Learning Objectives
In this section, you will:
- Solve linear trigonometric equations in sine and cosine.
- Solve equations involving a single trigonometric function.
- Solve trigonometric equations using a calculator.
- Solve trigonometric equations that are quadratic in form.
- Solve trigonometric equations using fundamental identities.
- Solve trigonometric equations with multiple angles.
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. 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.
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. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid.
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. The period of both the sine function and the cosine function is [latex]2\pi.[/latex] In other words, every [latex]2\pi[/latex] units, the y-values repeat, so [latex]\mathrm{sin}\left(\theta\right)=\mathrm{sin}\left(\theta \pm2k\pi\right)[/latex]. If we need to find all possible solutions, then we must add [latex]2\pi k,[/latex] where [latex]k[/latex] is an integer, to the initial solution.
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.
Example 1: Solving a Linear Trigonometric Equation Involving the Cosine Function
Find all possible exact solutions for the equation [latex]\mathrm{cos}\left(\theta\right) =\frac{1}{2}.[/latex]
Show Solution
From the unit circle, we know that there will be two angles where [latex]\mathrm{cos}\left(\theta\right) =\frac{1}{2}[/latex] in one complete revolution, i.e. [latex]0\le\theta\le{2\pi}.[/latex] They will occur in the first and fourth quadrants. We recognize [latex]\frac{1}{2}[/latex] as a value from one of our special right triangles. We can identify the acute angle as [latex]\frac{\pi}{3}.[/latex] We can then use this as the reference angle to find the angle in the fourth quadrant by computing [latex]2\pi-\frac{\pi}{3}.[/latex]
[latex]\begin{align*}\mathrm{cos}\left(\theta\right)&=\frac{1}{2}\\ \theta&=\frac{\pi }{3},\frac{5\pi }{3}\end{align*}[/latex][latex]\\[/latex]
These are the solutions in the interval [latex]\left[0,2\pi \right].[/latex] All possible solutions are given by [latex]\frac{\pi }{3}\pm2k\pi[/latex] and [latex]\frac{5\pi }{3}\pm2k\pi [/latex] where [latex]k[/latex] is an integer.
Example 2: Solving a Linear Equation Involving the Sine Function
Find all possible exact solutions for the equation [latex]\mathrm{sin}\left(t\right)=\frac{1}{2}.[/latex]
Show Solution
Solving for all possible values of t means that solutions include angles beyond the period of [latex]2\pi .[/latex] From previous work with the unit circle, we know that there are two solutions in one revolution. Since the value is positive, we know these solutions are in the first and second quadrants. From our special right triangles, we know that the angle in the first quadrant is [latex]\frac{\pi }{6}[/latex] and using this as the reference angle, the solution in the second quadrant is [latex]\frac{5\pi }{6}.[/latex] But the problem is asking for all possible values that solve the equation.
Therefore, the answer is [latex]\frac{\pi }{6}\pm2\pi k[/latex] and [latex]\frac{5\pi }{6}\pm2\pi k[/latex] where [latex]k[/latex] is an integer.
Example 3: Solve the Trigonometric Equation in Linear Form
Solve the equation exactly: [latex]2\text{ }\mathrm{cos}\left(\theta\right) -3=-5,\text{ }\text{ }0\le \theta <2\pi .[/latex]
Show Solution
Use algebraic techniques to solve the equation.
[latex]\begin{align*}2\text{ }\mathrm{cos}\left(\theta\right) -3&=-5\\ 2\mathrm{cos}\left(\theta\right) &=-2 &&\text{Added 3 to both sides. }\\ \mathrm{cos}\left(\theta\right) &=-1&&\text{Divided both sides by 2. }\\ \theta &=\pi\end{align*}[/latex]
Thinking of the unit circle, we can see that there is only one place where the cosine value is -1 in one complete revolution.
Try it #1
Solve exactly the following linear equation on the interval [latex]\left[0,2\pi \right):\text{ }2\text{ }\mathrm{sin}\left(x\right)+1=0.[/latex]
Show Solution
[latex]x=\frac{7\pi }{6},\text{ }\frac{11\pi }{6}[/latex]
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 information we know from the unit circle. 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 equation in terms of the reciprocal function, and solve for the angles using the functions we are most familiar with. 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 [latex]\pi,[/latex] not [latex]2\pi.[/latex] Further, the domain of tangent is all real numbers with the exception of odd integer multiples of [latex]\frac{\pi }{2},[/latex] unless, of course, a problem places its own restrictions on the domain.
Example 4: Solving a Trigonometric Equation Involving Cosecant
Solve the following equation exactly: [latex]\mathrm{csc}\left(\theta\right) =-2,\text{ }0\le \theta <4\pi .[/latex]
Show Solution
We want all values of [latex]\theta [/latex] for which [latex]\mathrm{csc}\left(\theta\right) =-2[/latex] over the interval [latex]0\le \theta <4\pi.[/latex]
[latex]\begin{align*}\mathrm{csc}\left(\theta\right) &=-2 \\ \frac{1}{\mathrm{sin}\left(\theta\right) }&=-2 \\ \mathrm{sin}\left(\theta\right) &=-\frac{1}{2}\\&=\frac{7\pi }{6},\text{ }\frac{11\pi }{6},\text{ }\frac{19\pi }{6},\text{ }\frac{23\pi }{6} \end{align*}[/latex][latex]\\[/latex]
As [latex]\mathrm{sin}\left(\theta\right) =-\frac{1}{2},[/latex] notice that all four solutions are in the third and fourth quadrants.
Example 5: Solving an Equation Involving Tangent
Solve the equation exactly: [latex]\mathrm{tan}\left(\theta -\frac{\pi }{2}\right)=1,\text{ } 0\le \theta <2\pi .[/latex]
Show Solution
Recall that the tangent function has a period of [latex]\pi.[/latex] On the interval [latex]\left[0,\pi \right),[/latex] and at the angle of [latex]\frac{\pi }{4},[/latex] the tangent has a value of 1. However, the angle we want is [latex]\left(\theta -\frac{\pi }{2}\right).[/latex] Thus, if [latex]\mathrm{tan}\left(\frac{\pi }{4}\right)=1,[/latex] then
[latex]\begin{align*}\theta -\frac{\pi }{2}&=\frac{\pi }{4}\\ \theta &=\frac{3\pi }{4}\pm k\pi \end{align*}[/latex][latex]\\[/latex]
Over the interval [latex]\left[0,2\pi \right),[/latex] we have two solutions:
[latex]\frac{3\pi }{4}[/latex] and [latex]\frac{3\pi}{4}+\pi =\frac{7\pi}{4}[/latex]
Try it #2
Find all solutions for [latex]\mathrm{tan}\left(x\right)=\sqrt[\leftroot{1}\uproot{2} ]{3}.[/latex]
Show Solution
[latex]\frac{\pi }{3}\pm\pi k[/latex]
Example 6: Identify all Solutions to the Equation Involving Tangent
Identify all exact solutions to the equation [latex]2\left(\mathrm{tan}\left(x\right)+3\right)=5+\mathrm{tan}\left(x\right),\text{ }0\le x<2\pi .[/latex]
Show Solution
We can solve this equation using only algebra. Isolate the expression [latex]\mathrm{tan}\left(x\right)[/latex] on the left side of the equals sign.
[latex]\begin{align*}2\text{ }\left(\mathrm{tan}\left(x\right)\right)+2\text{ }\left(3\right)& =5+\mathrm{tan}\left(x\right)&&\text{Distribute the 2 on the left hand side. }\\ 2\text{ }\mathrm{tan}\left(x\right)+6&=5+\mathrm{tan}\left(x\right) \\ 2\text{ }\mathrm{tan}\left(x\right)-\mathrm{tan}\left(x\right)& =5-6 &&\text{Isolate the tangent on one side.}\\ \mathrm{tan}\left(x\right)& =-1 \end{align*}[/latex][latex]\\[/latex]
There are two angles on the unit circle that have a tangent value of [latex]-1:\theta =\frac{3\pi }{4}[/latex] and [latex]\theta =\frac{7\pi }{4}.[/latex]
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.
Example 7: Using a Calculator to Solve a Trigonometric Equation Involving Sine
Use a calculator to solve the equation [latex]\mathrm{sin}\left(\theta\right) =0.8,[/latex] where [latex]\theta[/latex] is in radians.
Show Solution
Make sure mode is set to radians. To find [latex]\theta,[/latex] use the inverse sine function. On most calculators, you will need to push the 2ND button and then the SIN button to bring up the [latex]\mathrm{sin}^{-1}[/latex] function. What is shown on the screen is [latex]\mathrm{sin}^{-1}\left(\text{ }\right).[/latex] The calculator is ready for the input within the parentheses. For this problem, we enter [latex]\mathrm{sin}^{-1}\left(0.8\right),[/latex] and press ENTER. Thus, to four decimals places,
[latex]{\mathrm{sin}}^{-1}\left(0.8\right)\approx 0.9273[/latex]
The solution is [latex]0.9273\pm2\pi k.[/latex]
Keep in mind that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. Since the sine value is positive, there will be another solution in quadrant 2. The other angle is obtained by using [latex]\pi -\theta.[/latex]
This gives us a second set of values in the form [latex]2.2143\pm2\pi k.[/latex]
The angle measurements in degrees are based on the first revolution angles of
[latex]\theta\approx {53.1}^{\circ}\text{ or } \theta\approx {180}^{\circ}-{53.1}^{\circ}\approx {126.9}^{\circ} [/latex]
Example 8: Using a Calculator to Solve a Trigonometric Equation Involving Secant
Use a calculator to solve the equation [latex]\mathrm{sec}\left(\theta\right) =-4,[/latex] giving your answer in radians.
Show Solution
We can begin with some algebra.
[latex]\begin{align*}\mathrm{sec}\left(\theta\right) &=-4\\ \frac{1}{\mathrm{cos}\left(\theta\right)}&=-4\\\mathrm{cos}\left(\theta\right) &=-\frac{1}{4}\end{align*}[/latex]
Check that the MODE is in radians. Now use the inverse cosine function.
[latex]\begin{align*}\mathrm{cos}^{-1}\left(-\frac{1}{4}\right)&\approx 1.8235\\ \theta& \approx 1.8235+2\pi k\\ \end{align*}[/latex][latex]\\[/latex]
Since [latex]\frac{\pi }{2}\approx 1.57[/latex] and [latex]\pi \approx 3.14,[/latex] we know that 1.8235 is between these two numbers, thus [latex]\theta \approx \text{1}\text{.8235}[/latex] 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 Figure 1.
So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\theta^{\prime} \approx \pi -\text{1}\text{.8235}\approx \text{1}\text{.3181}\text{.}[/latex] The other solution in quadrant III is [latex]\pi +\text{1}\text{.3181}\approx \text{4}\text{.4597.}[/latex]
The solutions are [latex]1.8235\pm2\pi k[/latex] and [latex]4.4597\pm2\pi k.[/latex]
Try it #3
Solve [latex]\mathrm{csc}\left(\theta\right) =3.[/latex]
Show Solution
[latex]\theta \approx 0.33984\pm2\pi k[/latex] and [latex]\theta \approx 2.80176\pm2\pi k[/latex]
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 [latex]x[/latex] or [latex]u.[/latex] 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.
How To
Given a trigonometric equation, solve using algebra.
- Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
- Substitute the trigonometric expression with a single variable, such as [latex]x[/latex] or [latex]u.[/latex]
- Solve the equation the same way an algebraic equation would be solved.
- Substitute the trigonometric expression back in for the variable in the resulting expressions.
- Solve for the angle.
Example 9: Solving a Trigonometric Equation in Quadratic Form Using the Square Root Property
Solve the problem exactly: [latex]2\text{ }{\mathrm{sin}}^{2}\left(\theta\right) -1=0, \text{ }0\le \theta <2\pi .[/latex]
Show Solution
As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\mathrm{sin}\left(\theta\right).[/latex] Then we will find the angles.
[latex]\begin{align*}2\mathrm{sin}^{2}\left(\theta\right) -1&=0\\ 2\mathrm{sin}^{2}\left(\theta\right)&=1&&\text{Add 1 to both sides.} \\ \mathrm{sin}^{2}\left(\theta\right) &=\frac{1}{2}&&\text{Divide both sides by 2.}\\ \sqrt{\mathrm{sin}^{2}\left(\theta\right)}&=\pm\sqrt{\frac{1}{2}}&&\text{Consider both positive and negative square root values.} \\ \mathrm{sin}\left(\theta\right) &=\pm\frac{1}{\sqrt{2}}&&\text{Recognize the value from a special right triangle.}\\ \theta &=\frac{\pi }{4},\text{ }\frac{3\pi }{4},\text{ }\frac{5\pi }{4},\text{ }\frac{7\pi }{4}&&\text{Since we have + and -, we have answers in all 4 quadrants}\end{align*}[/latex]
Example 10: Solving a Trigonometric Equation in Quadratic Form Using the Quadratic Equation
Solve the equation exactly: [latex]\mathrm{cos}^{2}\left(\theta\right)+3\mathrm{cos}\left(\theta\right)-1=0,\text{ }0\le \theta <2\pi .[/latex]
Show Solution
We begin by using substitution and replacing [latex]\mathrm{cos}\left(\theta\right)[/latex] with [latex]u.[/latex] It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\mathrm{cos}\left(\theta\right) =u.[/latex] We have
[latex]{u}^{2}+3u-1=0[/latex][latex]\\[/latex]
The equation cannot be factored, so we will use the quadratic formula [latex]u=\frac{-b\pm\sqrt{{b}^{2}-4ac}}{2a},[/latex] where [latex]a=1,\text{ }b=3[/latex] and [latex]c=-1.[/latex]
[latex]\begin{align*} \\ u&=\frac{-3\pm\sqrt{{\left(3\right)}^{2}-4\left(1\right)\left(-1\right)}}{2}\\ &=\frac{-3\pm\sqrt{13}}{2}\end{align*}[/latex][latex]\\[/latex]
Replace [latex]u[/latex] with [latex]\mathrm{cos}\left(\theta\right),[/latex] and solve. Thus,
[latex]\begin{align*} \mathrm{cos}\left(\theta\right)&=\frac{-3\pm\sqrt{13}}{2}\\\theta&=\mathrm{cos}^{-1}\left(\frac{-3+\sqrt{13}}{2}\right) \end{align*}[/latex][latex]\\[/latex]
Note that only the + sign is used. This is because we get an error when we solve [latex]\theta =\mathrm{cos}^{-1}\left(\frac{-3-\sqrt{13}}{2}\right)[/latex] on a calculator, since the domain of the inverse cosine function is [latex]\left[-1,1\right].[/latex] Therefore the solution is
[latex]\begin{align*}\mathrm{cos}^{-1}\left(\frac{-3+\sqrt{13}}{2}\right)\approx 1.26\end{align*}[/latex][latex]\\[/latex]
This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is
[latex]\begin{align*}2\pi -{\mathrm{cos}}^{-1}\left(\frac{-3+\sqrt{13}}{2}\right)\approx 5.02 \end{align*}[/latex][latex]\\[/latex]
Example 11: Solving a Trigonometric Equation in Quadratic Form by Factoring
Solve the equation exactly: [latex]2\text{ }\mathrm{sin}^{2}\left(\theta\right) -5\text{ }\mathrm{sin}\left(\theta\right)+3=0,\text{ }0\le \theta \le 2\pi .[/latex]
Show Solution
Using grouping, this quadratic can be factored. Either make the real substitution, [latex]\mathrm{sin}\left(\theta\right) =u[/latex] and factor [latex]2u^2-5u+3=0,[/latex] or imagine it, as we factor:
[latex]\begin{align*}\textrm{Without the substitution }&& &&\text{With the substitution}\\2\text{ }\mathrm{sin}^{2}\left(\theta\right) -5\text{ }\mathrm{sin}\left(\theta\right) +3&=0&& &2u^2-5u+3=0\\ \left(2\text{ }\mathrm{sin}\left(\theta\right)-3\right)\left(\mathrm{sin}\left(\theta\right) -1\right)&=0 &&&\left(2u-3\right)\left(u-1\right)\end{align*}[/latex][latex]\text{ }[/latex][latex]\\[/latex]
Now set each factor equal to zero and solve the two equations [latex]2\text{ }\mathrm{sin}\left(\theta\right) -3=0[/latex] and [latex]\mathrm{sin}\left(\theta\right) -1=0.[/latex]
For the first equation,
[latex]\begin{align*}2\text{ }\mathrm{sin}\left(\theta\right) -3&=0\\ 2\text{ }\mathrm{sin}\left(\theta\right)& =3&&\textrm{Add 3 to both sides.}\\ \mathrm{sin}\left(\theta\right)& =\frac{3}{2}&&\textrm{Divide both sides by 2.} \\ \mathrm{sin}\left(\theta\right)&\ne \frac{3}{2}&&\frac{3}{2}\textrm{ is not in the domain of the sine function.}\end{align*}[/latex][latex]\text{ }[/latex][latex]\\[/latex]
The first equation did not have any solution. To solve the second equation,
[latex]\begin{align*}\mathrm{sin}\left(\theta\right) -1&=0\\ \mathrm{sin}\left(\theta\right)&=1&&\textrm{Add 1 to both sides.}\\\theta&=\frac{\pi}{2}&&\textrm{Sine equals one only for the quadrantal angle.}\end{align*}[/latex][latex]\text{ }[/latex][latex]\\[/latex]
The only solution for this equation is [latex]\theta=\frac{\pi}{2}.[/latex]
Analysis
Make sure to check all solutions on the given domain as some factors have no solution. This is because the range of the sine function is [latex]\left[-1,1\right].[/latex]
Example 12: Solving an Equation Using an Identity
Solve the equation exactly using an identity: [latex]3\text{ }\mathrm{cos}\left(\theta\right)+3=2\text{ }{\mathrm{sin}}^{2}\left(\theta\right) ,\text{ }0\le \theta <2\pi .[/latex]
Show Solution
If we rewrite the right side, we can write the equation in terms of cosine. Recall that the Pythagorean Identity is [latex]\mathrm{sin}^2\left(\theta\right)+\mathrm{cos}^2\left(\theta\right)=1[/latex] and can be solved for [latex]\mathrm{sin}^{2}\left(\theta\right)[/latex] so [latex]\mathrm{sin}^{2}\left(\theta\right)=1-\mathrm{cos}^{2}\left(\theta\right).[/latex]
[latex]\begin{align*}3\mathrm{cos}\left(\theta\right)+3&=2\mathrm{sin}^{2}\left(\theta\right)\\ 3\mathrm{cos}\left(\theta\right)+3&=2\left(1-\mathrm{cos}^{2}\left(\theta\right)\right)&&\text{Substitute for }\mathrm{sin}^{2}\left(\theta\right).\\\end{align*}[/latex]
Recognize that we now have a quadratic function in [latex]\mathrm{cos}\left(\theta\right).[/latex] Factoring can be used to solve this quadratic equation. Below, the same steps used on the left are demonstrated on the right using the substitution method since the quadratic equation in that form may be easier to work with. For the substitution method, let [latex]u=\mathrm{cos}\left(\theta\right).[/latex][latex]\\[/latex]
[latex]\begin{align*}\textrm{Without the substitution }&& &\text{With the substitution}\\3\mathrm{cos}\left(\theta\right)+3&=2-2\mathrm{cos}^{2}\left(\theta\right)&& 3u+3=2-2u^2\\ 2\mathrm{cos}^{2}\left(\theta\right)+3\mathrm{cos}\left(\theta\right)+1&=0&&2u^2+3u+1=0\\ \left(2\mathrm{cos}\left(\theta\right)+1\right)\left(\mathrm{cos}\left(\theta\right)+1\right)&=0&&\left(2u+1\right)\left(u+1\right)=0\end{align*}[/latex][latex]\\[/latex]
Working with the factors using cosine, set each factor equal to zero and solve.[latex]\\[/latex]
[latex]\begin{align*}2\mathrm{cos}\left(\theta\right)+1&=0 &\mathrm{cos}\left(\theta\right)+1&=0\\ \mathrm{cos}\left(\theta\right)&=-\frac{1}{2}& \mathrm{cos}\left(\theta\right)&=-1\\\theta&=\frac{2\pi}{3},\text{ }\frac{4\pi}{3}\text{ }\text{ }\text{ }\text{ }\text{ }&\theta&=\pi\end{align*}[/latex][latex]\\[/latex]
Again, remember that there are two quadrants where the [latex]\mathrm{cos}\left(\theta\right)=-\frac{1}{2}.[/latex] These are quadrants 2 and 3. We can find the acute reference angle of [latex]\frac{\pi}{3}[/latex] and use that to generate the solutions in these quadrants.
Our solutions are [latex]\frac{2\pi}{3},\text{ }\frac{4\pi }{3},\text{ }\pi.[/latex]
Try it #4
Solve [latex]\mathrm{sin}^{2}\left(\theta\right)=2\mathrm{cos}\left(\theta\right)+2,\text{ }0\le\theta \le2\pi [/latex] [Hint: Make a substitution to express the equation only in terms of cosine.]
Show Solution
[latex]\mathrm{cos}\left(\theta\right) =-1,\text{ }\theta =\pi [/latex]
Try it #5
Solve the quadratic equation [latex]2\mathrm{cos}^{2}\left(\theta\right) +\mathrm{cos}\left(\theta\right) =0.[/latex]
Show Solution
[latex]\frac{\pi }{2},\text{ }\frac{2\pi }{3},\text{ }\frac{4\pi }{3},\text{ }\frac{3\pi }{2}[/latex]
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.
We will need to use some new identities in this section. They are called the double-angle identities. We will not take the time to show where the following identities come from. Keep in mind that there are other trigonometric identities that we have not covered in this material. If you are interested, you can look up the sum and difference formulas for sine and cosine, and use those to generate some other identities, including the ones shown below.
Definition
The double angle identities are:
[latex]\mathrm{sin}\left(2\theta\right)=2\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right)[/latex]
[latex]\\[/latex]
[latex]\begin{align*}\mathrm{cos}\left(2\theta\right)&=\mathrm{cos}^{2}\left(\theta\right)-\mathrm{sin}^{2}\left(\theta\right)\text{ or,}\\&=1-2\mathrm{sin}^{2}\left(\theta\right)\text{ or,}\\&=2\mathrm{cos}^{2}\left(\theta\right)-1\end{align*}[/latex]
Example 13: Solving the Equation Using a Double-Angle Formula
Solve the equation exactly using a double-angle formula: [latex]\mathrm{cos}\left(2\theta\right)=\mathrm{cos}\left(\theta\right).[/latex]
Show Solution
We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:
[latex]\begin{align*}\mathrm{cos}\left(2\theta\right)&=\mathrm{cos}\left(\theta\right)\\ 2\mathrm{cos}^{2}\left(\theta\right)-1&=\mathrm{cos}\left(\theta\right)&&\text{Replace left hand side with identity.}\\ 2\mathrm{cos}^{2}\left(\theta\right)-\mathrm{cos}\left(\theta\right)-1&=0&&\text{Move all terms to one side.}\\ \left(2\mathrm{cos}\left(\theta\right)+1\right)\left(\mathrm{cos}\left(\theta\right)-1\right)&=0&&\text{Factor the left hand side.}\end{align*}[/latex]
[latex]\\[/latex]
Set the factors equal to zero and solve.
[latex]\begin{align*}2\mathrm{cos}\left(\theta\right)+1&=0&\mathrm{cos}\left(\theta\right)-1&=0\\\mathrm{cos}\left(\theta\right)&=-\frac{1}{2}&\mathrm{cos}\left(\theta\right)&=1\\\theta&=\frac{2\pi}{3}\pm2\pi k&\theta&=0\pm2\pi k\\\theta&=\frac{4\pi}{3}\pm2\pi k\end{align*}[/latex]
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 [latex]\mathrm{sin}\left(2x\right)[/latex] or [latex]\mathrm{cos}\left(3x\right).[/latex] When confronted with these equations, recall that [latex]y=\mathrm{sin}\left(2x\right)[/latex] is a horizontal compression by a factor of 2 of the function [latex]y=\mathrm{sin}\left(x\right).[/latex] On an interval of [latex]2\pi,[/latex] we can graph two periods of [latex]y=\mathrm{sin}\left(2x\right),[/latex] as opposed to one cycle of [latex]y=\mathrm{sin}\left(x\right).[/latex] This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to [latex]\mathrm{sin}\left(2x\right)=0[/latex] compared to [latex]\mathrm{sin}\left(x\right)=0.[/latex] This information will help us solve the similar type of equation shown in the example.
Example 14: Solving a Multiple Angle Trigonometric Equation
Solve exactly: [latex]\mathrm{cos}\left(2x\right)=\frac{1}{2}[/latex] on [latex]\left[0,2\pi \right).[/latex]
Show Solution
We can see that this equation is the standard equation with a multiple of an angle. If [latex]\mathrm{cos}\left(\theta\right)=\frac{1}{2},[/latex] we know [latex]\theta[/latex] is in quadrants I and IV. While [latex]\theta =\mathrm{cos}^{-1}\left(\frac{1}{2}\right)[/latex] will only yield solutions in quadrants I and II because of the range of the inverse cosine function, we recognize that the solutions to the equation [latex]\mathrm{cos}\left(\theta\right)=\frac{1}{2}[/latex] will be in quadrants I and IV using ideas from our unit circle.
Therefore, the possible angles are [latex]\theta=\frac{\pi }{3}[/latex] and [latex]\theta=\frac{5\pi}{3}.[/latex] So, [latex]2x=\frac{\pi}{3}[/latex] or [latex]2x=\frac{5\pi}{3},[/latex] which means that [latex]x=\frac{\pi}{6}[/latex] or [latex]x=\frac{5\pi}{6}.[/latex] [latex]\\[/latex]Does this make sense? Yes, because [latex]\mathrm{cos}\left(2\left(\frac{\pi}{6}\right)\right)=\mathrm{cos}\left(\frac{\pi}{3}\right)=\frac{1}{2}.[/latex]
Are there any other possible answers? Let us return to our first step.
In quadrant I, [latex]2x=\frac{\pi}{3},[/latex] so [latex]x=\frac{\pi}{6}[/latex] as noted. Let us revolve around the circle again:
[latex]\begin{align*}2x&=\frac{\pi}{3}+2\pi \\ 2x&=\frac{\pi }{3}+\frac{6\pi}{3}\\ 2x&=\frac{7\pi }{3}\\x&=\frac{7\pi}{6}\end{align*}[/latex][latex]\\[/latex]
One more rotation yields
[latex]\begin{align*}2x&=\frac{\pi }{3}+4\pi \\2x&=\frac{\pi}{3}+\frac{12\pi}{3}\\ 2x&=\frac{13\pi}{3}\\x&=\frac{13\pi}{6}\end{align*}[/latex][latex]\\[/latex]
[latex]x=\frac{13\pi}{6}>2\pi,[/latex] so this value for [latex]x[/latex] is larger than [latex]2\pi,[/latex] so it is not a solution on [latex]\left[0,2\pi \right).[/latex]
In quadrant IV, [latex]2x=\frac{5\pi }{3},[/latex] so [latex]x=\frac{5\pi }{6}[/latex] as noted. Let us revolve around the circle again:
[latex]\begin{align*}2x&=\frac{5\pi}{3}+2\pi \\2x&=\frac{5\pi }{3}+\frac{6\pi }{3} \\ 2x&=\frac{11\pi }{3}\\x&=\frac{11\pi}{6} \end{align*}[/latex][latex]\\[/latex]
One more rotation yields
[latex]\begin{align*}2x&=\frac{5\pi }{3}+4\pi \\ 2x&=\frac{5\pi }{3}+\frac{12\pi }{3} \\ 2x&=\frac{17\pi }{3}\\x&=\frac{17\pi}{6}\end{align*}[/latex][latex]\\[/latex]
[latex]x=\frac{17\pi }{6}>2\pi ,[/latex] so this value for [latex]x[/latex] is larger than [latex]2\pi,[/latex] which means it is not a solution on [latex]\left[0,2\pi \right).[/latex]
Our solutions are [latex]\frac{\pi}{6},\text{ }\frac{5\pi }{6},\text{ }\frac{7\pi }{6},[/latex] and [latex]\frac{11\pi }{6}.[/latex] Note that whenever we solve a problem in the form of [latex]\mathrm{sin}\left(nx\right)=c[/latex] or [latex]\mathrm{cos}\left(nx\right)=c,[/latex] we must go around the unit circle [latex]{n}[/latex] times.
We can see this easily if we first write the solution for [latex]2x[/latex] in the general form. One of these equations is shown below:
[latex]\begin{align*}2x&=\frac{\pi}{3}+2\pi\cdot k\\x&=\frac{\pi}{6}+\pi \cdot k&&\text{Divide both sides by 2.}\end{align*}[/latex][latex]\\[/latex]
We can now see that adding [latex]\pi[/latex] to [latex]\frac{\pi}{6}[/latex] gives us [latex]\frac{7\pi}{6}[/latex] and that adding [latex]2\cdot\pi[/latex] would give us [latex]\frac{13\pi}{6}[/latex] which is larger than [latex]2\pi.[/latex]
Key Concepts
- When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution.
- Equations involving a single trigonometric function can be solved or verified using the unit circle.
- We can also solve trigonometric equations using a graphing calculator.
- Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc.
- We can also use the identities to solve trigonometric equation.
- We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval.
