5.7 solving trigonometric equations homework answers

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5.7 Solving trigonometric equations

5.7 solving trigonometric equations (ema3t).

In this section we will first look at finding unknown lengths in right-angled triangles and then we will look at finding unknown angles in right-angled triangles. Finally we will look at how to solve more general trigonometric equations.

Finding lengths (EMA3V)

From the definitions of the trigonometric ratios and what we have learnt about determining the values of these ratios for any angle we can now use this to help us find unknown lengths in right-angled triangles. The following worked examples will show you how.

Worked example 4: Finding lengths

Find the length of \(x\) in the following right-angled triangle using the appropriate trigonometric ratio (round your answer to two decimal places).

Identify the opposite and adjacent sides and the hypotenuse with reference to the given angle

Remember that the hypotenuse side is always opposite the right angle, it never changes position. The opposite side is opposite the angle we are interested in and the adjacent side is the remaining side.

Rearrange the equation to solve for \(x\)

Use your calculator to find the answer, worked example 5: finding lengths.

Worked example 6: Finding lengths

Find the length of \(x\) and \(y\) in the following right-angled triangle using the appropriate trigonometric ratio (round your answers to two decimal places).

Rearrange the equations to solve for \(x\) and \(y\)

Use your calculator to find the answers.

The following video shows an example of finding unknown lengths in a triangle using the trigonometric ratios.

Video: 2FQT

In each triangle find the length of the side marked with a letter. Give your answers correct to \(\text{2}\) decimal places.

Write down two ratios for each of the following in terms of the sides: \(AB; BC; BD; AD; DC \text{ and } AC\).

\(\sin \hat{B}\)

We note that triangles \(ABC\) and \(ABD\) both contain angle \(B\) so we can use these triangles to write down the ratios:

\(\cos \hat{D}\)

We note that triangles \(ACD\) and \(ABD\) both contain angle \(D\) so we can use these triangles to write down the ratios:

\(\tan \hat{B}\)

In \(\triangle MNP\), \(\hat{N} = 90°\), \(MP=20\) and \(\hat{P} = 40°\). Calculate \(NP\) and \(MN\) (correct to \(\text{2}\) decimal places).

Sketch the triangle:

To find \(MN\) we use the sine ratio:

To find \(NP\) we can use the cosine ratio:

Therefore \(MN = \text{12,86} \text{ and } NP = \text{15,32}\)

Calculate \(x\) and \(y\) in the following diagram.

To find \(x\) we use \(\triangle ABC\) and the tangent ratio. To find \(y\) we use \(\triangle ABD\) and the tangent ratio.

Therefore \(x = \text{29,82} \text{ and } y = \text{31,98}\).

Finding an angle (EMA3W)

If the length of two sides of a triangle are known, the angles can be calculated using trigonometric ratios. In this section, we are finding angles inside right-angled triangles using the ratios of the sides.

Worked example 7: Finding angles

Find the value of \(\theta\) in the following right-angled triangle using the appropriate trigonometric ratio.

Identify the opposite and adjacent sides with reference to the given angle and the hypotenuse

In this case you have the opposite side and the adjacent side for angle \(\theta\).

Use your calculator to solve for \(\theta\)

To solve for \(\theta\), you will need to use the inverse tangent function on your calculator. This works backwards by using the ratio of the sides to determine the angle which resulted in that ratio.

Press \(\boxed{\text{SHIFT}} \enspace \boxed{\text{tan}} \enspace \boxed{50} \enspace \boxed{\div} \enspace \boxed{100} \enspace \boxed{)} \enspace \boxed{=} \enspace \text{26,56505...} \approx \text{26,6}\)

Write the final answer

Determine \(\alpha\) in the following right-angled triangles:

We have now seen how to solve trigonometric equations in right-angled triangles. We can use the same techniques to help us solve trigonometric equations when the triangle is not shown.

Worked example 8: Solving trigonometric equations

Find the value of \(\theta\) if \(\cos \theta = \text{0,2}\).

To solve for \(\theta\), you will need to use the inverse cosine function on your calculator. This works backwards by using the ratio of the sides to determine the angle which resulted in that ratio.

Press \(\boxed{\text{SHIFT}} \enspace \boxed{\text{cos}} \enspace \boxed{0} \enspace \boxed{.} \enspace \boxed{2} \enspace \boxed{)} \enspace \boxed{=} \enspace \text{78,46304} \approx \text{78,46}\)

Worked example 9: Solving trigonometric equations

Find the value of \(\theta\) if \(3\sin \theta = \text{2,4}\).

Rearrange the equation

We need to rearrange the equation so that \(\sin \theta\) is on one side of the equation.

To solve for \(\theta\), you will need to use the inverse sine function on your calculator. This works backwards by using the ratio of the sides to determine the angle which resulted in that ratio.

Press \(\boxed{\text{SHIFT}} \enspace \boxed{\text{sin}} \enspace \boxed{(} \enspace \boxed{2} \enspace \boxed{.} \enspace \boxed{4} \enspace \boxed{\div} \enspace \boxed{3} \enspace \boxed{)} \enspace \boxed{=} \enspace \text{53,1301...} \approx \text{53,13}\).

When you are solving trigonometric equations you might find that you get an error when you try to calculate \(\sin\) or \(\cos\) (remember that both the sine and cosine functions have a maximum value of 1). For these cases there is no solution to the equation.

If learners get a math error on their calculator encourage them to think about what might have happened. It is also important to ensure that they know they must write down no solution rather than math error when this happens.

Worked example 10: Solving trigonometric equations

Solve for \(\alpha\): \(3 \sec \alpha = \text{1,4}\).

Convert \(\sec\) to \(\cos\)

There is no “\(\sec\)” button on the calculator and so we need to convert \(\sec\) to \(\cos\) so we can find \(\alpha\).

We need to rearrange the equation so that we have \(\cos \alpha\) on one side of the equation.

Use your calculator to solve for \(\alpha\)

To solve for \(\alpha\), you will need to use the inverse cosine function on your calculator. This works backwards by using the ratio of the sides to determine the angle which resulted in that ratio.

Press \(\boxed{\text{SHIFT}} \enspace \boxed{\text{cos}} \enspace \boxed{3} \enspace \boxed{\div} \enspace \boxed{1} \enspace \boxed{.} \enspace \boxed{4} \enspace \boxed{)} \enspace \boxed{=}\) math error

In this case we get an error when we try to do the calculation. This is because \(\dfrac{3}{\text{1,4}}\) is greater than 1 and the maximum value of the cosine function is 1. Therefore there is no solution. It is important in this case to write no solution and not math error.

There is no solution.

Determine the angle (correct to \(\text{1}\) decimal place):

\(\tan \theta = \text{1,7}\)

\(\sin \theta = \text{0,8}\)

\(\cos \alpha = \text{0,32}\)

\(\tan \beta = \text{4,2}\)

\(\tan \theta = 5\frac{3}{4}\)

\(\sin \theta = \frac{2}{3}\)

\(\cos \beta = \text{1,2}\)

\(4\cos \theta = 3\)

\(\cos 4\theta = \text{0,3}\)

\(\sin \beta + 2 = \text{2,65}\)

\(2\sin \theta + 5 = \text{0,8}\)

\(3 \tan \beta = 1\)

\(\sin 3 \alpha = \text{1,2}\)

\(\tan \frac{\theta}{3} = \sin 48°\)

\(\frac{1}{2}\cos 2\beta = \text{0,3}\)

\(2 \sin 3\theta + 1 = \text{2,6}\)

If \(x = 16°\) and \(y = 36°\), use your calculator to evaluate each of the following, correct to 3 decimal places.

\(\sin(x - y)\)

\(3\sin x\)

\(\tan x - \tan y\)

\(\cos x + \cos y\)

\(\frac{1}{3} \tan y\)

\(\text{cosec }(x - y)\)

\(2\cos x + \cos 3y\)

\(\tan(2x - 5y)\)

In each of the following find the value of \(x\) correct to two decimal places.

\(\sin x = \text{0,814}\)

\(\sin x = \tan \text{45}°\)

\(\tan 2x = \text{3,123}\)

\(\tan x = 3 \sin \text{41}°\)

\(\sin(2x + 45) = \text{0,123}\)

\(\sin(x - 10°) = \cos\text{57}°\)

Chapter 5: Equations and Identities

Exercises: Chapter 5 Review Problems

Suggested problems.

Exercises for Chapter 5 Review

Exercise group.

For Problems 1–4, evaluate the expressions for [latex]x = 120°,~ y = 225°,[/latex] and [latex]z = 90°{.}[/latex] Give exact values for your answers.

[latex]\sin^2 x \cos y[/latex]

[latex]\sin z - \dfrac{1}{2} \sin y[/latex]

[latex]\tan (z - x) \cos (y - z)[/latex]

[latex]\dfrac{\tan^2 x}{2 \cos y}[/latex]

For Problems 5–8, evaluate the expressions using a calculator. Are they equal?

• [latex]\sin (20° + 40°)[/latex]
• [latex]\sin 20° + \sin 40°[/latex]
• [latex]\cos^2 70° - \sin^2 70°[/latex]
• [latex]\cos (2\cdot 70°)[/latex]
• [latex]\dfrac{\sin 55°}{\cos 55°}[/latex]
• [latex]\tan 55°[/latex]
• [latex]\tan 80° - \tan 10°[/latex]
• [latex]\tan (80° - 10°)[/latex]

For Problems 9–12, simplify the expression.

[latex]3\sin x - 2\sin x \cos y + 2\sin x - \cos y[/latex]

[latex]\cos t + 3\cos 3t - 3\cos t - 2\cos 3t[/latex]

[latex]6 \tan^2 \theta + 2\tan \theta - (4\tan \theta )^2[/latex]

[latex]\sin \theta (2\cos \theta - 2) + \sin \theta (1 - \sin \theta)[/latex]

For Problems 13–16, decide whether or not the expressions are equivalent. Explain.

[latex]\cos \theta + \cos 2\theta;~~\cos 3\theta[/latex]

[latex]1 + \sin^2 x;~~(1 + \sin x)^2[/latex]

[latex]3\tan^2 t - \tan^2 t;~~2\tan^2 t[/latex]

[latex]\cos 4\theta;~~2\cos 2\theta[/latex]

For Problems 17–20, multiply or expand.

[latex](\cos \alpha + 2)(2\cos \alpha - 3)[/latex]

[latex](1 - 3\tan \beta)^2[/latex]

[latex](\tan \phi - \cos \phi)^2 = 0[/latex]

[latex](\sin \rho - 2\cos \rho)(\sin \rho + \cos \rho)[/latex]

For Problems 21–24, factor the expression.

[latex]12\sin 3x - 6\sin 2x[/latex]

[latex]2\cos^2 \beta + \cos \beta[/latex]

[latex]1 - 9\tan^2 \theta[/latex]

[latex]\sin^2 \phi - \sin \phi \tan \phi - 2\tan^2 \phi[/latex]

For Problems 25–30, reduce the fraction.

[latex]\dfrac{\cos^2 \alpha - \sin^2 \alpha}{\cos \alpha - \sin \alpha}[/latex]

[latex]\dfrac{1 - \tan^2 \theta}{1 - \tan \theta}[/latex]

[latex]\dfrac{3\cos x + 9}{2\cos x + 6}[/latex]

[latex]\dfrac{5\sin \theta - 10}{\sin^2 \theta - 4}[/latex]

[latex]\dfrac{3\tan^2 C - 12}{\tan^2 C - 4\tan C + 4}[/latex]

[latex]\dfrac{\tan^2 \beta - \tan \beta - 6}{\tan \beta - 3}[/latex]

For Problems 31–32, use a graph to solve the equation for [latex]0° \le x \lt 360°{.}[/latex] Check your solutions by substitution.

[latex]8\cos x - 3 = 2[/latex]

[latex]6\tan x - 2 = 8[/latex]

For Problems 33–40, find all solutions between [latex]0°[/latex] and [latex]360°{.}[/latex] Give exact answers.

[latex]2\cos^2 \theta + \cos \theta = 0[/latex]

[latex]\sin^2 \alpha - \sin \alpha = 0[/latex]

[latex]2\sin^2 x - \sin x - 1 = 0[/latex]

[latex]\cos^2 B + 2\cos B + 1 = 0[/latex]

[latex]\tan^2 x = \dfrac{1}{3}[/latex]

[latex]\tan^2 t - \tan t = 0[/latex]

[latex]6\cos^2 \alpha - 3\cos \alpha - 3 = 0[/latex]

[latex]2\sin^2 \theta + 4\sin \theta + 2 = 0[/latex]

For Problems 41–44, solve the equation for [latex]0° \le x \lt 360°{.}[/latex] Round your answers to two decimal places.

[latex]2 - 5\tan \theta = -6[/latex]

[latex]3 + 5\cos \theta = 4[/latex]

[latex]3\cos^2 x + 7\cos x = 0[/latex]

[latex]8 - 9\sin^2 x = 0[/latex]

A light ray passes from glass to water, with a [latex]37°[/latex] angle of incidence. What is the angle of refraction? The index of refraction from water to glass is 1.1.

A light ray passes from glass to water, with a [latex]76°[/latex] angle of incidence. What is the angle of refraction? The index of refraction from water to glass is 1.1.

For Problems 47–50, decide which of the following equations are identities. Explain your reasoning.

[latex]\cos x\tan x = \sin x[/latex]

[latex]\sin \theta = 1 - \cos \theta[/latex]

[latex]\tan \phi + \tan \phi = \tan 2\phi[/latex]

[latex]\tan^2 x = \dfrac{\sin^2 x}{1 - \sin^2 x}[/latex]

For Problems 51–54, use graphs to decide which of the following equations are identities.

[latex]\cos 2\theta = 2 \cos \theta[/latex]

[latex]\cos (x - 90°) = \sin x[/latex]

[latex]\sin 2x = 2\sin x \cos x[/latex]

[latex]\cos (\theta + 90°) = \cos \theta - 1[/latex]

For Problems 55–58, show that the equation is an identity by transforming the left side into the right side.

[latex]\dfrac{1 - \cos^2 \alpha}{\tan \alpha} = \sin \alpha \cos \alpha[/latex]

[latex]\cos^2 \beta \tan^2 \beta + \cos^2 \beta = 1[/latex]

[latex]\dfrac{\tan \theta - \sin \theta \cos \theta}{\sin \theta \cos \theta} = \sin \theta[/latex]

[latex]\tan \phi - \dfrac{\sin^2 \phi}{\tan \phi} = \tan \phi \sin^2 \phi[/latex]

For Problems 59–62, simplify, using identities as necessary.

[latex]\tan \theta + \dfrac{\cos \theta}{\sin \theta}[/latex]

[latex]\dfrac{1 - 2\cos^2 \beta}{\sin \beta \cos \beta} + \dfrac{\cos \beta}{\sin \beta}[/latex]

[latex]\dfrac{1}{1 - \sin^2 v} - \tan^2 v[/latex]

[latex]\cos u + (\sin u)(\tan u)[/latex]

For Problems 63–66, evaluate the expressions without using a calculator.

[latex]\sin 137° - \tan 137° \cdot \cos 137°[/latex]

[latex]\cos^2 8° + \cos 8° \cdot \tan 8° \cdot \sin 8°[/latex]

[latex]\dfrac{1}{\cos^2 54°} - \tan^2 54°[/latex]

[latex]\dfrac{2}{\cos^2 7°} - 2\tan^2 7°[/latex]

For Problems 67–70, use identities to rewrite each expression.

Write [latex]\tan^2 \beta + 1[/latex] in terms of [latex]\cos^2 \beta{.}[/latex]

Write [latex]2\sin^2 t + \cos t[/latex] in terms of [latex]\cos t{.}[/latex]

Write [latex]\dfrac{\cos x}{\tan x}[/latex] in terms of [latex]\sin x{.}[/latex]

For Problems 71–74, find the values of the three trigonometric functions.

[latex]7\tan \beta - 4 = 2, ~~ 180° \lt \beta \lt 270°[/latex]

[latex]3\tan C + 5 = 3, ~~-90° \lt C \lt 0°[/latex]

[latex]5\cos \alpha + 3 = 1, ~~ 90° \lt \alpha \lt 180°[/latex]

[latex]3\sin \theta + 2 = 4, ~~ 90° \lt \beta \lt 180°[/latex]

For Problems 75–82, solve the equation for [latex]0° \le x \lt 360°{.}[/latex] Round angles to three decimal places if necessary.

[latex]\sin w + 1 = \cos^2 w[/latex]

[latex]\cos^2 \phi - \cos \phi - \sin^2 \phi = 0[/latex]

[latex]\cos x + \sin x = 0[/latex]

[latex]3\sin \theta = \sqrt{3} \cos \theta[/latex]

[latex]2\sin \beta - \tan \beta = 0[/latex]

[latex]6\tan \theta \cos \theta + 6 = 0[/latex]

[latex]\cos^2 t - \sin^2 t = 1[/latex]

[latex]5\cos^2 \beta - 5\sin^2 \beta = -5[/latex]

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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9.5 Solving Trigonometric Equations

Learning objectives.

• 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.
• Solve right triangle problems.

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. 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 π . 2 π . In other words, every 2 π 2 π units, the y- values repeat. If we need to find all possible solutions, then we must add   2 π k ,   2 π k , where   k     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 π : 2 π :

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.

Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation cos θ = 1 2 . cos θ = 1 2 .

From the unit circle , we know that

These are the solutions in the interval [ 0 , 2 π ] . [ 0 , 2 π ] . All possible solutions are given by

where k k is an integer.

Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation sin t = 1 2 . sin t = 1 2 .

Solving for all possible values of t means that solutions include angles beyond the period of 2 π . 2 π . From Figure 2 , we can see that the solutions are t = π 6 t = π 6 and t = 5 π 6 . t = 5 π 6 . But the problem is asking for all possible values that solve the equation. Therefore, the answer is

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 x x or u . u .
• 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.

Solve the Linear Trigonometric Equation

Solve the equation exactly: 2 cos θ − 3 = − 5 , 0 ≤ θ < 2 π . 2 cos θ − 3 = − 5 , 0 ≤ θ < 2 π .

Use algebraic techniques to solve the equation.

Solve exactly the following linear equation on the interval [ 0 , 2 π ) : 2 sin x + 1 = 0. [ 0 , 2 π ) : 2 sin x + 1 = 0.

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 Figure 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   π ,   π , not   2 π .     2 π .   Further, the domain of tangent is all real numbers with the exception of odd integer multiples of π 2 , π 2 , unless, of course, a problem places its own restrictions on the domain.

Solving a Problem Involving a Single Trigonometric Function

Solve the problem exactly: 2 sin 2 θ − 1 = 0 , 0 ≤ θ < 2 π . 2 sin 2 θ − 1 = 0 , 0 ≤ θ < 2 π .

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate sin θ . sin θ . Then we will find the angles.

Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: csc θ = − 2 , 0 ≤ θ < 4 π . csc θ = − 2 , 0 ≤ θ < 4 π .

We want all values of θ θ for which csc θ = − 2 csc θ = − 2 over the interval 0 ≤ θ < 4 π . 0 ≤ θ < 4 π .

As sin θ = − 1 2 , sin θ = − 1 2 , notice that all four solutions are in the third and fourth quadrants.

Solving an Equation Involving Tangent

Solve the equation exactly: tan ( θ − π 2 ) = 1 , 0 ≤ θ < 2 π . tan ( θ − π 2 ) = 1 , 0 ≤ θ < 2 π .

Recall that the tangent function has a period of π . π . On the interval [ 0 , π ) , [ 0 , π ) , and at the angle of π 4 , π 4 , the tangent has a value of 1. However, the angle we want is ( θ − π 2 ) . ( θ − π 2 ) . Thus, if tan ( π 4 ) = 1 , tan ( π 4 ) = 1 , then

Over the interval [ 0 , 2 π ) , [ 0 , 2 π ) , we have two solutions:

Find all solutions for tan x = 3 . tan x = 3 .

Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation 2 ( tan x + 3 ) = 5 + tan x , 0 ≤ x < 2 π . 2 ( tan x + 3 ) = 5 + tan x , 0 ≤ x < 2 π .

We can solve this equation using only algebra. Isolate the expression tan x tan x on the left side of the equals sign.

There are two angles on the unit circle that have a tangent value of −1 : θ = 3 π 4 −1 : θ = 3 π 4 and θ = 7 π 4 . θ = 7 π 4 .

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.

Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation sin θ = 0.8 , sin θ = 0.8 , where   θ     θ   is in radians.

Make sure mode is set to radians. To find θ , θ , use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the sin − 1 sin − 1 function. What is shown on the screen is sin − 1 ( . sin − 1 ( . The calculator is ready for the input within the parentheses. For this problem, we enter sin − 1 ( 0.8 ) , sin − 1 ( 0.8 ) , and press ENTER. Thus, to four decimals places,

The solution is

The angle measurement in degrees is

Note 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. The other angle is obtained by using π − θ . π − θ . Thus, the additional solution is ≈ 2.2143 ± 2 π k ≈ 2.2143 ± 2 π k

Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation sec θ = −4 , sec θ = −4 , giving your answer in radians.

We can begin with some algebra.

Check that the MODE is in radians. Now use the inverse cosine function.

Since π 2 ≈ 1.57 π 2 ≈ 1.57 and π ≈ 3.14 , π ≈ 3.14 , 1.8235 is between these two numbers, thus θ ≈ 1 .8235   θ ≈ 1 .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 Figure 2 .

So, we also need to find the measure of the angle in quadrant III. In quadrant II, the reference angle is θ ​ ​ ' ≈ π − 1 .8235 ≈ 1 .3181 .   θ ​ ​ ' ≈ π − 1 .8235 ≈ 1 .3181 .   The other solution in quadrant III is θ ​ ​ ' ≈ π + 1 .3181 ≈ 4 .4597 . θ ​ ​ ' ≈ π + 1 .3181 ≈ 4 .4597 .

The solutions are θ ≈ 1.8235 ± 2 π k θ ≈ 1.8235 ± 2 π k and θ ≈ 4.4597 ± 2 π k . θ ≈ 4.4597 ± 2 π k .

Solve cos θ = − 0.2. cos θ = − 0.2.

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 x or u . 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.

Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: cos 2 θ + 3 cos θ − 1 = 0 , 0 ≤ θ < 2 π . cos 2 θ + 3 cos θ − 1 = 0 , 0 ≤ θ < 2 π .

We begin by using substitution and replacing cos θ θ with x . x . It is not necessary to use substitution, but it may make the problem easier to solve visually. Let cos θ = x . cos θ = x . We have

The equation cannot be factored, so we will use the quadratic formula x = − b ± b 2 − 4 a c 2 a . x = − b ± b 2 − 4 a c 2 a .

Replace x x with cos θ , cos θ , and solve.

Note that only the + sign is used. This is because we get an error when we solve θ = cos − 1 ( − 3 − 13 2 ) θ = cos − 1 ( − 3 − 13 2 ) on a calculator, since the domain of the inverse cosine function is [ − 1 , 1 ] . [ − 1 , 1 ] . However, there is a second solution:

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: 2 sin 2 θ − 5 sin θ + 3 = 0 , 0 ≤ θ ≤ 2 π . 2 sin 2 θ − 5 sin θ + 3 = 0 , 0 ≤ θ ≤ 2 π .

Using grouping, this quadratic can be factored. Either make the real substitution, sin θ = u , sin θ = u , or imagine it, as we factor:

Now set each factor equal to zero.

Next solve for θ : sin θ ≠ 3 2 , θ : sin θ ≠ 3 2 , as the range of the sine function is [ −1 , 1 ] . [ −1 , 1 ] . However, sin θ = 1 , sin θ = 1 , giving the solution θ = π 2 . θ = π 2 .

Make sure to check all solutions on the given domain as some factors have no solution.

Solve sin 2 θ = 2 cos θ + 2 , 0 ≤ θ ≤ 2 π . sin 2 θ = 2 cos θ + 2 , 0 ≤ θ ≤ 2 π . [Hint: Make a substitution to express the equation only in terms of cosine.]

Solving a Trigonometric Equation Using Algebra

Solve exactly:

This problem should appear familiar as it is similar to a quadratic. Let sin θ = x . sin θ = x . The equation becomes 2 x 2 + x = 0. 2 x 2 + x = 0. We begin by factoring:

Set each factor equal to zero.

Then, substitute back into the equation the original expression sin θ sin θ for x . x . Thus,

The solutions within the domain 0 ≤ θ < 2 π 0 ≤ θ < 2 π are θ = 0 , π , 7 π 6 , 11 π 6 . θ = 0 , π , 7 π 6 , 11 π 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.

We can see the solutions on the graph in Figure 3 . On the interval 0 ≤ θ < 2 π , 0 ≤ θ < 2 π , 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.

We can verify the solutions on the unit circle in Figure 2 as well.

Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: 2 sin 2 θ − 3 sin θ + 1 = 0 , 0 ≤ θ < 2 π . 2 sin 2 θ − 3 sin θ + 1 = 0 , 0 ≤ θ < 2 π .

We can factor using grouping. Solution values of θ θ can be found on the unit circle.

Solve the quadratic equation 2 cos 2 θ + cos θ = 0. 2 cos 2 θ + cos θ = 0.

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.

Use Identities to Solve an Equation

Use identities to solve exactly the trigonometric equation over the interval 0 ≤ x < 2 π . 0 ≤ x < 2 π .

Notice that the left side of the equation is the difference formula for cosine.

From the unit circle in Figure 2 , we see that cos x = 3 2 cos x = 3 2 when x = π 6 , 11 π 6 . x = π 6 , 11 π 6 .

Solving the Equation Using a Double-Angle Formula

Solve the equation exactly using a double-angle formula: cos ( 2 θ ) = cos θ . cos ( 2 θ ) = cos θ .

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:

So, if cos θ = − 1 2 , cos θ = − 1 2 , then θ = 2 π 3 ± 2 π k θ = 2 π 3 ± 2 π k and θ = 4 π 3 ± 2 π k ; θ = 4 π 3 ± 2 π k ; if cos θ = 1 , cos θ = 1 , then θ = 0 ± 2 π k . θ = 0 ± 2 π k .

Solving an Equation Using an Identity

Solve the equation exactly using an identity: 3 cos θ + 3 = 2 sin 2 θ , 0 ≤ θ < 2 π . 3 cos θ + 3 = 2 sin 2 θ , 0 ≤ θ < 2 π .

If we rewrite the right side, we can write the equation in terms of cosine:

Our solutions are θ = 2 π 3 , 4 π 3 , π . θ = 2 π 3 , 4 π 3 , π .

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 ( 2 x ) sin ( 2 x ) or cos ( 3 x ) . cos ( 3 x ) . When confronted with these equations, recall that y = sin ( 2 x ) y = sin ( 2 x ) is a horizontal compression by a factor of 2 of the function y = sin x . y = sin x . On an interval of 2 π , 2 π , we can graph two periods of y = sin ( 2 x ) , y = sin ( 2 x ) , as opposed to one cycle of y = sin x . y = sin x . This compression of the graph leads us to believe there may be twice as many x -intercepts or solutions to sin ( 2 x ) = 0 sin ( 2 x ) = 0 compared to sin x = 0. sin x = 0. This information will help us solve the equation.

Solving a Multiple Angle Trigonometric Equation

Solve exactly: cos ( 2 x ) = 1 2 cos ( 2 x ) = 1 2 on [ 0 , 2 π ) . [ 0 , 2 π ) .

We can see that this equation is the standard equation with a multiple of an angle. If cos ( α ) = 1 2 , cos ( α ) = 1 2 , we know α α is in quadrants I and IV. While θ = cos − 1 1 2 θ = cos − 1 1 2 will only yield solutions in quadrants I and II, we recognize that the solutions to the equation cos θ = 1 2 cos θ = 1 2 will be in quadrants I and IV.

Therefore, the possible angles are θ = π 3 θ = π 3 and θ = 5 π 3 . θ = 5 π 3 . So, 2 x = π 3 2 x = π 3 or 2 x = 5 π 3 , 2 x = 5 π 3 , which means that x = π 6 x = π 6 or x = 5 π 6 . x = 5 π 6 . Does this make sense? Yes, because cos ( 2 ( π 6 ) ) = cos ( π 3 ) = 1 2 . cos ( 2 ( π 6 ) ) = cos ( π 3 ) = 1 2 .

Are there any other possible answers? Let us return to our first step.

In quadrant I, 2 x = π 3 , 2 x = π 3 , so x = π 6 x = π 6 as noted. Let us revolve around the circle again:

so x = 7 π 6 . x = 7 π 6 .

One more rotation yields

x = 13 π 6 > 2 π , x = 13 π 6 > 2 π , so this value for x x is larger than 2 π , 2 π , so it is not a solution on [ 0 , 2 π ) . [ 0 , 2 π ) .

In quadrant IV, 2 x = 5 π 3 , 2 x = 5 π 3 , so x = 5 π 6 x = 5 π 6 as noted. Let us revolve around the circle again:

so x = 11 π 6 . x = 11 π 6 .

x = 17 π 6 > 2 π , x = 17 π 6 > 2 π , so this value for x x is larger than 2 π , 2 π , so it is not a solution on [ 0 , 2 π ) . [ 0 , 2 π ) .

Our solutions are x = π 6 , 5 π 6 , 7 π 6 , and  11 π 6 x = π 6 , 5 π 6 , 7 π 6 , and  11 π 6 . Note that whenever we solve a problem in the form of sin ( n x ) = c , sin ( n x ) = c , we must go around the unit circle n n times.

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 , a 2 + b 2 = c 2 , and model an equation to fit a situation.

Using the Pythagorean Theorem to Model an Equation

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 Figure 4 .

Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.

The angle of elevation is θ , θ , 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.

The angle of elevation is approximately 71.7° , 71.7° , and the length of the cable is 73.2 meters.

Using the Pythagorean Theorem to Model an Abstract Problem

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.

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 Figure 5 .

The side adjacent to θ θ is a and the hypotenuse is 4 a . 4 a . Thus,

The elevation of the ladder forms an angle of 75.5° 75.5° with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:

Thus, the ladder touches the wall at a 15 a 15 feet from the ground.

Access these online resources for additional instruction and practice with solving trigonometric equations.

• Solving Trigonometric Equations I
• Solving Trigonometric Equations II
• Solving Trigonometric Equations III
• Solving Trigonometric Equations IV
• Solving Trigonometric Equations V
• Solving Trigonometric Equations VI

9.5 Section Exercises

Will there always be solutions to trigonometric function equations? If not, describe an equation that would not have a solution. Explain why or why not.

When solving a trigonometric equation involving more than one trig function, do we always want to try to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not?

When solving linear trig equations in terms of only sine or cosine, how do we know whether there will be solutions?

For the following exercises, find all solutions exactly on the interval 0 ≤ θ < 2 π . 0 ≤ θ < 2 π .

2 sin θ = − 2 2 sin θ = − 2

2 sin θ = 3 2 sin θ = 3

2 cos θ = 1 2 cos θ = 1

2 cos θ = − 2 2 cos θ = − 2

tan θ = −1 tan θ = −1

tan x = 1 tan x = 1

cot x + 1 = 0 cot x + 1 = 0

4 sin 2 x − 2 = 0 4 sin 2 x − 2 = 0

csc 2 x − 4 = 0 csc 2 x − 4 = 0

For the following exercises, solve exactly on [ 0 , 2 π ) . [ 0 , 2 π ) .

2 cos θ = 2 2 cos θ = 2

2 cos θ = −1 2 cos θ = −1

2 sin θ = −1 2 sin θ = −1

2 sin θ = − 3 2 sin θ = − 3

2 sin ( 3 θ ) = 1 2 sin ( 3 θ ) = 1

2 sin ( 2 θ ) = 3 2 sin ( 2 θ ) = 3

2 cos ( 3 θ ) = − 2 2 cos ( 3 θ ) = − 2

cos ( 2 θ ) = − 3 2 cos ( 2 θ ) = − 3 2

2 sin ( π θ ) = 1 2 sin ( π θ ) = 1

2 cos ( π 5 θ ) = 3 2 cos ( π 5 θ ) = 3

For the following exercises, find all exact solutions on [ 0 , 2 π ) . [ 0 , 2 π ) .

sec ( x ) sin ( x ) − 2 sin ( x ) = 0 sec ( x ) sin ( x ) − 2 sin ( x ) = 0

tan ( x ) − 2 sin ( x ) tan ( x ) = 0 tan ( x ) − 2 sin ( x ) tan ( x ) = 0

2 cos 2 t + cos ( t ) = 1 2 cos 2 t + cos ( t ) = 1

2 tan 2 ( t ) = 3 sec ( t ) 2 tan 2 ( t ) = 3 sec ( t )

2 sin ( x ) cos ( x ) − sin ( x ) + 2 cos ( x ) − 1 = 0 2 sin ( x ) cos ( x ) − sin ( x ) + 2 cos ( x ) − 1 = 0

cos 2 θ = 1 2 cos 2 θ = 1 2

sec 2 x = 1 sec 2 x = 1

tan 2 ( x ) = −1 + 2 tan ( − x ) tan 2 ( x ) = −1 + 2 tan ( − x )

8 sin 2 ( x ) + 6 sin ( x ) + 1 = 0 8 sin 2 ( x ) + 6 sin ( x ) + 1 = 0

tan 5 ( x ) = tan ( x ) tan 5 ( x ) = tan ( x )

For the following exercises, solve with the methods shown in this section exactly on the interval [ 0 , 2 π ) . [ 0 , 2 π ) .

sin ( 3 x ) cos ( 6 x ) − cos ( 3 x ) sin ( 6 x ) = −0.9 sin ( 3 x ) cos ( 6 x ) − cos ( 3 x ) sin ( 6 x ) = −0.9

sin ( 6 x ) cos ( 11 x ) − cos ( 6 x ) sin ( 11 x ) = −0.1 sin ( 6 x ) cos ( 11 x ) − cos ( 6 x ) sin ( 11 x ) = −0.1

cos ( 2 x ) cos x + sin ( 2 x ) sin x = 1 cos ( 2 x ) cos x + sin ( 2 x ) sin x = 1

6 sin ( 2 t ) + 9 sin t = 0 6 sin ( 2 t ) + 9 sin t = 0

9 cos ( 2 θ ) = 9 cos 2 θ − 4 9 cos ( 2 θ ) = 9 cos 2 θ − 4

sin ( 2 t ) = cos t sin ( 2 t ) = cos t

cos ( 2 t ) = sin t cos ( 2 t ) = sin t

cos ( 6 x ) − cos ( 3 x ) = 0 cos ( 6 x ) − cos ( 3 x ) = 0

For the following exercises, solve exactly on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Use the quadratic formula if the equations do not factor.

tan 2 x − 3 tan x = 0 tan 2 x − 3 tan x = 0

sin 2 x + sin x − 2 = 0 sin 2 x + sin x − 2 = 0

sin 2 x − 2 sin x − 4 = 0 sin 2 x − 2 sin x − 4 = 0

5 cos 2 x + 3 cos x − 1 = 0 5 cos 2 x + 3 cos x − 1 = 0

3 cos 2 x − 2 cos x − 2 = 0 3 cos 2 x − 2 cos x − 2 = 0

5 sin 2 x + 2 sin x − 1 = 0 5 sin 2 x + 2 sin x − 1 = 0

tan 2 x + 5 tan x − 1 = 0 tan 2 x + 5 tan x − 1 = 0

cot 2 x = − cot x cot 2 x = − cot x

− tan 2 x − tan x − 2 = 0 − tan 2 x − tan x − 2 = 0

For the following exercises, find exact solutions on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Look for opportunities to use trigonometric identities.

sin 2 x − cos 2 x − sin x = 0 sin 2 x − cos 2 x − sin x = 0

sin 2 x + cos 2 x = 0 sin 2 x + cos 2 x = 0

sin ( 2 x ) − sin x = 0 sin ( 2 x ) − sin x = 0

cos ( 2 x ) − cos x = 0 cos ( 2 x ) − cos x = 0

2 tan x 2 − sec 2 x − sin 2 x = cos 2 x 2 tan x 2 − sec 2 x − sin 2 x = cos 2 x

1 − cos ( 2 x ) = 1 + cos ( 2 x ) 1 − cos ( 2 x ) = 1 + cos ( 2 x )

sec 2 x = 7 sec 2 x = 7

10 sin x cos x = 6 cos x 10 sin x cos x = 6 cos x

−3 sin t = 15 cos t sin t −3 sin t = 15 cos t sin t

4 cos 2 x − 4 = 15 cos x 4 cos 2 x − 4 = 15 cos x

8 sin 2 x + 6 sin x + 1 = 0 8 sin 2 x + 6 sin x + 1 = 0

8 cos 2 θ = 3 − 2 cos θ 8 cos 2 θ = 3 − 2 cos θ

6 cos 2 x + 7 sin x − 8 = 0 6 cos 2 x + 7 sin x − 8 = 0

12 sin 2 t + cos t − 6 = 0 12 sin 2 t + cos t − 6 = 0

tan x = 3 sin x tan x = 3 sin x

cos 3 t = cos t cos 3 t = cos t

For the following exercises, algebraically determine all solutions of the trigonometric equation exactly, then verify the results by graphing the equation and finding the zeros.

6 sin 2 x − 5 sin x + 1 = 0 6 sin 2 x − 5 sin x + 1 = 0

8 cos 2 x − 2 cos x − 1 = 0 8 cos 2 x − 2 cos x − 1 = 0

100 tan 2 x + 20 tan x − 3 = 0 100 tan 2 x + 20 tan x − 3 = 0

2 cos 2 x − cos x + 15 = 0 2 cos 2 x − cos x + 15 = 0

20 sin 2 x − 27 sin x + 7 = 0 20 sin 2 x − 27 sin x + 7 = 0

2 tan 2 x + 7 tan x + 6 = 0 2 tan 2 x + 7 tan x + 6 = 0

130 tan 2 x + 69 tan x − 130 = 0 130 tan 2 x + 69 tan x − 130 = 0

For the following exercises, use a calculator to find all solutions to four decimal places.

sin x = 0.27 sin x = 0.27

sin x = −0.55 sin x = −0.55

tan x = −0.34 tan x = −0.34

cos x = 0.71 cos x = 0.71

For the following exercises, solve the equations algebraically, and then use a calculator to find the values on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Round to four decimal places.

tan 2 x + 3 tan x − 3 = 0 tan 2 x + 3 tan x − 3 = 0

6 tan 2 x + 13 tan x = −6 6 tan 2 x + 13 tan x = −6

tan 2 x − sec x = 1 tan 2 x − sec x = 1

sin 2 x − 2 cos 2 x = 0 sin 2 x − 2 cos 2 x = 0

2 tan 2 x + 9 tan x − 6 = 0 2 tan 2 x + 9 tan x − 6 = 0

4 sin 2 x + sin ( 2 x ) sec x − 3 = 0 4 sin 2 x + sin ( 2 x ) sec x − 3 = 0

For the following exercises, find all solutions exactly to the equations on the interval [ 0 , 2 π ) . [ 0 , 2 π ) .

csc 2 x − 3 csc x − 4 = 0 csc 2 x − 3 csc x − 4 = 0

sin 2 x − cos 2 x − 1 = 0 sin 2 x − cos 2 x − 1 = 0

sin 2 x ( 1 − sin 2 x ) + cos 2 x ( 1 − sin 2 x ) = 0 sin 2 x ( 1 − sin 2 x ) + cos 2 x ( 1 − sin 2 x ) = 0

3 sec 2 x + 2 + sin 2 x − tan 2 x + cos 2 x = 0 3 sec 2 x + 2 + sin 2 x − tan 2 x + cos 2 x = 0

sin 2 x − 1 + 2 cos ( 2 x ) − cos 2 x = 1 sin 2 x − 1 + 2 cos ( 2 x ) − cos 2 x = 1

tan 2 x − 1 − sec 3 x cos x = 0 tan 2 x − 1 − sec 3 x cos x = 0

sin ( 2 x ) sec 2 x = 0 sin ( 2 x ) sec 2 x = 0

sin ( 2 x ) 2 csc 2 x = 0 sin ( 2 x ) 2 csc 2 x = 0

2 cos 2 x − sin 2 x − cos x − 5 = 0 2 cos 2 x − sin 2 x − cos x − 5 = 0

1 sec 2 x + 2 + sin 2 x + 4 cos 2 x = 4 1 sec 2 x + 2 + sin 2 x + 4 cos 2 x = 4

Real-World Applications

An airplane has only enough gas to fly to a city 200 miles northeast of its current location. If the pilot knows that the city is 25 miles north, how many degrees north of east should the airplane fly?

If a loading ramp is placed next to a truck, at a height of 4 feet, and the ramp is 15 feet long, what angle does the ramp make with the ground?

If a loading ramp is placed next to a truck, at a height of 2 feet, and the ramp is 20 feet long, what angle does the ramp make with the ground?

A woman is watching a launched rocket currently 11 miles in altitude. If she is standing 4 miles from the launch pad, at what angle is she looking up from horizontal?

An astronaut is in a launched rocket currently 15 miles in altitude. If a man is standing 2 miles from the launch pad, at what angle is she looking down at him from horizontal? (Hint: this is called the angle of depression.)

A woman is standing 8 meters away from a 10-meter tall building. At what angle is she looking to the top of the building?

A man is standing 10 meters away from a 6-meter tall building. Someone at the top of the building is looking down at him. At what angle is the person looking at him?

A 20-foot tall building has a shadow that is 55 feet long. What is the angle of elevation of the sun?

A 90-foot tall building has a shadow that is 2 feet long. What is the angle of elevation of the sun?

A spotlight on the ground 3 meters from a 2-meter tall man casts a 6 meter shadow on a wall 6 meters from the man. At what angle is the light?

A spotlight on the ground 3 feet from a 5-foot tall woman casts a 15-foot tall shadow on a wall 6 feet from the woman. At what angle is the light?

For the following exercises, find a solution to the following word problem algebraically. Then use a calculator to verify the result. Round the answer to the nearest tenth of a degree.

A person does a handstand with his feet touching a wall and his hands 1.5 feet away from the wall. If the person is 6 feet tall, what angle do his feet make with the wall?

A person does a handstand with her feet touching a wall and her hands 3 feet away from the wall. If the person is 5 feet tall, what angle do her feet make with the wall?

A 23-foot ladder is positioned next to a house. If the ladder slips at 7 feet from the house when there is not enough traction, what angle should the ladder make with the ground to avoid slipping?

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5.0: Algebra with Trigonometric Ratios

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• Page ID 112428

• Katherine Yoshiwara
• Los Angeles Pierce College

In this chapter we apply some techniques from algebra to analyze more complicated trigonometric expressions. Before we begin, let’s review some algebraic terminology.

• An algebraic expression is any meaningful collection of numbers, variables, and operation symbols. For example, the height of a golf ball is given in feet by the expression \(−16t^2 + 64t\), where \(t\) is the number of seconds after the ball is hit.
• We evaluate an expression by substituting a specific value for the variable or variables involved. Thus, after 1 second, the height of the golf ball is

\(-16(1)^2+64(11)=-16+64=48 \text { feet }\)

and after 2 seconds, the height is

\(-16(2)^2+64(2)=-64+128=64 \text { feet }\)

Evaluating Trigonometric Expressions

Trigonometric ratios represent numbers, and they may appear as part of an algebraic expression. Expressions containing trig ratios can be simplified or evaluated like other algebraic expressions.

Example 5.1

Evaluate each expression for \(X=30^{\circ}\) and \(Y=135^{\circ}\).

a \(2 \tan Y+3 \sin X\) b \(6 \tan X \cos Y\)

a Substituting the values for \(X\) and \(Y\), we get

\(2 \tan 135^{\circ}+3 \sin 30^{\circ}\)

Next, we evaluate each trig ratio and follow the order of operations.

\begin{aligned} 2 \tan 135^{\circ}+3 \sin 30^{\circ} & =2(-1)+3\left(\dfrac{1}{2}\right) \\ & =-2+\dfrac{3}{2}=\dfrac{-1}{2} \end{aligned}

b This expression includes the product of two trig ratios \(6(\tan X)(\cos Y)\), where the parentheses indicate multiplication.

\begin{aligned} 6 \tan X \cos Y & =6\left(\tan 30^{\circ}\right)\left(\cos 135^{\circ}\right) \\ & =6\left(\dfrac{1}{\sqrt{3}}\right)\left(\dfrac{-1}{\sqrt{2}}\right) \\ & =\dfrac{-6}{\sqrt{6}}=-\sqrt{6} \end{aligned}

Checkpoint 5.2

Evaluate each expression for \(X=30^{\circ}, Y=60^{\circ}\).

a \(4 \sin \left(3 X+45^{\circ}\right)\) b \(1-\cos (4 Y)\)

a \(2 \sqrt{2}\) b \(\dfrac{3}{2}\)

Caution 5.3

In the previous Exercise, \(\sin \left(3 X+45^{\circ}\right)\) is not equal to \(\sin 3 X+\sin 45^{\circ}\). That is,

\(\sin \left(90^{\circ}+45^{\circ}\right) \neq \sin 90^{\circ}+\sin 45^{\circ}\)

(You can check this for yourself.) We must follow the order of operations and evaluate the expression \(3 X+45^{\circ}\) inside parentheses before applying the sine function.

Simplifying Trigonometric Expressions

When we simplify an algebraic expression, we obtain a new expression that has the same values as the old one, but is easier to work with. For example, we can apply the distributive law and combine like terms to simplify

\begin{aligned} 2 x(x-6)+3(x+2) & =2 x^2-12 x+3 x+6 \\ & =2 x^2-9 x+6 \end{aligned}

The new expression is equivalent to the old one, that is, the expressions have the same value when we evaluate them at any value of \(\boldsymbol{x}\). For instance, you can check that, at \(x=\mathbf{3}\),

\begin{aligned} 2(3)(3-6)+3(3+2)&=6(-3)+3(5)=-3 \\ 2(3)^2-9(3)+6&=18-27+6=-3 \end{aligned}

To simplify an expression containing trig ratios, we treat each ratio as a single variable. Compare the two calculations below:

\(\begin{array}{ll} 8 x y&-6 x y & =2 x y \\ 8 \cos \theta \sin \theta&-6 \cos \theta \sin \theta & =2 \cos \theta \sin \theta \end{array}\)

Both calculations are examples of combining like terms. In the second calculation, we treat \(\cos \theta\) and \(\sin \theta\) as variables, just as we treat \(x\) and \(y\) in the first calculation.

Example 5.4

a \(3 \tan A+4 \tan A-2 \cos A\) b \(2-\sin B+2 \sin B\)

a Combine like terms.

\(3 \tan A+4 \tan A-2 \cos A=7 \tan A-2 \cos A\)

Note that \(\tan A\) and \(\cos A\) are not like terms.

b Combine like terms.

\(2-\sin B+2 \sin B=2+\sin B\)

Note that \(-\sin B\) means \(-1 \cdot \sin B\).

Checkpoint 5.5

Simplify \(2 \cos t-4 \cos w \sin w+3 \cos t-2 \cos w\)

\(5 \cos t-4 \cos w \sin w-2 \cos w\)

Caution 5.6

In the previous Exercise, note that \(\cos t\) and \(\cos w\) are not like terms. (We can choose values for \(t\) and \(w\) so that \(\cos t\) and \(\cos w\) have different values.)

Example 5.7

Simplify, and evaluate for \(z=40^{\circ}\).

\(3 \sin z-\sin z \tan z+3 \sin z\)

We can combine like terms to get

\(6 \sin z-\sin z \tan z\)

Because \(40^{\circ}\) is not one of the angles for which we know exact trig values, we use a calculator to evaluate the expression.

\begin{aligned} 6 \sin 40^{\circ}-\left(\sin 40^{\circ}\right)\left(\tan 40^{\circ}\right) & =6(0.6428)(0.8391) \\ & =3.3174 \end{aligned}

Checkpoint 5.8

Simplify, and evaluate for \(x=25^{\circ}, y=70^{\circ}\)

\(3 \cos x+\cos y-2 \cos y+\cos x\)

Powers of Trigonometric Ratios

Compare the two expressions

\((\cos \theta)^2 \quad \text{and} \quad\cos \left(\theta^2\right)\)

They are not the same.

• The first expression, \((\cos \theta)^2\), says to compute \(\cos \theta\) and then square the result.
• \(\cos \left(\theta^2\right)\) says to square the angle first, and then compute the cosine.

For example, if \(\theta=30^{\circ}\), then

\begin{aligned} & \left(\cos 30^{\circ}\right)^2=\left(\dfrac{\sqrt{3}}{2}\right)^2=\dfrac{3}{4} \\ & \text { but } \quad \cos \left(30^2\right)^{\circ}=\cos 900^{\circ}=\cos 180^{\circ}=-1 \\ & \end{aligned}

We usually write \(\cos ^2 \theta\) instead of \((\cos \theta)^2\), and \(\cos \theta^2\) for \(\cos \left(\theta^2\right)\). You must remember that

The square of cosine.

\(\cos ^2 \theta \quad \text { means } \quad(\cos \theta)^2\)

The same notation applies to the other trig ratios, so that

\(\sin ^2 \theta=(\sin \theta)^2 \quad \text { and } \quad \tan ^2 \theta=(\tan \theta)^2\)

Example 5.9

Evaluate \(\sin ^2 45^{\circ}\).

\(\sin ^2 45^{\circ}=\left(\sin 45^{\circ}\right)^2=\left(\dfrac{1}{\sqrt{2}}\right)^2=\dfrac{1}{2}\)

Other powers are written in the same fashion. Thus, for example, \(\sin ^3 \theta=(\sin \theta)^3\).

Checkpoint 5.10

Evaluate \(\tan ^4 60^{\circ}\)

We can multiply together trigonometric expressions, just as we multiply algebraic expressions. Recall that we use the distributive law in computing products such as

\(x(3 x-2)=3 x^2-2 x\)

\((x-3)(x+5)=x^2+2 x-15\)

Example 5.11

Using the distributive law, multiply \(\quad \cos t(3 \cos t-2)\).

Think of \(\cos t\) as a single variable, and multiply by each term inside parentheses. (The algebraic form of the calculation is shown on the right in blue).

\begin{aligned} \cos t(3\cos t - 2) &= (\cos t)(3\cos t) - (\cos t) \cdot 2) \quad &&\boldsymbol{x(3x-2) =x \cdot 3x - x \cdot 2} \\ &= 3\cos^2 t - 2\cos t \quad &&=\boldsymbol{3x^2 - 2x} \end{aligned}

Notice that we write \((\cos t)^2\) as \(\cos ^2 t\).

Checkpoint 5.12

Multiply \(2 \tan \beta\left(4 \tan ^2 \beta+\tan \alpha\right)\)

\(8 \tan ^3 \beta+2 \tan \beta \tan \alpha\)

We can also use the distributive law to multiply binomials that include trig ratios. You may have used the acronym \(\boldsymbol{F O I L}\) to remember the four multiplications in a product of binomials: \(\boldsymbol{F}\)irst terms, Outside terms, \(\boldsymbol{I}\)nside terms, and \(\boldsymbol{L}\)ast terms.

Example 5.13

Multiply \(\quad(4 \sin C-1)(3 \sin C+2)\).

This calculation is similar to the product \((4 x-1)(3 x+2)\), except that the variable \(x\) has been replaced by \(\sin C\). Compare the calculations for the two products; first the familiar algebraic product:

\begin{aligned} (4 x-1)(3 x+2) & =4 x \cdot 3 x+4 x \cdot 2-1 \cdot 3 x-1 \cdot 2 \\ & =12 x^2+8 x-3 x-2=12 x^2+5 x-2 \end{aligned}

We compute the product in this example in the same way, but replacing \(x\) by \(\sin C\).

\begin{aligned} (4 \sin C-1)(\sin C+2) & =(4 \sin C)(3 \sin C)+(4 \sin C) \cdot 2-1(3 \sin C)-1 \cdot 2 \\ & =12 \sin ^2 C+5 \sin C-2 \end{aligned}

Checkpoint 5.14

Expand \((4 \cos \alpha+3)^2\)

\(16 \cos ^2 \alpha+24 \cos \alpha+9\)

We can factor trigonometric expressions with the same techniques we use for algebraic expressions. In the next two Examples, compare the familiar algebraic factoring with a similar trigonometric expression.

Example 5.15

a \(6 w^2-9 w\) b \(6 \sin ^2 \theta-9 \sin \theta\)

a We factor out the common factor, \(3 w\).

\(6 w^2-9 w=3 w(2 w-3)\)

b We factor out the common factor, \(3 \sin \theta\).

\(6 \sin ^2 \theta-9 \sin \theta=3 \sin \theta(2 \sin \theta-3)\)

Checkpoint 5.16

a \(2 a^2-a b\) b \(2 \cos ^2 \phi-\cos \phi \sin \phi\)

a \(a(2 a-b)\) b \(\cos \phi(2 \cos \phi-\sin \phi)\)

We can also factor quadratic trinomials.

Example 5.17

a \(t^2-3 t-10\) b \(\tan ^2 \alpha-3 \tan \alpha-10\)

a We look for numbers \(p\) and \(q\) so that \((t+p)(t+q)=t^2-3 t-10\). Their product is \(p q=-10\), and their sum is \(p+q=-3\). By checking the factors of \(-10\) for the correct sum, we find \(p=-5\) and \(q=2\). Thus,

\(t^2-3 t-10=(t-5)(t+2)\)

b Now replace \(t\) by \(\tan \alpha\) to find

\(\tan ^2 \alpha-3 \tan \alpha-10=(\tan \alpha-5)(\tan \alpha+2)\)

Checkpoint 5.18

a \(3 z^2-2 z-1\) b \(3 \sin ^2 \beta-2 \sin \beta-1\)

a \((3 z+1)(z-1)\) b \((3 \sin \beta+1)(\sin \beta-1)\)

Review the following skills you will need for this section.

Algebra Refresher 5.1

1 \(2x^2 + 5x - 12\)

2 \(3x^2 - 2x - 8\)

3 \(12x^2 + x - 1\)

4 \(6x^2 - 13x - 15\)

5 \(6x^2 - x - 12\)

6 \(24x^2 + 10x - 21\)

1 \((2x-3)(x+4)\)

2 \((3x+4)(x-2)\)

3 \((4x-1)(3x+1)\)

4 \((6x+5)(x-3)\)

5 \((2x-3)(3x+4)\)

6 \((6x+7)(4x-3)\)

Section 5.1 Summary

• Expression

• Evaluate

• Binomial

• Trinomial

• Simplify

• Equivalent expression

• Like terms

• Distributive law

• Factor

1 Expression containing trig ratios can be simplified or evaluated like other algebraic expressions. To simplify an expression containing trig ratios, we treat each ratio as a single variable.

2 \(\sin (X+Y)\) is not equal to \(\sin X+\sin Y\) (and the same holds for the other trig ratios). Remember that the parentheses indicate function notation, not multiplication.

3 We write \(\cos ^2 \theta\) to denote \((\cos \theta)^2\), and \(\cos ^n \theta\) to denote \((\cos \theta)^n\). (Similarly for the other trig ratios.)

4 We can factor trigonometric expressions with the same techniques we use for algebraic expressions.

Study Questions

1 To evaluate \(\cos ^2 30^{\circ}\), Delbert used the keystrokes

\(\text { COS } 30 \quad x^2 \text { ENTER }\)

and got the answer 1. Were his keystrokes correct? Why or why not?

2 Make up an example to show that \(\tan (\theta+\phi) \neq \tan \theta+\tan \phi\).

3 Factor each expression, if possible.

\(\begin{array}{ll} \text { a } x^2-4 x \quad \quad&& \text { e } x^2-4 x+4 \\ \text { b } x^2-4 && \text { f } x^2+4 x \\ \text { c } x^2+4 && \text { g } x^2-4 x-4 \\ \text { d } x^2+4 x+4 && \text { h }-x^2+4 \end{array}\)

1 Evaluate trigonometric expressions #1–22

2 Simplify trigonometric expressions #23–34

3 Recognize equivalent expressions #35–44

4 Multiply or expand trigonometric expressions #45–56

5 Factor trigonometric expressions #57–70

Homework 5.1

For Problems 1-8, evaluate the expressions, using exact values for the trigonometric ratios.

1. \(5 \tan 135^{\circ}+6 \cos 60^{\circ}\) 2. \(3 \tan 240^{\circ}+8 \sin 300^{\circ}\) 3. \(\sin \left(15^{\circ}+30^{\circ}\right)\) 4. \(\cos \left(2 \cdot 75^{\circ}\right)\) 5. \(8 \cos ^2\left(30^{\circ}\right)\) 6. \(12 \sin ^2\left(315^{\circ}\right)\) 7. \(3 \tan ^2\left(150^{\circ}\right)-\sin ^2\left(45^{\circ}\right)\) 8. \(1+\tan ^2\left(120^{\circ}\right)\)

For Problems 9-16, evaluate the expressions for \(x=30^{\circ}, y=45^{\circ}\), and \(z=60^{\circ}\). Give exact values for your answers.

9. \(3 \sin x+5 \cos y\) 10. \(4 \tan y+6 \cos y\) 11. \(-2 \tan 3 y\) 12. \(\sin (3 x-2 x)\) 13. \(\cos ^2 x+\sin ^2 x\) 14. \(7 \sin ^2 y+7 \cos ^2 y\) 15. \(\cos x \cos x-\sin x \sin z\) 16. \(\tan \left(180^{\circ}-x\right) \tan x\)

For Problems 17–22, evaluate the expressions using a calculator.

a \(\sin (10^{\circ} + 40^{\circ})\)

b \(\sin 10^{\circ} + \sin 40^{\circ}\)

c \(\sin 10^{\circ} \cos 40^{\circ} + \cos 10^{\circ} \sin 40^{\circ}\)

a \(\cos \left(20^{\circ}+50^{\circ}\right)\)

b \(\cos 20^{\circ}+\cos 50^{\circ}\)

c \(\cos 20^{\circ} \cos 50^{\circ}-\sin 20^{\circ} \sin 50^{\circ}\)

a \(\cos \left(2 \cdot 24^{\circ}\right)\)

b \(2 \cos 24^{\circ}\)

c \(2 \cos ^2\left(24^{\circ}\right)-1\)

a \(\cos \left(3 \cdot 49^{\circ}\right)\)

b \(3 \cos 49^{\circ}\)

c \(4 \cos ^3\left(49^{\circ}\right)-3 \cos 49^{\circ}\)

a \(\cos ^2\left(17^{\circ}\right)+\sin ^2\left(17^{\circ}\right)\)

b \(\cos ^2\left(86^{\circ}\right)+\sin ^2\left(86^{\circ}\right)\)

c \(\cos ^2\left(111^{\circ}\right)+\sin ^2\left(111^{\circ}\right)\)

a \(\dfrac{1}{\cos ^2\left(25^{\circ}\right)}-\tan^2\left(25^{\circ}\right)\)

b \(\dfrac{1}{\cos ^2\left(100^{\circ}\right)}-\tan^2\left(100^{\circ}\right)\)

c \(\dfrac{1}{\cos ^2\left(8^{\circ}\right)}-\tan^2\left(8^{\circ}\right)\)

For Problems 23–28, combine like terms.

a \(3x^2 - x - 5x^2\)

b \(3 \cos ^2 \theta-\cos \theta-5 \cos ^2 \theta\)

a \(4 x+5 y-y\)

b \(4 \cos \theta+5 \sin \theta-\sin \theta\)

a \(-3 S C+7 S C\)

b \(-3 \sin \theta \cos \theta+7 \sin \theta \cos \theta\)

a \(4 S C+11 S C-17 S C\)

b \(4 \sin \theta \cos \theta+11 \sin \theta \cos \theta-17 \sin \theta \cos \theta\)

a \(-C^2 S^3+6 C^2 S^3\)

b \(-\cos ^2 \theta \sin ^3 \theta+6 \cos ^2 \theta \sin ^3 \theta\)

a \(7 C^2 S^2-(2 C S)^2\)

b \(7 \cos ^2 \theta \sin ^2 \theta-(2 \cos \theta \sin \theta)^2\)

For Problems 29-34, simplify the expression, then evaluate.

29. \(\cos t+2 \cos t \sin t-3 \cos t\), for \(t=143^{\circ}\) 30. \(11 \tan w-4 \tan w+6 \tan w \cos w\), for \(w=8^{\circ}\) 31. \(7 \tan \theta-4 \tan \phi+3 \tan \phi-6 \tan \theta\), for \(\theta=21^{\circ}, \phi=89^{\circ}\) 32. \(5 \sin A+\sin B-6 \sin B+6 \sin A\), for \(A=111^{\circ}, B=26^{\circ}\) 33. \(-\sin x \cos x-\sin (2 x)+3 \sin x \cos x\), for \(x=107^{\circ}\) 34. \(4 \cos u \sin u+3 \sin 2 u-10 \sin u \cos u\), for \(u=2^{\circ}\)

For Problems 35–44, decide whether or not the expressions are equivalent. Explain.

35. \(\cos (x+y), \quad \cos x+\cos y\) 36. \(\sin (\theta-\phi), \quad \sin \theta-\sin \phi\) 37. \(\tan (2 A), 2 \tan A\) 38. \(\cos \left(\frac{1}{2} \beta\right), \quad \frac{1}{2} \cos \beta\) 39. \((\sin \alpha)^2, \quad \sin ^2 \alpha\) 40. \((\tan B)^2, \quad \tan \left(B^2\right)\) 41. \(\sin 3 t+\sin 5 t, \quad \sin 8 t\) 42. \(3 \cos 2 x+5 \cos 2 x, 8 \cos 2 x\) 43. \(\tan \left(\theta+45^{\circ}\right)-\tan \theta, \quad \tan 45^{\circ}\) 44. \(\sin \left(30^{\circ}+z\right)+\sin \left(30^{\circ}-z\right), \quad \sin 60^{\circ}\)

For Problems 45–56, multiply or expand.

a \(x(2 x-1)\)

b \(\sin A(2 \sin A-1)\)

a \(q(5 r-q)\)

b \(\cos \theta(5 \sin \theta-\cos \theta)\)

a \(a(b-3 a)\)

b \(\tan A(\tan B-3 \tan A)\)

a \(3 w(2 w-z)\)

b \(3 \sin \alpha(2 \sin \alpha-\sin \beta)\)

a \((C+1)(2 C-1)\)

b \((\cos \phi+1)(2 \cos \phi-1)\)

a \((3 S-2)(S+1)\)

b \((3 \sin B-2)(\sin B+1)\)

a \((a+b)(a-b)\)

b \((\cos \theta+\cos \phi)(\cos \theta-\cos \phi)\)

a \((t-w)(t+4 w)\)

b \((\tan \alpha-\tan \beta)(\tan \alpha+4 \tan \beta)\)

a \((1-T)^2\)

b \((1-\tan \theta)^2\)

a \((2+3 S)^2\)

b \((2+3 \sin \theta)^2\)

a \(\left(T^2+2\right)\left(T^2-2\right)\)

b \(\left(\tan ^2 \theta+2\right)\left(\tan ^2 \theta-2\right)\)

a \(\left(2 c^2-3\right)\left(2 c^2+3\right)\)

b \(\left(2 \cos ^2 \phi-3\right)\left(2 \cos ^2 \phi+3\right)\)

For Problems 57–70, factor.

a \(9 m+15 n\)

b \(9 \cos \alpha+15 \cos \beta\)

a \(12 p-20 q\)

b \(12 \sin \theta-20 \sin \phi\)

a \(5 r^2-10 q r\)

b \(5 \tan ^2 C-10 \tan B \tan C\)

a \(2 x^2-6 x y\)

b \(2 \sin ^2 a-6 \sin A \cos A\)

a \(9 C^2-1\)

b \(9 \cos ^2 \beta-1\)

a \(25 S^2-16 S\)

b \(25 \sin ^2 \beta-16 \sin \beta\)

a \(6 T^3-8 T^2\)

b \(6 \tan ^3 A-8 \tan ^2 A\)

a \(9 T^2-15 T^3\)

b \(9 \tan ^2 B-15 \tan ^3 B\)

a \(t^2-t-20\)

b \(\tan ^2 \theta-\tan \theta-20\)

a \(s^2+5 s-6\)

b \(\sin ^2 A+5 \sin A-6\)

a \(3 c^2+2 c-1\)

b \(3 \cos ^2 B+2 \cos B-1\)

a \(8 S^2-6 S+1\)

b \(8 \sin ^2 \phi-6 \sin \phi+1\)

a \(6 S^2-5 S-1\)

b \(6 \sin ^2 \alpha-5 \sin \alpha-1\)

a \(T^2-4 T-12\)

b \(\tan ^2 \alpha-4 \tan \alpha-12\)