6.4 Solving Trigonometric Equations
Figure 6.4.1: Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill)
Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. He reasoned that when the height of his staff’s shadow was exactly equal to the actual height of the staff, then the height of the nearby pyramid’s shadow must also be equal to the height of the actual pyramid. Since the structures and their shadows were creating a right triangle with two equal sides, they were similar triangles. By measuring the length of the pyramid’s shadow at that moment, he could obtain the height of the pyramid. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.
In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.
6.4.1 Solving Linear Trigonometric Equations in Sine and Cosine
Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is \(2\pi.\) In other words, every \(2\pi\) units, the y-values repeat. If we need to find all possible solutions, then we must add \(2\pi k,\) where \(k\) is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is \(2\pi\text{:}\)
\[\sin\;\theta = \sin(\theta \pm 2k\pi)\]
There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.
6.4.2 Solving Equations Involving a Single Trigonometric Function
When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see the unit circle in §6.2). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is \(\pi,\) not \(2\pi.\) Further, the domain of tangent is all real numbers with the exception of odd integer multiples of \(\frac{\pi}{2},\) unless, of course, a problem places its own restrictions on the domain.
6.4.3 Solve Trigonometric Equations Using a Calculator
Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.
Use a calculator to solve the equation \(\sec\;\theta = -4,\) giving your answer in radians.
Solution
We can begin with some algebra.
\[\begin{array}{rcl} {\sec\;\theta} & = & {- 4} \\ \frac{1}{\cos\;\theta} & = & {- 4} \\ {\cos\;\theta} & = & {- \frac{1}{4}} \end{array}\]
Check that the MODE is in radians. Now use the inverse cosine function.
\[\begin{array}{rcl} {\cos^{- 1}\left( {- \frac{1}{4}} \right)} & \approx & 1.8235 \\ \theta & \approx & {1.8235 + 2\pi k} \end{array}\]
Since \(\frac{\pi}{2} \approx 1.57\) and \(\pi \approx 3.14,\) 1.8235 is between these two numbers, thus \(\theta \approx \text{1}\text{.8235}\) is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See the figure below.
Figure: The two solutions of \(\cos\theta = -\tfrac{1}{4}\) on \([0,\, 2\pi)\) share the same reference angle \(\approx 1.3181\).
So, we also need to find the measure of the angle in quadrant III. In quadrant II, the reference angle is \(\theta' \approx \pi - \text{1}\text{.8235} \approx \text{1}\text{.3181}\text{.}\) The other solution in quadrant III is \(\theta' \approx \pi + \text{1}\text{.3181} \approx \text{4}\text{.4597}\text{.}\)
The solutions are \(\theta \approx 1.8235 \pm 2\pi k\) and \(\theta \approx 4.4597 \pm 2\pi k.\)
6.4.4 Solving Trigonometric Equations in Quadratic Form
Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as \(x\) or \(u.\) If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.
Solve exactly:
\[2\;\sin^{2}\theta + \sin\;\theta = 0;0 \leq \theta < 2\pi\]
Solution
This problem should appear familiar as it is similar to a quadratic. Let \(\;\sin\;\theta = x.\) The equation becomes \(2x^{2} + x = 0.\) We begin by factoring:
\[\begin{array}{rcl} {2x^{2} + x} & = & 0 \\ {x(2x + 1)} & = & 0 \end{array}\]
Set each factor equal to zero.
\[\begin{array}{rcl} x & = & 0 \\ {(2x + 1)} & = & 0 \\ x & = & {- \frac{1}{2}} \end{array}\]
Then, substitute back into the equation the original expression \(\;\sin\;\theta\) for \(x.\) Thus,
\[\begin{array}{rcl} {\sin\;\theta} & = & 0 \\ \theta & = & {0,\pi} \\ {\sin\;\theta} & = & {- \frac{1}{2}} \\ \theta & = & {\frac{7\pi}{6},\frac{11\pi}{6}} \end{array}\]
The solutions within the domain \(0 \leq \theta < 2\pi\) are \(\theta = 0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}.\)
If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.
\[\begin{matrix} {2\;\sin^{2}\theta + \sin\;\theta} & = & 0 \\ {\sin\;\theta(2\sin\;\theta + 1)} & = & 0 \\ {\sin\;\theta} & = & 0 \\ \theta & = & {0,\pi} \\ & & \\ {2\;\sin\;\theta + 1} & = & 0 \\ {2\sin\;\theta} & = & {- 1} \\ {\sin\;\theta} & = & {- \frac{1}{2}} \\ \theta & = & {\frac{7\pi}{6},\frac{11\pi}{6}} \end{matrix}\]
Analysis. We can see the solutions on the graph below. On the interval \(0 \leq \theta < 2\pi,\) the graph crosses the x-axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.
Figure: The graph of \(y = 2\sin^2\theta + \sin\theta\) on \([0,\, 2\pi)\). Its four zeros match the algebraic solutions.
We can verify the solutions on the unit circle in §6.2 as well.
6.4.5 Solving Trigonometric Equations Using Fundamental Identities
While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.
6.4.6 Solving Trigonometric Equations with Multiple Angles
Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as \(\sin\left( {2x} \right)\) or \(\cos\left( {3x} \right).\) When confronted with these equations, recall that \(y = \sin\left( {2x} \right)\) is a horizontal compression by a factor of 2 of the function \(y = \sin\; x.\) On an interval of \(2\pi,\) we can graph two periods of \(y = \sin\left( {2x} \right),\) as opposed to one cycle of \(y = \sin\; x.\) This compression of the graph leads us to believe there may be twice as many \(x\)-intercepts or solutions to \(\sin\left( {2x} \right) = 0\) compared to \(\;\sin\; x = 0.\) This information will help us solve the equation.
6.4.7 Solving Right Triangle Problems
We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem. We begin with the familiar Pythagorean Theorem, \(a^{2} + b^{2} = c^{2},\) and model an equation to fit a situation.
Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem.
One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See the figure below.
Figure 6.4.2: A Ferris wheel anchored by a support cable. The cable, ground, and central support form a right triangle.
Solution
Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.
\[\begin{array}{rcl} {a^{2} + b^{2}} & = & c^{2} \\ {{(23)}^{2} + {(69.5)}^{2}} & \approx & 5359 \\ \sqrt{5359} & \approx & {73.2\text{~m}} \end{array}\]
The angle of elevation is \(\theta,\) formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places.
\[\begin{matrix} {\tan\;\theta} & = & \frac{69.5}{23} \\ & & \\ & & \\ {\tan^{-1}\left( \frac{69.5}{23} \right)} & \approx & 1.2522 \\ & \approx & {71.69{^\circ}\operatorname{}} \end{matrix}\]
The angle of elevation is approximately \(71.7{^\circ},\) and the length of the cable is 73.2 meters.
OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.
Solution
For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “a” feet from the wall, the length of the ladder will be 4\(a\) feet. See the figure below.
Figure: A ladder against a wall. The ladder’s length is \(4a\) (per the OSHA \(1{:}4\) rule); the base sits \(a\) feet from the wall.
The side adjacent to \(\theta\) is \(a\) and the hypotenuse is \(4a.\) Thus,
\[\begin{array}{rcl} {\cos\;\theta} & = & {\frac{a}{4a} = \frac{1}{4}} \\ {\cos^{-1}\left( \frac{1}{4} \right)} & \approx & 75.5{^\circ} \end{array}\]
The elevation of the ladder forms an angle of \(75.5{^\circ}\) with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:
\[\begin{array}{rcl} {a^{2} + b^{2}} & = & {(4a)}^{2} \\ b^{2} & = & {{(4a)}^{2} - a^{2}} \\ b^{2} & = & {16a^{2} - a^{2}} \\ b^{2} & = & {15a^{2}} \\ b & = & {a\sqrt{15}} \end{array}\]
Thus, the ladder touches the wall at \(a\sqrt{15}\) feet from the ground.
6.4.8 Summary
- Period and general solutions. \(\sin\theta\) and \(\cos\theta\) have period \(2\pi\), so any solution \(\theta = \alpha\) on a single period generalises to \(\theta = \alpha \pm 2k\pi\) for integer \(k\). \(\tan\theta\) has period \(\pi\), so its solutions generalise as \(\theta = \alpha \pm k\pi\).
- Linear and single-function equations. Isolate the trigonometric function, then read off the angles whose value matches.
- Calculator-aided equations. When the value isn’t a unit-circle value, use the inverse trigonometric function on a calculator and use symmetry / period to find the other solutions.
- Quadratic-form equations. Substitute a single variable for the trigonometric expression, solve the resulting algebraic equation, then back-substitute. Discard any value of the substituted variable that lies outside the function’s range.
- Identity-driven equations. Apply Pythagorean, sum-and-difference, and double-angle identities to rewrite the equation in a single trigonometric function of a single angle.
- Multiple angles. A solution \(\alpha\) of \(\sin(n\theta) = c\) (or its cosine analogue) generates \(n\) values of \(\theta\) in one period.
- Right-triangle problems. Identify the right triangle, label the sides, and use the appropriate ratio combined with the Pythagorean theorem.