Trigonometric functions have many applications in periodic phenomena such as tides, pendulums, certain animal populations and analysing markets with seasonal changes. Just as with exponential functions before where we looked at some applications in further detail, let's look at how to solve some trigonometric equations using technology. Recall that solving trigonometric equations algebraically was explored in the previous lesson.
Solving trigonometric equations
To find unknown angles in a right-angled triangles we can use the functions $\sin^{-1}$sin−1, $\cos^{-1}$cos−1 and $\tan^{-1}$tan−1. When looking to find unknown angles in problems with trigonometric functions defined for angles beyond $90^\circ$90° or $\frac{\pi}{2}$π2, the difficulty is there can be infinitely many solutions.
We will focus on cases where the domain in restricted, so there will be a finite number of solutions. But how do we find them? Let's look at two main methods: finding solutions graphically and finding solutions algebraically.
Graphical solutions
To solve an equation graphically, such as $2\sin x+1=0$2sinx+1=0 where the right-hand side is zero, we are essentially finding the $x$x-intercepts of the graph of $y=2\sin x-1$y=2sinx−1. If the right-hand side was not equal to zero, such as $2\sin x=-1$2sinx=−1, we can move all terms to the left hand side and again find the $x$x-intercepts of the graph. Alternatively, we can graph both sides of the equation as separate functions, for our example $y=2\sin x$y=2sinx and $y=-1$y=−1, and then find the $x$x-coordinates of points of intersection of the two curves. We can see these methods are equivalent, in fact the first method is simply finding the intersection of the curve with the line $y=0$y=0.
In general, solving an equation can be thought of as finding the $x$x-values of the points of intersection of two curves representing the left- and right-hand side of the equation.
Worked example
Example 1
Find the values of $x$x that solve the equation $\sin\left(x-\frac{\pi}{3}\right)=1$sin(x−π3)=1 for the interval $-2\pi\le x\le2\pi$−2π≤x≤2π.
Think: Graphically speaking, this is the same as finding the $x$x-coordinates that correspond to the points of intersection of the curves $y=\sin\left(x-\frac{\pi}{3}\right)$y=sin(x−π3) and $y=1$y=1. Generally, we will use technology to solve such equations graphically. Graphing both functions using technology and taking care to show the correct domain we have:
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| $y=\sin\left(x-\frac{\pi}{3}\right)$y=sin(x−π3) (green) and $y=1$y=1 (blue). |
We can see in the region given by $\left(-2\pi,2\pi\right)$(−2π,2π) that there are two points where the two functions meet.
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| Points indicating where the two functions meet. |
Since we are fortunate enough to have gridlines that coincide with the intersections, the $x$x-values for these points of intersection can be easily deduced. Each grid line is separated by $\frac{\pi}{6}$π6, which means that the solution to the equation $\sin\left(x-\frac{\pi}{3}\right)=1$sin(x−π3)=1 in the region $\left(-2\pi,2\pi\right)$(−2π,2π) is given by:
$x=-\frac{7\pi}{6},\frac{5\pi}{6}$x=−7π6,5π6
- Ensure your calculator is in the correct mode of either radians or degrees, depending on the question.
- Using knowledge of transformations that have occurred and the interval given for solutions, select an appropriate viewing window before trying to find points of intersection.
- The calculator may not give the points of intersection in the exact form. Some problems may not have 'nice' solutions and sometimes we only need the answer correct to a certain number of decimal places. However, if we do require exact answers and you suspect the solutions are multiples of $\pi$π, you can find exact form by using the solve function of a CAS calculator, the numerical solve function in most graphics calculators or by dividing the $x$x-coordinate of the point of intersection by $\pi$π in either calculator to find what multiple of $\pi$π the solutions are appearing at.
Practice questions
question 1
Consider the function $y=3\sin x$y=3sinx.
Graph this function.
Loading Graph...Add the line $y=3$y=3 to your graph.
Loading Graph...Hence, state all solutions to the equation $3\sin x=3$3sinx=3 over the domain $\left[-2\pi,2\pi\right]$[−2π,2π]. Give your answers as exact values separated by commas.
question 2
Consider the functions $y=-\cos\left(x-\frac{\pi}{4}\right)-2$y=−cos(x−π4)−2 and $y=-3$y=−3.
Draw the functions $y=-\cos\left(x-\frac{\pi}{4}\right)-2$y=−cos(x−π4)−2 and $y=-3$y=−3.
Loading Graph...Hence, state all solutions to the equation $-\cos\left(x-\frac{\pi}{4}\right)-2=-3$−cos(x−π4)−2=−3 over the domain $\left(-2\pi,2\pi\right)$(−2π,2π). Give your answers as exact values separated by commas.
question 3
Consider the function $y=\tan\left(x-\frac{\pi}{4}\right)$y=tan(x−π4).
Graph this function.
Loading Graph...Add the line $y=1$y=1 to your graph.
Loading Graph...Hence, state all solutions to the equation $\tan\left(x-\frac{\pi}{4}\right)=1$tan(x−π4)=1 over the domain $\left[-2\pi,2\pi\right)$[−2π,2π). Give your answers as exact values separated by commas.
Applications
| SITUATION | Details of the real world situation being modelled. |
| EQUATION | The equation describing the situation. |
| GRAPH | The graph describing the situation. |
| QUESTIONS | Questions about specific features of the situation, such as the period, amplitude, certain values, maxima or minima. |
Sometimes the equation or the graph will be given, and sometimes you will be asked to find them.
Finding the equation of a model
Let's review the key features of a sine graph and how we might calculate the parameters given a graph or information.
| Feature | How to find |
|---|---|
| Amplitude, $a$a |
This is half the vertical distance between a maximum and a minimum, hence can be calculated as: $a=\frac{y_{max}-y_{min}}{2}$a=ymax−ymin2 |
| Period, $P$P |
This is the time taken to repeat one cycle. This can be found as the horizontal distance between two successive maximums or minimums. $P=x_{max,2}-x_{max,1}$P=xmax,2−xmax,1 |
| $b$b |
The parameter $b$b impacts the period as it causes a horizontal dilation by a factor of $\frac{1}{b}$1b. It can be calculated as follows: $b=\frac{2\pi}{P}$b=2πP |
| Principal axis, $y=d$y=d |
The principal axis or mid-line can be found as the average of the $y$y-coordinates for a maximum and minimum. Hence, can be calculated using: $d=\frac{y_{max}+y_{min}}{2}$d=ymax+ymin2 |
| Phase shift, $c$c |
This is how far the model has been horizontally shifted. For a sine model this can be found visually as the $x$x-value where the graph crosses the mid-line between a minimum and maximum. Or calculated as the average of the $x$x-coordinates of a minimum followed by a maximum: (note order is important) $c=\frac{x_{min}+x_{max}}{2}$c=xmin+xmax2 For a cosine model this can be found as the $x$x-coordinate of the first maximum: $c=x_{max}$c=xmax |
The only difference between finding a sine model or a cosine model is the phase shift. We could in fact model all such graphs as sine models but if a graph starts at a maximum or minimum we can model using a cosine function and not require a phase shift.
Worked example
Example 2
Find the equation of the form $y=a\sin\left(b\left(x-c\right)\right)+d$y=asin(b(x−c))+d to fit the following model:
Think: Do we have enough information to find the model? In this case three points were labelled for us but if they were not we could label them or extract key information from a written question.
Do: Find or list each of the parameters. You may be able to easily read them from the graph or information given. In this case we will calculate each from the given points.
Amplitude:
| $a$a | $=$= | $\frac{y_{max}-y_{min}}{2}$ymax−ymin2 |
| $=$= | $\frac{2-\left(-4\right)}{2}$2−(−4)2 | |
| $=$= | $3$3 |
Period:
| $P$P | $=$= | $x_{max,2}-x_{max,1}$xmax,2−xmax,1 |
| $=$= | $6.5-2.5$6.5−2.5 | |
| $=$= | $4$4 |
b:
| $b$b | $=$= | $\frac{2\pi}{P}$2πP |
| $=$= | $\frac{2\pi}{4}$2π4 | |
| $=$= | $\frac{\pi}{2}$π2 |
Principal axis:
| $d$d | $=$= | $\frac{y_{max}+y_{min}}{2}$ymax+ymin2 |
| $=$= | $\frac{2+\left(-4\right)}{2}$2+(−4)2 | |
| $=$= | $-1$−1 |
Phase shift:
| $c$c | $=$= | $\frac{x_{min}+x_{max}}{2}$xmin+xmax2 |
| $=$= | $\frac{0.5+2.5}{2}$0.5+2.52 | |
| $=$= | $1.5$1.5 |
Hence, the model that suits the graph is $y=3\sin\left(\frac{\pi}{2}\left(x-1.5\right)\right)-1$y=3sin(π2(x−1.5))−1
Reflect: Do the transformations fit the model shown? Check direction of shifts and period.
Practice question
Question 4
Determine the equation of the graph given that it is of the form $y=a\cos\left(x-c\right)$y=acos(x−c), where $c$c is the least positive value and $x$x is in radians.