Hong Kong
Stage 4 - Stage 5

# Find solutions from a graph for sine and cosine equations including phase shift (degrees)

Lesson

The points of intersection of two curves, $y=f\left(x\right)$y=f(x) and $y=g\left(x\right)$y=g(x) can be obtained by first setting the expressions $f\left(x\right)$f(x) and $g\left(x\right)$g(x) to be equal and then solving for $x$x. More specifically, we want to find the values of $x$x that satisfy the equation:

$f(x)=g(x)$f(x)=g(x)

Then, naturally, we would substitute our $x$x-values into either $y=f\left(x\right)$y=f(x) or $y=g\left(x\right)$y=g(x) to find the corresponding $y$y-values of the points of intersection. In general, solving an equation can be thought of as finding the $x$x-values of the points of intersection of two curves.

#### Exploration

Say we wanted to find the values of $x$x that solve the equation $\sin\left(x-60^\circ\right)=1$sin(x60°)=1. Graphically speaking, this is the same as finding the $x$x-values that correspond to the points of intersection of the curves $y=\sin\left(x-60^\circ\right)$y=sin(x60°) and $y=1$y=1.

 $y=\sin\left(x-60^\circ\right)$y=sin(x−60°) (green) and $y=1$y=1 (blue).

We can see in the region given by $\left(-360^\circ,360^\circ\right)$(360°,360°) that there are two points where the two functions meet.

 Points indicating where the two functions meet.

Since we are fortunate enough to have gridlines, the $x$x-values for these points of intersection can be easily deduced. Each grid line is separated by $30^\circ$30°, which means that the solution to the equation $\sin\left(x-60^\circ\right)=1$sin(x60°)=1 in the region $\left(-360^\circ,360^\circ\right)$(360°,360°) is given by:

$x=-210^\circ,150^\circ$x=210°,150°

Careful!

We can only solve equations graphically if the curves are drawn accurately and to scale. You won't be expected to solve equations graphically if it requires drawing the curves by hand.

#### Practice questions

##### question 1

The functions $y=\sin\left(x-60^\circ\right)+4$y=sin(x60°)+4 and $y=\frac{9}{2}$y=92 are drawn below.

1. State all solutions to the equation $\sin\left(x-60^\circ\right)+4=\frac{9}{2}$sin(x60°)+4=92 over the domain $\left[-360^\circ,360^\circ\right)$[360°,360°). Give your answers in degrees, separated by commas.

##### question 2

Consider the functions $y=-\cos\left(x-45^\circ\right)-2$y=cos(x45°)2 and $y=-3$y=3.

1. Draw the functions $y=-\cos\left(x-45^\circ\right)-2$y=cos(x45°)2 and $y=-3$y=3.

2. Hence, state all solutions to the equation $-\cos\left(x-45^\circ\right)-2=-3$cos(x45°)2=3 over the domain $\left(-360^\circ,360^\circ\right)$(360°,360°). Give your answers as exact values separated by commas.

##### question 3

Consider the functions $y=2\sin\left(x-60^\circ\right)+3$y=2sin(x60°)+3 and $y=2$y=2.

1. Draw the functions $y=2\sin\left(x-60^\circ\right)+3$y=2sin(x60°)+3 and $y=2$y=2.

2. Hence, state all solutions to the equation $2\sin\left(x-60^\circ\right)+3=2$2sin(x60°)+3=2 over the domain $\left[-360^\circ,360^\circ\right)$[360°,360°). Give your answers in degrees, separated by commas.