Hong Kong
Stage 4 - Stage 5

# Applications of sine and cosine functions including phase shift (radians)

Lesson

Behaviour in the real world that involves rotation or repetitive motion can be modelled using trigonometric equations like $y=\sin x$y=sinx and $y=\cos x$y=cosx.

We can develop more accurate models by transforming these equations. In this lesson we will focus on phase shifts of the form $y=\sin\left(x-c\right)$y=sin(xc), and also look at other transformations.

#### Exploration

A small disc sitting on a flat surface begins to roll slowly with a constant angular velocity of $1$1 radian per second when a fan is switched on.

As the disc rolls, the height $h$h in centimetres of a point on the disc is described by the equation $h=\cos t+1$h=cost+1, where $t$t is in seconds.

A graph of the equation is shown below. Notice that the period of the equation is $2\pi$2π seconds, or just over $6$6 seconds. This is the time it takes the point on the disc to complete one full revolution.

Graph of the function $h=\cos t+1$h=cost+1.

Two seconds after the first disc starts rolling along the surface, a second disc of the same size is placed in front of the fan. It begins to roll in a similar way, and the height of a point on its surface is described by the equation $h=\cos\left(t-2\right)+1$h=cos(t2)+1.

The second disc has the same behaviour as the first disc, except it has been translated in time. The graph below shows that the phase shift of $2$2 seconds corresponds to a horizontal translation of the graph of $h=\cos t+1$h=cost+1 to the graph of $h=\cos\left(t-2\right)+1$h=cos(t2)+1.

Graph of the functions $h=\cos t+1$h=cost+1 and $h=\cos\left(t-2\right)+1$h=cos(t2)+1.

Suppose now that we want the discs to roll a bit faster. If we crank up the speed of the fan, we can increase the angular velocity of the disc to $4$4 radians per second. The motion of the point on the first disc is then described by the equation $h=\cos\left(4t\right)+1$h=cos(4t)+1, and the point on the second disc has the equation $h=\cos\left(4\left(t-2\right)\right)+1$h=cos(4(t2))+1.

In the image below we see the consequence of increasing the rolling speed is to decrease the period of each equation. The new period is $\frac{2\pi\text{ radians}}{4\text{ radians per second}}=\frac{\pi}{2}$2π radians4 radians per second=π2 seconds, or about $1.57$1.57 seconds.

Graph of the functions $h=\cos\left(4t\right)+1$h=cos(4t)+1 and $h=\cos\left(4\left(t-2\right)\right)+1$h=cos(4(t2))+1.

#### Worked example

The height of a point on a disc from the scenario above is described by the equation $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t4))+8. The graph of the function is shown below.

Graph of the function $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t4))+8.

a. Find the initial height of the point.

Think: The initial point in time is the moment when $t=0$t=0 seconds.

Do: Substitute $t=0$t=0 into the equation $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t4))+8.

 $h$h $=$= $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$8sin(π5​(t−4))+8 $=$= $8\sin\left(\frac{\pi}{5}\left(0-4\right)\right)+8$8sin(π5​(0−4))+8 (Substitute the value of $t$t) $=$= $8\sin\left(-\frac{4\pi}{5}\right)+8$8sin(−4π5​)+8 (Simplify the product) $=$= $3.30$3.30 cm (2 d.p.) (Evaluate the expression)

The initial height of the point is $3.30$3.30 cm above the surface.

b. Find the maximum height of the point.

Think: The sine function takes values between $-1$1 and $1$1. The maximum height will correspond to the times at which $\sin\left(\frac{\pi}{5}\left(t-4\right)\right)=1$sin(π5(t4))=1.

Do: By substituting $\sin\left(\frac{\pi}{5}\left(t-4\right)\right)=1$sin(π5(t4))=1 into the equation $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t4))+8 we see that the maximum height is $h=8\times1+8=16$h=8×1+8=16 cm above the surface.

c. Find the time when the point is first at a height of $12$12 cm.

Think: We have a value of $h$h, and we want to find the corresponding value of $t$t.

Do: Substitute $h=12$h=12 into the equation and rearrange to solve for $t$t.

 $h$h $=$= $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$8sin(π5​(t−4))+8 $12$12 $=$= $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$8sin(π5​(t−4))+8 (Substitute the value of $h$h) $4$4 $=$= $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)$8sin(π5​(t−4)) (Subtract $8$8 from both sides) $\frac{1}{2}$12​ $=$= $\sin\left(\frac{\pi}{5}\left(t-4\right)\right)$sin(π5​(t−4)) (Divide both sides by $8$8) $\frac{\pi}{6}$π6​ $=$= $\frac{\pi}{5}\left(t-4\right)$π5​(t−4) (Take the inverse $\sin$sin of both sides) $\frac{5}{6}$56​ $=$= $t-4$t−4 (Multiply both sides by $\frac{5}{\pi}$5π​) $t$t $=$= $4.83$4.83 seconds (2 d.p.) (Add $4$4 to both sides)

The point on the disk will first reach a height of $12$12 cm when it has rolled for $4.83$4.83 seconds.

#### Practice question

Tobias is jumping on a trampoline. Victoria watches him bounce at a regular rate and wants to try to model his height over time. When Victoria starts her stopwatch, Tobias is at a minimum height of $30$30 cm below the trampoline frame. A moment later Victoria records Tobias reaching a maximum height of $50$50 cm above the trampoline frame. She uses the function $H\left(s\right)=a\sin\left(2\pi\left(s-c\right)\right)+d$H(s)=asin(2π(sc))+d, where $H$H is the height in cm above the trampoline frame and $s$s is the time in seconds.

1. Find the value of $a$a, the amplitude of the function.

2. Find the value of $d$d.

3. Find the value of $H\left(0\right)$H(0).

4. Find the value of $c$c, if $00<c<1. 5. At what time does Tobias first reach a height of$30\$30 cm?