1.04 Trigonometric functions

Before looking at applications and calculus involving trigonometric functions, let's review the functions of sine, cosine and tangent.

 

$\sin\left(\theta\right)$sin(θ) and $\cos\left(\theta\right)$cos(θ)

We will begin by looking at the unit circle, that is a circle with the origin at the centre and radius of 1 unit.  A point on the circle at an angle of $\theta$θ anticlockwise from the $x$x-axis has coordinates $\left(\cos\theta,\sin\theta\right)$(cosθ,sinθ).

The unit circle

 

If we imagine the point moving around the unit circle, as we move through different values of $\theta$θ the value of $\sin\theta$sinθ and $\cos\theta$cosθ move accordingly between $-1$−1 and $1$1. If we plot the values of $\sin\theta$sinθ and $\cos\theta$cosθ according to different values of $\theta$θ on the unit circle, we get the following graphs:

 

$y=\sin\theta$y=sinθ

 

$y=\cos\theta$y=cosθ

 

Consequently, the graphs of $y=\sin\theta$y=sinθ and $y=\cos\theta$y=cosθ have many properties. Each graph demonstrates repetition. We call the graphs of $y=\sin\theta$y=sinθ and $y=\cos\theta$y=cosθ periodic and define the period as the length of one complete cycle. For both graphs, the period is $2\pi$2π.

An example of a cycle

 

Because of the oscillating behaviour, both graphs have regions where the curve is increasing and decreasing. Remember that we say the graph of a particular curve is increasing if the $y$y-values increase as the $x$x-values increase. Similarly, we say the graph is decreasing if the $y$y-values decrease as the $x$x-values increase.

An example of where $y=\sin x$y=sinx is decreasing

 

In addition, the height of each graph stays between $y=-1$y=−1 and $y=1$y=1 for all values of $\theta$θ, since each coordinate of a point on the unit circle can be at most $1$1 unit from the origin.

Worked example

Example 1

By using the graph of $y=\cos x$y=cosx, what is the sign of $\cos\frac{23\pi}{12}$cos23π12​?

Think: Using the graph of $y=\cos x$y=cosx, we can roughly estimate where the point $\left(\frac{23\pi}{12},\cos\frac{23\pi}{12}\right)$(23π12​,cos23π12​) lies and from this, determine the sign of $\cos\frac{23\pi}{12}$cos23π12​. We can write $\frac{23\pi}{12}$23π12​ as $\frac{24\pi}{12}-\frac{\pi}{12}$24π12​−π12​ , therefore $\frac{23\pi}{12}$23π12​ is a little to the left of $2\pi$2π. 

Do: We plot the point on the graph of $y=\cos x$y=cosx below.

The point $\left(\frac{23\pi}{12},\cos\frac{23\pi}{12}\right)$(23π12​,cos23π12​) drawn on the graph of $y=\cos x$y=cosx.

 

We can quickly observe that the height of the curve at this point is above the $x$x-axis, and observe that $\cos\frac{23\pi}{12}$cos23π12​ is positive.

 

Practice questions

question 1

question 2

question 3

 

Transformations of $y=\sin x$y=sinx and $y=\cos x$y=cosx

The impact of the parameters $a$a, $b$b, $c$c and $d$d, have on the graph of $y=af\left(b\left(x-c\right)\right)+d$y=af(b(x−c))+d, can be summarised as follows:

Let's look at these more closely in relation to the graphs $f\left(x\right)=\sin\left(x\right)$f(x)=sin(x) and $f\left(x\right)=\cos\left(x\right)$f(x)=cos(x)and the impact on key features of these graphs. Use the applet below to observe the impact of $a$a, $b$b, $c$c and $d$d on $f\left(x\right)=a\sin\left(b\left(x-c\right)\right)+d$f(x)=asin(b(x−c))+d and $f\left(x\right)=a\cos\left(b\left(x-c\right)\right)+d$f(x)=acos(b(x−c))+d:

Created with Geogebra

We can see the parameters had the expected impact on the graph and we can describe key features in relation to these.

 

Exploration

Consider the graphs of $y=\sin x$y=sinx and $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4​)+2 which are drawn below.

The graphs of $y=\sin x$y=sinx and $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4​)+2

Starting with the graph of $y=\sin x$y=sinx, we can work through a series of transformations so that it coincides with the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4​)+2.

We can first reflect the graph of $y=\sin x$y=sinx across the $x$x-axis. This is represented by applying a negative sign to the function (multiplying the $y$y values by $-1$−1).

The graph of $y=-\sin x$y=−sinx

Then we can increase the amplitude of the function to match. This is represented by multiplying the $y$y-value of every point on $y=-\sin x$y=−sinx by $2$2.

The graph of $y=-2\sin x$y=−2sinx

Next we can apply the period change that is the result of multiplying the $x$x-value inside the function by $3$3. This means that to get a particular $y$y-value, we must put in an $x$x-value that is $3$3 times smaller than before so every $x$x value is divided by $3$3. The result is a horizontal dilation by a factor of $\frac{1}{3}$13​, note the new graph will now cycle every $\frac{2\pi}{3}$2π3​.

The graph of $y=-2\sin3x$y=−2sin3x

 

Our next step will be to obtain the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)$y=−2sin(3x+π4​), and we can do so by applying a horizontal translation. In order to see what translation to apply, however, we first factorise the function into the form $y=-2\sin\left(3\left(x+\frac{\pi}{12}\right)\right)$y=−2sin(3(x+π12​)). Remember we do this because usually, in the form presented, we need to translate horizontally before we can horizontally dilate. But for trigonometry graphs, performing the horizontal dilation first is far easier when drawing.

In this form, we can see that the $x$x-values are increased by $\frac{\pi}{12}$π12​ inside the function. This means that to get a particular $y$y-value, we must put in an $x$x-value that is $\frac{\pi}{12}$π12​ smaller than before. Graphically, this corresponds to shifting the entire function to the left by $\frac{\pi}{12}$π12​ units (subtract $\frac{\pi}{12}$π12​ from every $x$x value).

The graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)$y=−2sin(3x+π4​)

 

Lastly, we translate the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)$y=−2sin(3x+π4​) upwards by $2$2 units, to obtain the final graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4​)+2.

The graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4​)+2

 

 

Practice questions

Question 4

Question 5

Question 6

Question 7

 

$Tan\left(\theta\right)$Tan(θ)

Three ways to define the tangent of an angle:

• The tangent of the angle can be geometrically defined to be $y$y-coordinate of point $Q$Q, where $Q$Q is the intersection of the extension of the line $OP$OP and the tangent of the circle at $\left(1,0\right)$(1,0)
• Using similar triangles we can also define this algebraically as the ratio: $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ​
• This also represents the gradient of the line that forms the angle $\theta$θ to the positive $x$x-axis

 

The graph of $y=\tan\theta$y=tanθ

 

The asymptotes of the function are where the angle $\theta$θ would cause the line $OP$OP to be vertical and hence the gradient is undefined. We can also see this through the definition $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ​. The function is undefined where $\cos\theta=0$cosθ=0 and the graph approaches the vertical lines at these values forming asymptotes.

The fact that the tangent function repeats at intervals of $\pi$π can be verified by considering the unit circle diagram. Either by considering the gradient of the line $OP$OP and how the gradient will be the same for a point at an angle of $\theta$θ or at at angle of $\theta+\pi$θ+π. Or we can consider this algebraically by looking at the ratio $\frac{\sin\theta}{\cos\theta}$sinθcosθ​. If $\pi$π is added to an angle $\theta$θ, then the diagram below shows that $\sin(\theta+\pi)$sin(θ+π) has the same magnitude as $\sin\theta$sinθ but opposite sign. The same relation holds between $\cos(\theta+\pi)$cos(θ+π) and $\cos\theta$cosθ.

We make use of the definition: $$

$\tan(\theta+\pi)$tan(θ+π)	$=$=	$\frac{\sin(\theta+\pi)}{\cos(\theta+\pi)}$sin(θ+π)cos(θ+π)​
	$=$=	$\frac{-\sin\theta}{-\cos\theta}$−sinθ−cosθ​
	$=$=	$\tan\theta$tanθ

 

Practice question

question 8

 

Transformations of $y=\tan x$y=tanx

Just as we transformed the trigonometric functions $y=\sin x$y=sinx and $y=\cos x$y=cosx we can apply parameters to the equation $y=\tan x$y=tanx to transform it to $y=a\tan\left(b\left(x-c\right)\right)+d$y=atan(b(x−c))+d.

Use the geogebra applet below to adjust the parameters in $y=a\tan\left(b\left(x-c\right)\right)+d$y=atan(b(x−c))+d and observe how it affects the graph. Try to answer the following questions.

• Which parameters affect the position of the vertical asymptotes? Which ones don't?
• Which parameters translate the graph, leaving the shape unchanged? Which ones affect the size?
• Which parameter changes the period of the graph? Does making this parameter larger make the period larger?

Created with Geogebra

The constants $a$a, $b$b, $c$c and $d$d transform the tangent graph. Let's summarise the impact of each:

 

Worked examples

Example 2

Illustrate the change in dilation of the graph $y=\tan x$y=tanx by sketching the graphs of $y=\tan x$y=tanx together with $y=3\tan x$y=3tanx and $y=\frac{2}{7}\tan x$y=27​tanx.

Think: Note since only a vertical dilation has been applied all three graphs will share $x$x-intercepts and vertical asymptotes. We first sketch the base graph of $y=\tan x$y=tanx, shown in blue in the graph below. This graph will go through the points $\left(\frac{\pi}{4},1\right)$(π4​,1), $\left(0,0\right)$(0,0) and $\left(\frac{-\pi}{4},-1\right)$(−π4​,−1). The asymptotes will be located at $\frac{\pi}{2}$π2​, $\frac{-\pi}{2}$−π2​, $\frac{3\pi}{2}$3π2​, $\frac{-3\pi}{2}$−3π2​,....

The graph of $y=3\tan x$y=3tanx has been vertically dilated by a factor of $3$3. So we can plot the points $\left(\frac{\pi}{4},3\right)$(π4​,3) and $\left(\frac{-\pi}{4},-3\right)$(−π4​,−3) to show this stretch clearly. Similarly the graph $y=\frac{2}{7}\tan x$y=27​tanx has been vertically dilated by a factor of $\frac{2}{7}$27​. To show this we can plot the points $\left(\frac{\pi}{4},\frac{2}{7}\right)$(π4​,27​) and $\left(\frac{-\pi}{4},\frac{-2}{7}\right)$(−π4​,−27​).

Do: The graphs together are shown below.

 

 

Example 3

Sketch the graph of $f(x)=\tan\left(x-\frac{\pi}{4}\right)$f(x)=tan(x−π4​).

Think: We see that the function $\tan x$tanx has been moved to the right by a distance of $\frac{\pi}{4}$π4​ units. We can sketch this by drawing the base graph of $y=\tan\theta$y=tanθ and shifting each point right by $\frac{\pi}{4}$π4​.

The phase shift will also move the asymptotes and since $\tan x$tanx is undefined at $x=\frac{\pi}{2}+n\pi$x=π2​+nπ for all integers $n$n, the undefined points for $\tan\left(x-\frac{\pi}{4}\right)$tan(x−π4​) must be $x=\frac{3\pi}{4}+n\pi$x=3π4​+nπ.

Do: The graph is shown below in purple.

 

Practice questions

question 9

question 10

question 11