta) = 1 .

What about when the opposite side is equal to 0 ?

\sin(\theta)=\frac{0}{10}=0

So when the opposite side is equal to 0, \; \sin(\theta) = 0.

If we plotted a graph to show the value of \sin(\theta) for each value of \theta between 0^o and 90^o , we get the following graph of the sine function:

Let us add the values of \sin(\theta) for the three triangles from earlier into the graph to show how they would look:

We can now use the graph to find the angle \theta for triangles 1, 2, and 3 :

This graph shows that when \sin(\theta) = 0.2 , \; \theta = 12^o so we have the triangle

This graph shows that when \sin(\theta) = 0.6 , \; \theta = 37^o so we have the triangle

This graph shows that when \sin(\theta) = 0.9 , \; \theta = 64^o so we have the triangle

This shows us that we can use the graph of the sine function to find missing angles in a triangle. More on this later as we have a large problem to resolve. How can the opposite side of a right angle triangle be the same as the hypotenuse, or equal to 0 ?

Unfortunately there is a limit to the use of trigonometric ratios to find angles between 0 and 90^o . For any larger or smaller angles, we need to look at the unit circle.

### Unit circle

The unit circle is a circle of radius 1 with the centre at the origin. We can label the values where the circle intersects the axes because we know that the radius of the unit circle is 1 unit.

We can still construct a triangle within the unit circle with the angle starting from the positive x axis.

Looking at the trigonometric ratios of sine and cosine, we can say that:

• \sin(\theta)=\frac{a}{1} \; \text{so} \; a=\sin(\theta)
• \cos(\theta)=\frac{b}{1} \; \text{so} \; b=\cos(\theta)
• \tan(\theta)=\frac{a}{b} \; \text{so} \; \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}

This means that the width of the triangle is equal to \cos(\theta) and the height of the triangle is equal to \sin(\theta) . The point on the unit circle is therefore the coordinate ( \cos(\theta) , \; \sin(\theta) ):

The y value of the coordinate on the unit circle gives us the value for \tan(\theta) so:

• At the point (1,0) we can say that \sin(0)=0 as the angle \theta would be 0 .
• At the point (0,1) we can say that \sin(90)=1 as the angle \theta is equal to 90^o .
• At the point (-1,0) we can say that \sin(180)=0 as the angle \theta from the positive x axis is now 180^o
• At the point (0,-1) we can say \sin(270)=-1 as the angle \theta is 270^o .
• Continuing the point around the unit circle back to the coordinate (1,0) , we get \sin(360)=0 as \theta has made a full turn back to the positive x axis.

\sin(\theta) is as periodic function as we could continue to turn around the origin from the positive x axis beyond 360^o . We can also turn anticlockwise which would give us the negative value for the value of \sin(\theta) . The period of the function is 360^o .

### When do we get positive and negative values for sin(θ)?

From the unit circle, we can see that the angle \theta gives a positive value for y when the angle \theta is between 0^o and 180^o .

Between 180^o and 360^o , the y value is negative, and so \sin(\theta) is negative.

Visualising this using a coordinate axes, we can show when \sin(\theta) is positive and negative:

We are no longer limited to the range of values for \theta as we can continue to turn any angle, in a positive or negative direction and obtain a value for \sin(\theta) . We can therefore expand the graph drawn previously to represent this new set of values on a graph for angles between -360^o and 360^o .

### The graph of the sine function

• The graph does not start at the origin but it does pass through it.
• The graph is continuous for both positive and negative values of \theta and has a period of 360^o (this means that the graph repeats every 360^o ).
• The function of \sin(\theta) can be described as a wave, namely the sine wave, with an amplitude of 1 because the function is limited between the values of -1 and 1 . The maximum value is therefore 1 , and the minimum value is -1 .
• The sine wave can have different amplitudes when we transform the function.
• Step by step guide: transformations of trigonometric functions.

You should be able to interpret the unit circle to determine the value of \theta .

Here are a couple of examples to help you.

Top tip: you can remember the coordinate order as alphabetical; x before y and cos before sin to get the coordinate (\cos(\theta), \; \sin(\theta))

Now that we have generated the graph of \sin(\theta) , we need to be able to interpret it to find values of \theta and \sin(\theta) for any value between the range 360^o \leq \theta \leq 360^o (values for \theta between -360^o and 360^o .

### The reciprocal to sin(θ)

The inverse function to \sin(\theta) is \sin^{-1}(\theta) .

The reciprocal to \sin(\theta) is called the cosecant or \cosec(\theta) where \cosec(\theta)=\frac{1}{\sin(\theta)} (not needed for GCSE).

### Comparing the graphs of sin(θ) and cos(θ)

The graphs of \sin \theta and \cos \theta are very similar. In fact, if you applied a phase shift to either graph, you will get the graph of the other function.

Graphing the functions of y=\sin(\theta) (in red) and y=\cos(\theta) (in blue):

If we moved the graph of \cos \theta by adding 90 degrees to every angle (cos(+90)) we get \cos(\theta) . This horizontal shift can be in either direction. We can also carry out a vertical shift, a horizontal stretch, a vertical stretch, a rotation, and more!

Step by step guide: transforming trigonometric graphs.

## How to use sine graphs

In order to use sine graphs:

1. Draw a straight line from the axis of the known value to the sine curve.
2. Draw a straight, perpendicular line at the intersection point to the other axis.
3. Read the value where the perpendicular line meets the other axis.

## Sine graph examples

### Example 1: 0° ≤ θ ≤ 90°

Use the graph of \sin(\theta) to estimate the value for \theta when \sin(\theta)=1 for 0^o \leq \theta \leq 90^o

1. Draw a straight line from the axis of the known value to the sine curve.

2Draw a straight, perpendicular line at the intersection point to the other axis.

3Read the value where the perpendicular line meets the other axis.

When \sin(\theta)=1, \; \theta=90^o

### Example 2: 90° ≤ θ ≤ 270°

Use the graph of \sin(\theta) to estimate the value for \theta when \sin(\theta)=-0.4 for 90^o \leq \theta \leq 270^o

When \sin(\theta)=-0.4, \; \theta=204^o

### Example 3: 270 ≤ θ ≤ 360°

Use the graph of sin(\theta) to estimate the value for \sin(\theta) when \theta=186^o .

When \theta=186^o, \; \sin(\theta)=-0.1 .

### Example 4: −180 ≤ θ ≤ 0°

Use the graph of \sin(\theta) to estimate the value for \theta when \sin(\theta)=0 for -180 \leq \theta \leq 0^o

When \sin(\theta)=0, \; \theta=180^o and \theta =0^o

### Example 5: −360° ≤ θ ≤ 0°

Use the graph of \sin(\theta) to estimate the value for \sin(\theta) when \theta=-307^o .

When \theta=-307^o, \; \sin(\theta)=0.8 .

### Example 6: −360° ≤ θ ≤ 360°

Use the graph of \sin(\theta) to estimate the value for \theta when \sin(\theta)=-0.7 for -360^o \leq \theta \leq 360^o

When \sin(\theta)=-0.6, \; \theta= 143^o, \; \theta=-37^o, \; \theta=217^o \; \text{and} \; \theta=323^o

### Common misconceptions

• Drawing the horizontal line only to one intersection point

When you are finding the value of \theta using a trigonometric graph, only one value is calculated when there are more points of intersection.

E.g. Only the value of 65^o is read off the graph.

• Sine and cosine graphs switched

The sine and cosine graphs are very similar and can easily be confused with one another. A tip to remember is that you “sine up” from 0 for the sine graph so the line is increasing whereas you “cosine down” from 1 so the line is decreasing for the cosine graph.

• Asymptotes are drawn incorrectly for the graph of the tangent function

The tangent function has an asymptote at 90^o because this value is undefined. As the curve repeats every 180^o , the next asymptote is at 270^o and so on.

• The graphs are sketched using a ruler

Each trigonometric graph is a curve and therefore the only time you are required to use a ruler is to draw a set of axes. Practice sketching each curve freehand and label important values on each axis.

• Value given out of range

When finding a value of \theta using a trigonometric graph, you must make sure that the value of \theta is within the range specified in the question.

For example, the range of values for \theta is given as 0^o \leq \theta \leq 360^o and only the value of \theta=240^o is written for the solution, whereas the solution \theta=300^o is also correct.

Sin graph is part of our series of lessons to support revision on trigonometry. You may find it helpful to start with the main trigonometry lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

### Practice sin graph questions

1. Use the graph of \sin(\theta) to calculate the value of \theta when \sin(\theta)=0.26 for 0^o \leq \theta \leq 180^o

\theta=15^o, \;\theta=165^o

\theta=15^o

\theta=165^o

\theta=195^o

2. Use the graph of \sin(\theta) to calculate the value of \theta when \sin(\theta)=-0.5 for -360^o \leq \theta \leq 90^o

\theta=-30^o

\theta=-30^o, \; \theta=-150^o

\theta=-150^o

\theta=-330^o, \; \theta=-210^o

3. Use the graph of \sin(\theta) to calculate the value of \theta when \sin(\theta)=0 for -180^o \leq \theta \leq 180^o

\theta=-180^o

\theta=0^o

\theta=180^o

\theta=180^o, \; 0^o, \; 180^o

4. Use the graph of \sin(\theta) to calculate the value of \sin(\theta) when \theta=200^o

\sin(200)=0.3

\sin(200)=-0.5

\sin(200)=-1

\sin(200)=-0.3

5. Use the graph of \sin(\theta) to calculate the value of \sin(\theta) when \theta=-225^o

\sin(-225)=-0.7

\sin(-225)=0.7

\sin(-225)=1

\sin(-225)=0.2

6. What value for \theta would return the same value for 270^o ?

\sin(270)=sin(90)

\sin(270)=sin(-90)

\sin(270)=sin(0)

\sin(270)=sin(-270)

### Sin graph GCSE questions

1. Below is a sketch of the graph of y=\sin(\theta) for 0 \leq \theta \leq 180^o .

If \sin(x)=0.7 , what other value for \theta would return the same value of \sin(\theta) ?

(1 mark)

180-x

(1)

2. (a) Write an equation in terms of \theta for the size of the angle CAB .

(b) Use the graph of y=\sin(\theta) to estimate the value of \theta in triangle ABC .

(4 marks)

(a)

\sin(\theta)=\frac{12}{13}

(1)

\theta =\sin^{-1}(\frac{12}{13})

(1)

(b)

(1)

\sin^{-1}(\frac{12}{13}) = [60-75^o]

(1)

3. Write down the solutions to the equation \sin(\theta)=-0.2 for 360 \leq \theta \leq 720^o .

(2 marks)

[-190 \; \text{to} \;-200^o] and [340 \; \text{to} \; 350^o] highlighted on graph

(1)

[550-560^o] and [700-710^o]

(1)

4. Below are the graphs of y=\sin(\theta) and y=-\cos(\theta). Estimate the solutions for \sin(\theta) = -\cos(\theta) for -360 \leq \theta \leq 360^o .

(3 marks)

\theta =-45^o, \; -225^o

(1)

\theta = 135^o, \; \theta = 315^o

(1)

## Learning checklist

You have now learned how to:

• recognise, sketch and interpret graphs of trigonometric functions (with arguments in degrees) y = sin \; x for angles of any size.

## Still stuck?

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#### FREE GCSE Maths Practice Papers - 2022 Topics

Practice paper packs based on the advanced information for the Summer 2022 exam series from Edexcel, AQA and OCR.

Designed to help your GCSE students revise some of the topics that will come up in the Summer exams.