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Stage 4 - Stage 5

Applications of sine and cosine functions including phase shift (degrees)


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.



A small disc sitting on a flat surface begins to roll slowly with a constant angular velocity of $1^\circ$1° 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 $360$360 seconds, or $6$6 minutes. 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.

Thirty 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 $h=\cos\left(t-30^\circ\right)+1$h=cos(t30°)+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 $30$30 seconds corresponds to a horizontal translation of the graph of $h=\cos t+1$h=cost+1 to the graph of $h=\cos\left(t-30^\circ\right)+1$h=cos(t30°)+1.

Graph of the functions $h=\cos t+1$h=cost+1 and $h=\cos\left(t-30^\circ\right)+1$h=cos(t30°)+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 $15^\circ$15° per second. The motion of the point on the first disc is then described by the equation $h=\cos\left(15t\right)+1$h=cos(15t)+1, and the point on the second disc has the equation $h=\cos\left(15\left(t-30\right)\right)+1$h=cos(15(t30))+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{360^{\circ}}{15^{\circ}\text{ per second}}=24$36015 per second=24 seconds.

Graph of the functions $h=\cos\left(15t\right)+1$h=cos(15t)+1 and $h=\cos\left(15\left(t-30\right)\right)+1$h=cos(15(t30))+1.


Worked example

The height of a point on a disc from the scenario above is described by the equation $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t7))+10. The graph of the function is shown below.

Graph of the function $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t7))+10.


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 $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t7))+10.

$h$h $=$= $10\sin\left(20\left(t-7\right)\right)+10$10sin(20(t7))+10  
  $=$= $10\sin\left(20\left(0-7\right)\right)+10$10sin(20(07))+10 (Substitute the value of $t$t)
  $=$= $10\sin\left(20\times\left(-7\right)\right)+10$10sin(20×(7))+10 (Simplify the product)
  $=$= $10\sin\left(-140^\circ\right)+10$10sin(140°)+10 (Evaluate the multiplication)
  $=$= $3.57$3.57 cm (2 d.p.) (Evaluate the expression)

The initial height of the point is $3.57$3.57 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 $\sin\left(20\left(t-7\right)\right)=1$sin(20(t7))=1.

Do: By substituting $\sin\left(20\left(t-7\right)\right)=1$sin(20(t7))=1 into the equation $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t7))+10 we see that the maximum height is $h=10\times1+10=20$h=10×1+10=20 cm above the surface.


c. Find the time when the point is first at a height of $15$15 cm.

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

Do: Substitute $h=15$h=15 into the equation and rearrange to solve for $t$t.

$h$h $=$= $10\sin\left(20\left(t-7\right)\right)+10$10sin(20(t7))+10  
$15$15 $=$= $10\sin\left(20\left(t-7\right)\right)+10$10sin(20(t7))+10 (Substitute the value of $h$h)
$5$5 $=$= $10\sin\left(20\left(t-7\right)\right)$10sin(20(t7)) (Subtract $10$10 from both sides)
$\frac{1}{2}$12 $=$= $\sin\left(20\left(t-7\right)\right)$sin(20(t7)) (Divide both sides by $10$10)
$30$30 $=$= $20\left(t-7\right)$20(t7) (Take the inverse $\sin$sin of both sides)
$\frac{3}{2}$32 $=$= $t-7$t7 (Divide both sides by $20$20)
$t$t $=$= $8.5$8.5 seconds (Add $7$7 to both sides)

The point on the disk will first reach a height of $15$15 cm when it has rolled for $8.5$8.5 seconds.


Practice question

The height above the ground of a rider on a Ferris wheel can be modelled by the function $h\left(t\right)=25\cos\left(3\left(t-60\right)\right)+30$h(t)=25cos(3(t60))+30, where $h\left(t\right)$h(t) is the height in metres, $t$t is the time in seconds, and the angular velocity of $3^\circ$3° per second is the speed at which the wheel rotates.

  1. Draw the graph of $y=h\left(t\right)$y=h(t).

    Loading Graph...

  2. What is the maximum height of the rider?

  3. What is the minimum height of the rider?

  4. At what height is the rider after $85$85 seconds? Give your answer to two decimal places.

  5. How long does it take the rider to complete one full revolution?

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