Learning Outcomes
- Identify reference angles for angles measured in both radians and degrees
- Evaluate trigonometric functions using the unit circle
- Solve trigonometric equations
- Apply reduction formulas
In the Area and Arc Length in Polar Coordinates section, rather than use the rectangular coordinate system to calculate area under a curve and arc length, we will use the polar coordinate system. Here we will review how to evaluate sine and cosine functions at specific angle measures, solve trigonometric equations, and use reduction formulas.
Find Reference Angles
(See Module 1, Skills Review for Polar Coordinates)
Evaluate Trigonometric Functions Using the Unit Circle
(See Module 1, Skills Review for Polar Coordinates)
Solve Trigonometric Equations
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 [latex]2\pi[/latex]. In other words, every [latex]2\pi[/latex] units, the y-values repeat. 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. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\pi :[/latex]
[latex]\sin \theta =\sin \left(\theta \pm 2k\pi \right)[/latex]
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: Solving a Linear Trigonometric Equation Containing Cosine
Find all possible exact solutions for the equation [latex]\cos \theta =\frac{1}{2}[/latex].
Show Solution
From the unit circle, we know that
[latex]\begin{gathered}\cos \theta =\frac{1}{2} \\ \theta =\frac{\pi }{3},\frac{5\pi }{3} \end{gathered}[/latex]
These are the solutions in the interval [latex]\left[0,2\pi \right][/latex]. All possible solutions are given by
[latex]\theta =\frac{\pi }{3}\pm 2k\pi \text{ and }\theta =\frac{5\pi }{3}\pm 2k\pi[/latex]
where [latex]k[/latex] is an integer.
EXAMPLE: Solving a Linear Trigonometric Equation Containing Sine
Find all possible exact solutions for the equation [latex]\sin t=\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 the unit circle, we can see that the solutions are [latex]t=\frac{\pi }{6}[/latex] and [latex]t=\frac{5\pi }{6}[/latex]. But the problem is asking for all possible values that solve the equation. Therefore, the answer is
[latex]t=\frac{\pi }{6}\pm 2\pi k\text{ and }t=\frac{5\pi }{6}\pm 2\pi k[/latex]
where [latex]k[/latex] is an integer.
EXAMPLE: Solving a Linear Trigonometric Equation Containing Cosine
Solve the equation exactly: [latex]2\cos \theta -3=-5,0\le \theta <2\pi[/latex].
Show Solution
[latex]\begin{gathered}2\cos \theta -3=-5 \\ \cos \theta =-2 \\ \cos \theta =-1 \\ \theta =\pi \end{gathered}[/latex]
Try It
Solve exactly the following linear equation on the interval [latex]\left[0,2\pi \right):2\sin x+1=0[/latex].
Show Solution
[latex]x=\frac{7\pi }{6},\frac{11\pi }{6}[/latex]
When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and 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 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 [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: Solving a Trigonometric Equation
Solve the problem exactly: [latex]2{\sin }^{2}\theta -1=0,0\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]\sin \theta[/latex]. Then we will find the angles.
[latex]\begin{gathered}2{\sin }^{2}\theta -1=0 \\ 2{\sin }^{2}\theta =1 \\ {\sin }^{2}\theta =\frac{1}{2} \\ \sqrt{{\sin }^{2}\theta }=\pm \sqrt{\frac{1}{2}} \\ \sin \theta =\pm \frac{1}{\sqrt{2}}=\pm \frac{\sqrt{2}}{2} \\ \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4} \end{gathered}[/latex]
Example: Solving a Trigonometric Equation
Solve the following equation exactly: [latex]\csc \theta =-2,0\le \theta <4\pi[/latex].
Show Solution
We want all values of [latex]\theta[/latex] for which [latex]\csc \theta =-2[/latex] over the interval [latex]0\le \theta <4\pi[/latex].
[latex]\begin{gathered}\csc \theta =-2 \\ \frac{1}{\sin \theta }=-2 \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6},\frac{19\pi }{6},\frac{23\pi }{6} \end{gathered}[/latex]
Analysis of the Solution
As [latex]\sin \theta =-\frac{1}{2}[/latex], notice that all four solutions are in the third and fourth quadrants.
Example: Solving a Trigonometric Equation
Solve the equation exactly: [latex]\tan \left(\theta -\frac{\pi }{2}\right)=1,0\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]\tan \left(\frac{\pi }{4}\right)=1[/latex], then
[latex]\begin{gathered}\theta -\frac{\pi }{2}=\frac{\pi }{4}\\ \theta =\frac{3\pi }{4}\pm k\pi \end{gathered}[/latex]
Over the interval [latex]\left[0,2\pi \right)[/latex], we have two solutions:
[latex]\theta =\frac{3\pi }{4}\text{ and }\theta =\frac{3\pi }{4}+\pi =\frac{7\pi }{4}[/latex]
Try It
Find all solutions for [latex]\tan x=\sqrt{3}[/latex].
Show Solution
[latex]\frac{\pi }{3}\pm \pi k[/latex]
Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as [latex]\sin \left(2x\right)[/latex] or [latex]\cos \left(3x\right)[/latex]. When confronted with these equations, recall that [latex]y=\sin \left(2x\right)[/latex] is a horizontal compression by a factor of 2 of the function [latex]y=\sin x[/latex]. On an interval of [latex]2\pi[/latex], we can graph two periods of [latex]y=\sin \left(2x\right)[/latex], as opposed to one cycle of [latex]y=\sin x[/latex]. This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to [latex]\sin \left(2x\right)=0[/latex] compared to [latex]\sin x=0[/latex]. This information will help us solve the equation.
Example: Solving a Multiple Angle Trigonometric Equation
Solve exactly: [latex]\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]\cos \left(\alpha \right)=\frac{1}{2}[/latex], we know [latex]\alpha[/latex] is in quadrants I and IV. While [latex]\theta ={\cos }^{-1}\frac{1}{2}[/latex] will only yield solutions in quadrants I and II, we recognize that the solutions to the equation [latex]\cos \theta =\frac{1}{2}[/latex] will be in quadrants I and IV.
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]. Does this make sense? Yes, because [latex]\cos \left(2\left(\frac{\pi }{6}\right)\right)=\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 \\ &=\frac{\pi }{3}+\frac{6\pi }{3} \\ &=\frac{7\pi }{3} \end{align}[/latex]
so [latex]x=\frac{7\pi }{6}[/latex].
One more rotation yields
[latex]\begin{align} 2x&=\frac{\pi }{3}+4\pi \\ &=\frac{\pi }{3}+\frac{12\pi }{3} \\ &=\frac{13\pi }{3} \end{align}[/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 \\ &=\frac{5\pi }{3}+\frac{6\pi }{3} \\ &=\frac{11\pi }{3}\end{align}[/latex]
so [latex]x=\frac{11\pi }{6}[/latex].
One more rotation yields
[latex]\begin{align}2x&=\frac{5\pi }{3}+4\pi \\ &=\frac{5\pi }{3}+\frac{12\pi }{3} \\ &=\frac{17\pi }{3} \end{align}[/latex]
[latex]x=\frac{17\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].
Our solutions are [latex]x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\text{and }\frac{11\pi }{6}[/latex]. Note that whenever we solve a problem in the form of [latex]\sin \left(nx\right)=c[/latex], we must go around the unit circle [latex]n[/latex] times.
Use Reduction Formulas
A General Note: Reduction Formulas
The reduction formulas are summarized as follows:
[latex]\begin{align}&{\sin }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{2} \\ &{\cos }^{2}\theta =\frac{1+\cos \left(2\theta \right)}{2} \\ &{\tan }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{1+\cos \left(2\theta \right)} \end{align}[/latex]
Example: Using reduction formulas to reduce powers
Write an equivalent expression for [latex]{\cos }^{4}x[/latex] that does not involve any powers of sine or cosine greater than 1.
Show Solution
We will apply the reduction formula for cosine twice.
[latex]\begin{align}{\cos }^{4}x&={\left({\cos }^{2}x\right)}^{2} \\ &={\left(\frac{1+\cos \left(2x\right)}{2}\right)}^{2}&& {\text{Substitute reduction formula for cos}}^{2}x. \\ &=\frac{1}{4}\left(1+2\cos \left(2x\right)+{\cos }^{2}\left(2x\right)\right) \\ &=\frac{1}{4}+\frac{1}{2}\cos \left(2x\right)+\frac{1}{4}\left(\frac{1+\cos \left(2\left(2x\right)\right)}{2}\right) && {\text{ Substitute reduction formula for cos}}^{2}x. \\ &=\frac{1}{4}+\frac{1}{2}\cos \left(2x\right)+\frac{1}{8}+\frac{1}{8}\cos \left(4x\right) \\ &=\frac{3}{8}+\frac{1}{2}\cos \left(2x\right)+\frac{1}{8}\cos \left(4x\right) \end{align}[/latex]
Analysis of the Solution
The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.
Example: Using Reduction formulas To Reduce Powers
Use the power-reducing formulas to prove
[latex]{\sin }^{3}\left(2x\right)=\left[\frac{1}{2}\sin \left(2x\right)\right]\left[1-\cos \left(4x\right)\right][/latex]
Show Solution
We will work on simplifying the left side of the equation:
[latex]\begin{align}{\sin }^{3}\left(2x\right)&=\left[\sin \left(2x\right)\right]\left[{\sin }^{2}\left(2x\right)\right] \\ &=\sin \left(2x\right)\left[\frac{1-\cos \left(4x\right)}{2}\right]&& \text{Substitute the power-reduction formula}. \\ &=\sin \left(2x\right)\left(\frac{1}{2}\right)\left[1-\cos \left(4x\right)\right] \\ &=\frac{1}{2}\left[\sin \left(2x\right)\right]\left[1-\cos \left(4x\right)\right] \end{align}[/latex]
Analysis of the Solution
Note that in this example, we substituted
[latex]\frac{1-\cos \left(4x\right)}{2}[/latex]
for [latex]{\sin }^{2}\left(2x\right)[/latex]. The formula states
[latex]{\sin }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{2}[/latex]
We let [latex]\theta =2x[/latex], so [latex]2\theta =4x[/latex].
Try It
Use the power-reducing formulas to prove that [latex]10{\cos }^{4}x=\frac{15}{4}+5\cos \left(2x\right)+\frac{5}{4}\cos \left(4x\right)[/latex].
Show Solution
[latex]\begin{align}10{\cos }^{4}x&=10{\left({\cos }^{2}x\right)}^{2} \\ &=10{\left[\frac{1+\cos \left(2x\right)}{2}\right]}^{2}&& {\text{Substitute reduction formula for cos}}^{2}x. \\ &=\frac{10}{4}\left[1+2\cos \left(2x\right)+{\cos }^{2}\left(2x\right)\right] \\ &=\frac{10}{4}+\frac{10}{2}\cos \left(2x\right)+\frac{10}{4}\left(\frac{1+\cos\left( 2\left(2x\right)\right)}{2}\right)&& {\text{Substitute reduction formula for cos}}^{2}x. \\ &=\frac{10}{4}+\frac{10}{2}\cos \left(2x\right)+\frac{10}{8}+\frac{10}{8}\cos \left(4x\right) \\ &=\frac{30}{8}+5\cos \left(2x\right)+\frac{10}{8}\cos \left(4x\right) \\ &=\frac{15}{4}+5\cos \left(2x\right)+\frac{5}{4}\cos \left(4x\right) \end{align}[/latex]
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