Before looking at applications and calculus involving trigonometric functions, let's review the functions of sine, cosine and tangent.
We will begin by looking at the unit circle, that is a circle with the origin at the centre and unit radius, the coordinates of any point on that circle can be described using trigonometry. Specifically 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 cyclical and define a cycle as any section of the graph that can be translated to complete the rest of the graph. We also define the period as the length of one 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.
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.
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.
Consider the equation $y=\sin x$y=sinx.
Using the fact that $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$sinπ3=√32, what is the value of $\sin\frac{2\pi}{3}$sin2π3?
Using the fact that $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$sinπ3=√32, what is the value of $\sin\frac{4\pi}{3}$sin4π3?
Using the fact that $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$sinπ3=√32, what is the value of $\sin\frac{5\pi}{3}$sin5π3?
Complete the table of values. Give your answers in exact form.
$x$x | $0$0 | $\frac{\pi}{3}$π3 | $\frac{\pi}{2}$π2 | $\frac{2\pi}{3}$2π3 | $\pi$π | $\frac{4\pi}{3}$4π3 | $\frac{3\pi}{2}$3π2 | $\frac{5\pi}{3}$5π3 | $2\pi$2π |
---|---|---|---|---|---|---|---|---|---|
$\sin x$sinx | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Draw the graph of $y=\sin x$y=sinx.
Consider the curve $y=\cos x$y=cosx drawn below and determine whether the following statements are true or false.
The graph of $y=\cos x$y=cosx is cyclic.
True
False
As $x$x approaches infinity, the height of the graph for $y=\cos x$y=cosx approaches infinity.
True
False
The graph of $y=\cos x$y=cosx is increasing between $x=-\frac{\pi}{2}$x=−π2 and $x=0$x=0.
False
True
Consider the curve $y=\sin x$y=sinx drawn below and answer the following questions.
If one cycle of the graph of $y=\sin x$y=sinx starts at $x=0$x=0, when does the next cycle start?
In which of the following regions is the graph of $y=\sin x$y=sinx decreasing? Select all that apply.
$-\frac{\pi}{2}
$\frac{\pi}{2}
$-\frac{5\pi}{2}
$-\frac{3\pi}{2}
What is the $x$x-value of the $x$x-intercept in the region $0
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:
To obtain the graph of $y=af\left(b\left(x-c\right)\right)+d$y=af(b(x−c))+d from the graph of $y=f\left(x\right)$y=f(x):
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:
We can see the parameters had the expected impact on the graph and we can describe key features in relation to these.
Key features of the graph of $y=a\sin\left(b\left(x-c\right)\right)+d$y=asin(b(x−c))+d or $y=a\cos\left(b\left(x-c\right)\right)+d$y=acos(b(x−c))+d are:
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 function 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 can put in an $x$x-value that is $3$3 times smaller than before. 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 can 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.
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 |
When we geometrically apply each transformation to the graph of $y=\sin x$y=sinx, it's important to consider the order of operations. If we had wanted to vertically translate the graph before reflecting across the $x$x-axis, we would have needed to translate the graph downwards first. Alternatively, you can apply the reflection last by reflecting across the new principal axis.
Consider the function $y=-3\cos x$y=−3cosx.
What is the maximum value of the function?
What is the minimum value of the function?
What is the amplitude of the function?
Select the two transformations that are required to turn the graph of $y=\cos x$y=cosx into the graph of $y=-3\cos x$y=−3cosx.
Vertical dilation.
Horizontal translation.
Reflection across the $x$x-axis.
Vertical translation.
Determine the equation of the graphed function given that it is of the form $y=\cos\left(x-c\right)$y=cos(x−c), where $c$c is the least positive value.
The functions $f\left(x\right)$f(x) and $g\left(x\right)=jf\left(kx\right)$g(x)=jf(kx) have been graphed on the same set of axes in grey and black respectively.
What transformations have occurred from $f\left(x\right)$f(x) to $g\left(x\right)$g(x)?
Select all that apply.
Vertical stretching by a factor of $4$4.
Horizontal compression by factor of $3$3.
Vertical compression by a factor of $3$3.
Horizontal stretching by factor of $4$4.
Determine the value of $j$j.
Determine the value of $k$k.
Consider the function $y=3\sin\left(x-\frac{\pi}{3}\right)+2$y=3sin(x−π3)+2.
Determine the period of the function, giving your answer in radians.
Determine the amplitude of the function.
Determine the maximum value of the function.
Determine the minimum value of the function.
Graph the function.
The tangent function can be defined as follows:
Key features of the graph of $y=\tan\left(\theta\right)$y=tan(θ) are:
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 |
Consider the graph of $y=\tan x$y=tanx for $-2\pi\le x\le2\pi$−2π≤x≤2π.
How would you describe the graph?
Periodic
Decreasing
Even
Linear
Which of the following is not appropriate to refer to in regard to the graph of $y=\tan x$y=tanx?
Amplitude
Range
Period
Asymptotes
The period of a periodic function is the length of $x$x-values that it takes to complete one full cycle.
Determine the period of $y=\tan x$y=tanx in radians.
State the range of $y=\tan x$y=tanx.
$-\infty
$y>0$y>0
$\frac{-\pi}{2}
$-\pi
As $x$x increases, what would be the next asymptote of the graph after $x=\frac{7\pi}{2}$x=7π2?
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.
|
The constants $a$a, $b$b, $c$c and $d$d transform the tangent graph. Let's summarise the impact of each:
To obtain the graph of $y=a\tan\left(b\left(x-c\right)\right)+d$y=atan(b(x−c))+d from the graph of $y=\tan\left(x\right)$y=tan(x):
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=27tanx.
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=27tanx 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.
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.
Consider the function $y=-4\tan\frac{1}{5}\left(x+\frac{\pi}{4}\right)$y=−4tan15(x+π4).
Determine the period of the function, giving your answer in radians.
Determine the phase shift of the function, giving your answer in radians.
Determine the range of the function.
$[-1,1]$[−1,1]
$(-\infty,0]$(−∞,0]
$[0,\infty)$[0,∞)
$(-\infty,\infty)$(−∞,∞)
The graph of $y=\tan x$y=tanx is shown below. On the same set of axes, draw the graph of $y=5\tan x$y=5tanx.
Select all functions that have the same graph as $y=-\tan x$y=−tanx.
$y=-\tan\left(x+\frac{3\pi}{4}\right)$y=−tan(x+3π4)
$y=-\tan\left(x+\pi\right)$y=−tan(x+π)
$y=-\tan\left(x+\frac{\pi}{2}\right)$y=−tan(x+π2)
$y=-\tan\left(x+2\pi\right)$y=−tan(x+2π)