Before looking at applications and calculus involving trigonometric functions, let's review the functions of sine, and cosine.
We will begin by looking at the unit circle, that is, a circle with the origin at the centre and a unit radius (radius equals one). 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 |
Exact values of the trigonometric ratios exist within both the unit circle and the trigonometric functions. The basic ratios in Quadrant 1 are as follows:
$\sin\theta$sinθ | $\cos\theta$cosθ | $\tan\theta$tanθ | |
$\frac{\pi}{6}$π6 | $\frac{1}{2}$12 | $\frac{\sqrt{3}}{2}$√32 | $\frac{1}{\sqrt{3}}$1√3 |
$\frac{\pi}{4}$π4 | $\frac{1}{\sqrt{2}}$1√2 | $\frac{1}{\sqrt{2}}$1√2 | $1$1 |
$\frac{\pi}{3}$π3 | $\frac{\sqrt{3}}{2}$√32 | $\frac{1}{2}$12 | $\sqrt{3}$√3 |
In order to use these Quadrant 1 ratios to find those in the other Quadrants, the following must be noted:
Quadrant | Domain | Rule | Sign of Solutions |
I | $0\le x\le\frac{\pi}{2}$0≤x≤π2 | $\theta$θ | All three functions are positive |
II | $\frac{\pi}{2}\le x\le\pi$π2≤x≤π | $\pi-\theta$π−θ |
Cosine and Tangent are negative Sine is positive |
III | $\pi\le x\le\frac{3\pi}{2}$π≤x≤3π2 | $\pi+\theta$π+θ |
Cosine and Sine are negative Tangent is positive |
IV | $\frac{3\pi}{2}\le x\le2\pi$3π2≤x≤2π | $2\pi-\theta$2π−θ |
Sine and Tangent are negative Cosine is positive |
Find the exact value of $\cos(\frac{2\pi}{3}).$cos(2π3).
$\frac{2\pi}{3}$2π3 is in the second quadrant. Therefore, we should think of $\frac{2\pi}{3}$2π3 as $\pi-\frac{\pi}{3}.$π−π3. As we know that $\cos(\frac{\pi}{3})=\frac{1}{2}$cos(π3)=12 and the cosine function is negative in the second quadrant, the solution is as follows:
$\cos(\frac{2\pi}{3})$cos(2π3) | $=$= | $-\cos(\frac{\pi}{3})$−cos(π3) |
$=$= | $-\frac{1}{2}$−12 |
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