# Trigonometric graphs - Higher

This circle has the centre at the origin and a radius of 1 unit.

The point P can move around the circumference of the circle. At point P the -coordinate is and the -coordinate is where is measured anti-clockwise from the positive -axis.

As the point P moves anti-clockwise round the circle from (1, 0), the angle increases until P returns to its starting position at (1, 0) when = 360°. If P continues moving past (1, 0), becomes greater than 360°, and the next time P is at (1, 0), will be 720°. And so on. Instead of P moving anticlockwise from (1, 0), if it goes clockwise then will be negative!

No matter where P is on the circle, the -coordinate gives the value of and the -coordinate gives the value of . Thus, the values of and will sometimes be positive and sometimes negative depending on the value of .

The graphs of and can be plotted.

## The graph of y = sin θ

The graph of has a maximum value of 1 and a minimum value of -1.

The graph has a period of 360°. This means that it repeats itself every 360°.

## The graph of y = cos θ

The graph of has a maximum value of 1 and a minimum value of -1.

The graph has a period of 360°.

## The graph of y = tan θ

This is defined as and from the circle and .

As the point P moves anticlockwise round the circle, the values of and change, therefore the value of will change.

The graph has a period of 180°.

## Calculating angles from trigonometric graphs

The symmetrical and periodic properties of the trigonometric graphs will give an number of solutions to any trigonometric equation.

### Example

Solve the equation for all values of between .

Using a calculator gives one solution:

Draw the horizontal line .

The line crosses the graph of four times in the interval so there are four solutions.

There is a line of symmetry at , so there is a solution at .

The period is 360° so to find the next solutions subtract 360°.

The solutions to the equation are:

= -330°, -210°, 30° and 150°.

Question

Solve the equation for all values of between . Give your answer to the nearest degree.

Using a calculator gives one solution:

(to the nearest degree)

Draw the horizontal line .

The line crosses the graph of four times in the interval so there are four solutions.

There is a line of symmetry at , so there is a solution at -41°.

The period is 360° so to find the other solutions add and subtract 360°.

The solutions to the equation are:

= -319°, -41°, 41° and 319°.

Question

Given that , calculate the other values of in the interval for which .

Using a calculator gives one solution:

(to the nearest degree)

The period of the graph is 180° so to calculate the other solutions keep adding 180°.

= 60°, 240°, 420° and 600°.