solving trigonometric equations assignment

7.5 Solving Trigonometric Equations

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.
• 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 π 6 π 6 and 5 π 6 . 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 Trigonometric Equation in Linear Form

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 .

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 III, the reference angle is θ ' ≈ π − 1 .8235 ≈ 1 .3181 . θ ' ≈ π − 1 .8235 ≈ 1 .3181 . The other solution in quadrant III is π + 1 .3181 ≈ 4 .4597 . π + 1 .3181 ≈ 4 .4597 .

The solutions are 1.8235 ± 2 π k 1.8235 ± 2 π k and 4.4597 ± 2 π k . 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. Thus,

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 π 2 . 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 0 , π , 7 π 6 , 11 π 6 . 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 2 π 3 , 4 π 3 , π . 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 π 6 , 5 π 6 , 7 π 6 , and  11 π 6 . 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 15 a 15 a 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

7.5 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?

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• Unit test Trigonometric equations and identities

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7.1: Solving Trigonometric Equations with Identities

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In the last chapter, we solved basic trigonometric equations. In this section, we explore the techniques needed to solve more complicated trig equations. Building from what we already know makes this a much easier task.

Consider the function \(f(x)=2x^{2} +x\). If you were asked to solve \(f(x)=0\), it requires simple algebra:

\[2x^{2} +x=0\nonumber\]Factor \[x(2x+1)=0\nonumber\]Giving solutions \[x = 0\text{ or }x = -\dfrac{1}{2}\nonumber\]

Similarly, for \(g(t)=\sin (t)\), if we asked you to solve \(g(t)=0\), you can solve this using unit circle values:

\[\sin (t)=0\text{ for }t=0, \pi , 2\pi\text{ and so on.}\nonumber\]

Using these same concepts, we consider the composition of these two functions:

\[f(g(t))=2(\sin (t))^{2} +(\sin (t))=2\sin ^{2} (t)+\sin (t)\nonumber\]

This creates an equation that is a polynomial trig function. With these types of functions, we use algebraic techniques like factoring and the quadratic formula, along with trigonometric identities and techniques, to solve equations.

As a reminder, here are some of the essential trigonometric identities that we have learned so far:

DefinitionS: IDENTITIES

Pythagorean Identities

\[\cos ^{2} (t)+\sin ^{2} (t)=1\quad 1+\cot ^{2} (t)=\csc ^{2} (t)\quad 1+\tan ^{2} (t)=\sec ^{2} (t)\]

Negative Angle Identities

\[\sin (-t)=-\sin (t)\quad \cos (-t)=\cos (t)\quad \tan (-t)=-\tan (t)\]

\[\csc (-t)=-\csc (t)\quad \sec (-t)=\sec (t)\quad \cot (-t)=-\cot (t)\]

Reciprocal Identities

\[\sec (t)=\dfrac{1}{\cos (t)}\quad \csc (t)=\dfrac{1}{\sin (t)}\quad \tan (t)=\dfrac{\sin (t)}{\cos (t)}\quad \cot (t)=\dfrac{1}{\tan (t)}\]

Example \(\PageIndex{1}\)

Solve \(2\sin ^{2} (t)+\sin (t)=0\) for all solutions with \(0\le t<2\pi\).

This equation kind of looks like a quadratic equation, but with sin( t ) in place of an algebraic variable (we often call such an equation “quadratic in sine”). As with all quadratic equations, we can use factoring techniques or the quadratic formula. This expression factors nicely, so we proceed by factoring out the common factor of sin(\(t\)):

\[\sin (t)\left(2\sin (t)+1\right)=0\nonumber\]

Using the zero product theorem, we know that the product on the left will equal zero if either factor is zero, allowing us to break this equation into two cases:

\[\sin (t)=0\text{ or }2\sin (t)+1=0\nonumber\]

\[\sin (t)=0\nonumber\] \[2\sin (t)+1=0\nonumber\] \[t = 0\text{ or }t = \pi\nonumber\] \[\sin (t)=-\dfrac{1}{2}\nonumber\] \[t=\dfrac{7\pi }{6}\text{ or }t=\dfrac{11\pi }{6}\nonumber\]

Together, this gives us four solutions to the equation on \(0\le t<2\pi\):

\[t=0,\pi ,\dfrac{7\pi }{6} ,\dfrac{11\pi }{6}\nonumber\]

We could check these answers are reasonable by graphing the function and comparing the zeros.

Example \(\PageIndex{2}\)

Solve \(3\sec ^{2} (t)-5\sec (t)-2=0\) for all solutions with \(0\le t<2\pi\).

Since the left side of this equation is quadratic in secant, we can try to factor it, and hope it factors nicely.

If it is easier to for you to consider factoring without the trig function present, consider using a substitution \(u=\sec (t)\), resulting in \(3u^{2} -5u-2=0\), and then try to factor:

\[3u^{2} -5u-2=(3u+1)(u-2)\nonumber\]

Undoing the substitution,

\[(3\sec (t)+1)(\sec (t)-2)=0\nonumber\]

Since we have a product equal to zero, we break it into the two cases and solve each separately.

\[3\sec (t)+1=0\nonumber\]Isolate the secant \[\sec (t)=-\dfrac{1}{3}\nonumber\]Rewrite as a cosine \[\dfrac{1}{\cos (t)} =-\dfrac{1}{3}\nonumber\]Invert both sides \[\cos (t)=-3\nonumber\]

Since the cosine has a range of [-1, 1], the cosine will never take on an output of -3. There are no solutions to this case.

Continuing with the second case,

These are the only two solutions on the interval.

By utilizing technology to graph \(f(t)=3\sec ^{2} (t)-5\sec (t)-2\), a look at a graph confirms there are only two zeros for this function on the interval [0, 2 \(\pi\)), which assures us that we didn’t miss anything.

Exercise \(\PageIndex{1}\)

Solve \(2\sin ^{2} (t)+3\sin (t)+1=0\) for all solutions with \(0\le t<2\pi\).

Factor as \[\left(2\sin (t)+1\right)\left(\sin (t)+1\right)=0\nonumber\] \[2\sin (t)+1=0\text{ at }t=\dfrac{7\pi }{6} ,\dfrac{11\pi }{6}\nonumber\] \[\sin (t)+1=0\text{ at }t=\dfrac{3\pi }{2}\nonumber\] \[t=\dfrac{7\pi }{6} ,\dfrac{3\pi }{2} ,\dfrac{11\pi }{6}\nonumber\]

When solving some trigonometric equations, it becomes necessary to first rewrite the equation using trigonometric identities. One of the most common is the Pythagorean Identity, \(\sin ^{2} (\theta )+\cos ^{2} (\theta )=1\) which allows you to rewrite \(\sin ^{2} (\theta )\) in terms of \(\cos ^{2} (\theta )\) or vice versa,

Alternate Forms of the Pythagorean Identity

\[\begin{array}{l} {\sin ^{2} (\theta )=1-\cos ^{2} (\theta )} \\ {\cos ^{2} (\theta )=1-\sin ^{2} (\theta )} \end{array}\]

These identities become very useful whenever an equation involves a combination of sine and cosine functions.

Example \(\PageIndex{3}\)

Solve \(2\sin ^{2} (t)-\cos (t)=1\) for all solutions with \(0\le t<2\pi\).

Since this equation has a mix of sine and cosine functions, it becomes more complicated to solve. It is usually easier to work with an equation involving only one trig function. This is where we can use the Pythagorean Identity.

\[2\sin ^{2} (t)-\cos (t)=1\nonumber\]Using \(\sin ^{2} (\theta )=1-\cos ^{2} (\theta )\) \[2\left(1-\cos ^{2} (t)\right)-\cos (t)=1\nonumber\]Distributing the 2 \[2-2\cos ^{2} (t)-\cos (t)=1\nonumber\]

Since this is now quadratic in cosine, we rearrange the equation so one side is zero and factor.

\[-2\cos ^{2} (t)-\cos (t)+1=0\nonumber\]Multiply by -1 to simplify the factoring \[2\cos ^{2} (t)+\cos (t)-1=0\nonumber\]Factor \[\left(2\cos (t)-1\right)\left(\cos (t)+1\right)=0\nonumber\]

This product will be zero if either factor is zero, so we can break this into two separate cases and solve each independently.

\[2\cos (t)-1=0\text{ or }\cos (t)+1=0\nonumber\] \[\cos (t)=\dfrac{1}{2}\text{ or }\cos (t)=-1\nonumber\] \[t=\dfrac{\pi }{3}\text{ or }t=\dfrac{5\pi }{3}\text{ or }t=\pi\nonumber\]

Exercise \(\PageIndex{3}\)

Solve \(2\sin ^{2} (t)=3\cos (t)\) for all solutions with \(0\le t<2\pi\).

\[2\left(1-\cos ^{2} (t)\right)=3\cos (t)\nonumber\] \[2\cos ^{2} (t)+3\cos (t)-2=0\nonumber\] \[\left(2\cos (t)-1\right)\left(\cos (t)+2\right)=0\nonumber\] \(\cos (t)+2=0\) has no solutions \(2\cos (t)-1=0\) at \(t=\dfrac{\pi }{3} ,\dfrac{5\pi }{3}\)

In addition to the Pythagorean Identity, it is often necessary to rewrite the tangent, secant, cosecant, and cotangent as part of solving an equation.

Example \(\PageIndex{4}\)

Solve \(\tan (x)=3\sin (x)\) for all solutions with \(0\le x<2\pi\).

With a combination of tangent and sine, we might try rewriting tangent

\[\tan (x)=3\sin (x)\nonumber\] \[\dfrac{\sin (x)}{\cos (x)} =3\sin (x)\nonumber\]Multiplying both sides by cosine \[\sin (x)=3\sin (x)\cos (x) \nonumber\]

At this point, you may be tempted to divide both sides of the equation by sin(\(x\)).

Resist the urge . When we divide both sides of an equation by a quantity, we are assuming the quantity is never zero. In this case, when sin(\(x\)) = 0 the equation is satisfied, so we’d lose those solutions if we divided by the sine.

To avoid this problem, we can rearrange the equation so that one side is zero (You technically can divide by sin( x ), as long as you separately consider the case where sin( x ) = 0. Since it is easy to forget this step, the factoring approach used in the example is recommended.).

\[\sin (x)-3\sin (x)\cos (x)=0\nonumber\]Factoring out sin(\(x\)) from both parts \[\sin (x)\left(1-3\cos (x)\right)=0 \nonumber\]

From here, we can see we get solutions when \(\sin (x)=0\) or \(1-3\cos (x)=0\).

Using our knowledge of the special angles of the unit circle,

\[\sin (x)=0\text{ when }x = 0\text{ or }x = \pi\nonumber\]

For the second equation, we will need the inverse cosine.

\[1-3\cos (x)=0\nonumber\] \[\cos (x)=\dfrac{1}{3}\nonumber\]Using our calculator or technology \[x=\cos ^{-1} \left(\dfrac{1}{3} \right)\approx 1.231\nonumber\]Using symmetry to find a second solution \[x=2\pi -1.231=5.052 \nonumber\]

We have four solutions on \(0 \le x<2\pi\):

\[x = 0, 1.231, \quad\pi , 5.052\nonumber\]

Solve \(\sec (\theta )=2\cos (\theta )\) to find the first four positive solutions.

\[\dfrac{1}{\cos (\theta )} =2\cos (\theta )\nonumber\] \[\dfrac{1}{2} =\cos ^{2} (\theta )\nonumber\] \[\cos (\theta )=\pm \sqrt{\dfrac{1}{2} } =\pm \dfrac{\sqrt{2} }{2}\nonumber\] \[\theta =\dfrac{\pi }{4} ,\dfrac{3\pi }{4} ,\dfrac{5\pi }{4} ,\dfrac{7\pi }{4}\nonumber\]

Example \(\PageIndex{5}\)

Solve \(\dfrac{4}{\sec ^{2} (\theta )} +3\cos \left(\theta \right)=2\cot \left(\theta \right)\tan \left(\theta \right)\) for all solutions with \(0\le \theta <2\pi\).

\[\dfrac{4}{\sec ^{2} (\theta )} +3\cos \left(\theta \right)=2\cot \left(\theta \right)\tan \left(\theta \right)\nonumber\] Using the reciprocal identities \[4\cos ^{2} (\theta )+3\cos (\theta )=2\dfrac{1}{\tan (\theta )} \tan (\theta )\nonumber\] Simplifying \[4\cos ^{2} \left(\theta \right)+3\cos \left(\theta \right)=2\nonumber\] Subtracting 2 from each side \[4\cos ^{2} \left(\theta \right)+3\cos \left(\theta \right)-2=0\nonumber\]

This does not appear to factor nicely so we use the quadratic formula, remembering that we are solving for cos( \(\theta\) ).

\[\cos (\theta )=\dfrac{-3\pm \sqrt{3^{2} -4(4)(-2)} }{2(4)} =\dfrac{-3\pm \sqrt{41} }{8}\nonumber\]

Using the negative square root first,

\[\cos (\theta )=\dfrac{-3-\sqrt{41} }{8} =-1.175\nonumber\]

This has no solutions, since the cosine can’t be less than -1.

Using the positive square root,

\[\cos (\theta )=\dfrac{-3+\sqrt{41} }{8} =0.425\nonumber\] \[\theta =\cos ^{-1} \left(0.425\right)=1.131\nonumber\] By symmetry, a second solution can be found \[\theta =2\pi -1.131=5.152\nonumber\]

Important Topics of This Section

• Review of Trig Identities
• Solving Trig Equations
• By Factoring
• Using the Quadratic Formula
• Utilizing Trig Identities to simplify

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This course is designed to introduce the student to the spirit, beauty and vitality of mathematics.  The emphasis is on development of ideas rather than problem solving skills.  Topics vary, but are typically chosen from diverse areas such as geometry, number theory, computation and graph theory.

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This course introduces statistics as a liberal discipline and applies the principles of statistics to problems of interest to health sciences professionals.  Students will gain an understanding of statistical ideas and methods, acquire the ability to deal critically with numerical arguments, and gain an understanding of the impact of statistical ideas on the health sciences, public policy and other areas of application.

MATH 1190 Fundamentals of Mathematics I (Fall: 3) MATH 1191 Fundamentals of Mathematics II (Spring: 3) Restricted to Lynch School of Education and Human Development students.

MATH 1190-1191 is a course sequence designed for those who plan to teach mathematics in grades K-8.  The emphasis is on building conceptual understanding of the mathematics present in the emerging K-8 curriculum and on deepening content knowledge.  Number and number systems through the real number system will be studied; functions and the structure of algebra and geometry will be developed.  Problem solving and reasoning, applications, and making connections will be featured.

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Note: Completion of both 1002 and 1003 satisfies the University Core Requirement in Mathematics, as well as any Calculus I requirement. 

Department permission is required; see the Assistant Chair for Undergraduates.

MATH 1100 Calculus I (Fall/Spring: 4)

MATH 1100 is a first course in the calculus of one variable intended for Biology,  Economics, Management, Psychology / Neuroscience and Premedical students.  It is open to others who are qualified and desire a more rigorous mathematics course at the core level.

Students contemplating majors in Chemistry, Computer Science, Geology/Geophysics, Mathematics, or Physics should enroll in MATH 1102 Calculus I for Math and Science Majors, rather than MATH 1100.

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 MATH 1101 is not open to students who have completed MATH1103. Students contemplating majors in Chemistry, Computer Science, Geology/Geophysics, Mathematics, or Physics should enroll in MATH 1103 Calculus II for Math and Science Majors rather than MATH 1101.

There are three main topics: integration (definition of integration, basic techniques for integration, and select applications); an introduction to differential equations (with applications to population modeling and other contexts); an introduction to multivariable functions and partial derivatives (with application to optimization in economics and other contexts).

MATH 1102 Calculus I for Math and Science Majors (Fall: 4)

MATH 1102 is a first course in the calculus of one variable intended for Chemistry, Computer Science, Geology/Geophysics, Mathematics, and Physics majors. It is open to others who are qualified and desire a more rigorous calculus course than MATH 1100.

Topics covered include the algebraic and analytic properties of the real number system, functions, limits, derivatives, and an introduction to integration.

MATH 1103 Calculus II for Math and Science Majors (Fall/Spring: 4) MATH 1103 is a continuation of MATH 1102. Topics covered in the course include several algebraic techniques of integration, many applications of integration, and infinite sequences and series.

MATH 2202 Multivariable Calculus (Fall/Spring: 4) and MATH 2203 Multivariable Calculus Honors (Fall: 4) Prerequisite: MATH 1101, MATH 1103 or equivalent (e.g., five on the BC Calculus Advanced Placement Exam).

This course is for students majoring in Chemistry, Computer Science BS, Geology-Geophysics, Mathematics and Physics as well as other students who have completed integral calculus.

Topics include vectors in two and three dimensions, analytic geometry of three dimensions, parametric curves, partial derivatives, the gradient, optimization in several variables, multiple integration with change of variables across different coordinate systems, line integrals, and Green's Theorem.

Prerequisite: A strong background in single variable calculus, as demonstrated, for example, by a score of 5 on the AP-BC Examination.

MATH 2203 is an honors version of MATH 2202 intended for students with strong preparation and high motivation. Topics covered include vector analysis, partial differentiation, multiple integration, line integrals, Green's theorem, Stokes's theorem, and the Divergence Theorem.

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These courses may be required or suggested in certain Schools or Programs.

This course introduces statistics as a liberal discipline and applies the principles of statistics to problems of interest to health sciences professionals. Students will gain an understanding of statistical ideas and methods, acquire the ability to deal critically with numerical arguments, and gain an understanding of the impact of statistical ideas on the health sciences, public policy and other areas of application.

MATH 1190 Fundamentals of Mathematics I (Fall: 3) Restricted to Lynch School of Education and Human Development students. Satisfies Mathematics Core Requirement.

MATH 1190-1191 is a course sequence designed for those who plan to teach mathematics in grades K-8. The emphasis is on building conceptual understanding of the mathematics present in the emerging K-8 curriculum and on deepening content knowledge. Number and number systems through the real number system will be studied; functions and the structure of algebra will be developed. Problem solving and reasoning, applications, and making connections will be featured.

MATH 1191 Fundamentals of Mathematics II (Spring: 3) Restricted to Lynch School of Education and Human Development students. Satisfies Mathematics Core Requirement.

As in MATH 1190, the course emphasizes building conceptual understanding of the mathematics present in the emerging K-8 curriculum and on deepening the content knowledge. Topics drawn from geometry and measurement, data analysis, statistics, and probability will be developed. Problem solving and reasoning, applications, and making connections will be featured.

MATH 2290 Number Theory for Teachers (Alternate Spring semesters: 3) Prerequisites: MATH 1190 and MATH 1191

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MATH 2291 Geometry for Teachers (Alternate Spring semesters: 3) Prerequisites: MATH 1190 and MATH 1191

This course is intended to fill a basic need of all teachers of grades K-9. Geometry now occupies a significant role in the elementary mathematics curriculum. The course will treat content, but ideas for presenting geometry as an activity-based program will also be stressed. Topics to be covered include the geoboard and other key manipulatives, elements of motion and Euclidean geometry, and suggestions for using Logo as a tool to enhance teaching geometry.

MATH 4453 Euclid's Elements (Spring: 3) Prerequisite: None

This course is a close reading of Euclid's Elements in seminar style, with careful attention to axiomatic reasoning and mathematical constructions that build on one another in a sequence of logical arguments.

We will also emphasize clear and creative communication on mathematical ideas, with some attention to the cultural background of the  Elements  and its place in a modern education.

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Permission of the instructor required for students outside the LSOE.

This course is designed to deepen students' mathematical knowledge through solving, explaining, and extending challenging and interesting problems. Students will work both individually and in groups on problems chosen from polynomials, trigonometry, analytic geometry, pre-calculus, one-variable calculus, probability, and numerical algorithms. The course will emphasize explanations and generalizations rather than formal proofs and abstract properties. Some pedagogical issues, such as composing good problems and expected points of confusion in explaining various topics will come up, but the primary goal is mathematical insight. The course will be of particular use to future secondary math teachers.

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Required and elective courses for majors and minors..

MATH 2210 Linear Algebra (Fall/Spring: 3) Corequisite: MATH 2202/3 Multivariable Calculus

This course is an introduction to the techniques of linear algebra in Euclidean space. Topics covered include matrices, determinants, systems of linear equations, vectors in n-dimensional space, complex numbers, and eigenvalues. The course is required of mathematics majors and minors, but is also suitable for students in the social sciences, natural sciences, and management.

MATH 2211 Linear Algebra (Honors) (Spring: 3) Prerequisite: MATH 2203 Multivariable Calculus (Honors) or permission of the Assistant Chair for Undergraduates

This course is intended for students with strong preparation and high motivation. Topics covered include matrices, linear equations, determinants, eigenvectors and eigenvalues, vector spaces and linear transformations, inner products, and canonical forms. The course will include significant work with proofs.

MATH 2216 Introduction to Abstract Mathematics (Fall/Spring: 3) This course is designed to develop the student's ability to do abstract mathematics through the presentation and development of the basic notions of logic and proof. Topics include elementary set theory, mappings, integers, rings, complex numbers, and polynomials.

MATH 3310 Introduction to Abstract Algebra (Fall/Spring: 3) Prerequisites: MATH 2210 Linear Algebra and MATH 2216 Introduction to Abstract Mathematics.

This course studies four fundamental algebraic structures: groups, including subgroups, cyclic groups, permutation groups, symmetry groups and Lagrange's Theorem; rings, including subrings, integral domains, and unique factorization domains; polynomials, including a discussion of unique factorization and methods for finding roots; fields, introducing the basic ideas of field extensions and ruler and compass constructions.

MATH 3311 Algebra I (Fall: 3) MATH 3312 Algebra II (Spring: 3)   Prerequisites: MATH 2210 Linear Algebra and MATH 2216 Introduction to Abstract Mathematics.

This year-long sequence studies the basic structures of abstract algebra. Topics include groups, subgroups, normal subgroups, factor groups, Lagrange's Theorem, the Sylow Theorems, rings, ideal theory, integral domains, field extensions, and Galois theory.

Note: Students may not take both MATH 3310 and MATH 3311. With the permission of the Assistant Chair for Undergraduates, students who have taken MATH 3310 may be allowed to take MATH 3312. However, they may need to do additional work on their own in order to make that transition. Students considering a B.S. in Mathematics are strongly encouraged to take MATH 3311.

MATH 3320 Introduction to Analysis (Fall/Spring: 3) Prerequisites: MATH 2202 Multivariable Calculus and MATH 2216 Introduction to Abstract Mathematics .

The purpose of this course is to give students the theoretical foundations for the topics taught in MATH 1102-1103. It will cover algebraic and order properties of the real numbers, the least upper bound axiom, limits, continuity, differentiation, the Riemann integral, sequences, and series. Definitions and proofs will be stressed throughout the course.

MATH 3321 Analysis I (Fall: 3) MATH 3322 Analysis II (Spring: 3) Prerequisites: MATH 2210 Linear Algebra and MATH 2216 Introduction to Abstract Mathematics.

This year-long sequence studies the basic structure of the real numbers. Topics include the least upper bound principle, compactness of closed intervals (the Heine-Borel theorem), sequences, convergence, the Bolzano-Weierstrass theorem, continuous functions, boundedness and intermediate value theorems, uniform continuity, differentiable functions, the mean value theorem, construction of the Riemann integral, the fundamental theorem of calculus, sequences and series of functions, uniform convergence, the Weierstrass approximation theorem, special functions (exponential and trig), and Fourier series. As time permits, other topics may include metric spaces, calculus of functions of several variables, and an introduction to measure and integration.

Note: Students may not take both MATH 3320 and MATH 3321. With the permission of the Assistant Chair for Undergraduates, students who have taken MATH 3320 may be allowed to take MATH 3322. However, they may need to do additional work on their own in order to make that transition. Students considering a B.S. in Mathematics are strongly encouraged to take MATH 3321.

MATH 4410 Differential Equations (Fall; sometimes Spring: 3) Prerequisites: MATH 2202 Multivariable Calculus and MT 210 Linear Algebra.

This course is a junior-senior elective intended primarily for the student who is interested in seeing applications of mathematics.

Among the topics covered will be the following: first order linear equations, higher order linear equations with constant coefficients, linear systems, Laplace transforms, and other topics as time permits.

MATH 4412 Partial Differential Equations (Offered Occasionally: 3) Prerequisite: MATH 4410 Differential Equations.

This course investigates the classical partial differential equations of applied mathematics (diffusion, Laplace/Poisson, and wave) and their methods of solution (separation of variables, Fourier series, transforms, Green's functions, and eigenvalue applications). Additional topics will be included as time permits.

MATH 4414 Numerical Analysis (Spring: 3) Prerequisites: MATH 2202 Multivariable Calculus, and MATH 2210 Linear Algebra.

Topics include the solution of linear and nonlinear algebraic equations, interpolation, numerical differentiation and integration, numerical solution of ordinary differential equations, approximation theory.

MATH 4426 Probability (Fall/Spring: 3) Prerequisites: MATH 2202 Multivariable Calculus and familiarity with using a computer.

This course provides a general introduction to modern probability theory.

Topics include probability spaces, discrete and continuous random variables, joint and conditional distributions, mathematical expectation, the central limit theorem, and the weak law of large numbers. Applications to real data will be stressed, and we will use the computer to explore many concepts.

MATH 4427 Mathematical Statistics (sometimes Fall; Spring: 3) Prerequisites: MT 426 Probability and familiarity with using a computer.

Topics studied include the following: sampling distributions, parametric point and interval estimation, hypothesis testing, goodness-of-fit, parametric and nonparametric two-sample analysis. Applications to real data will be stressed, and the computer will be used to explore concepts and analyze data.

MATH 4430 Introduction to Number Theory (Spring: 3) Prerequisite: MATH 2216 Introduction to Abstract Mathematics.

Topics covered include divisibility, unique factorization, congruences, number-theoretic functions, primitive roots, diophantine equations, continued fractions, quadratic residues, and the distribution of primes. An attempt will be made to provide historical background for various problems and to provide examples useful in the secondary school curriculum.

MATH 4435 Mathematical Programming (Fall: 3) Prerequisite: MATH 2210 Linear Algebra.

Topics studied from linear programming include a general discussion of linear optimization models, the theory and development of the simplex algorithm, degeneracy, duality, sensitivity analysis, and the dual simplex algorithm. Integer programming problems, and the transportation and assignment problems are considered,and algorithms are developed for their resolution.

MATH 4440 Dynamical Systems (Offered Occasionally: 3) Prerequisites: MATH 2202 Multivariable Calculus, MATH 2210 Linear Algebra, and MATH 2216 Introduction to Abstract Mathematics.

This course is an introduction to nonlinear dynamics and their applications, emphasizing qualitative methods for differential equations.

Topics include fixed and periodic points, stability, linearization, parameterized families and bifurcations, and existence and nonexistence theorems for closed orbits in the plane. The final part of the course is an introduction to chaotic systems and fractals, including the Lorenz system and the quadratic map.

MATH 4445 Combinatorics (Fall: 3) Prerequisites: MT 216 Introduction to Abstract Mathematics and MT 210 Linear Algebra. MT 210 may be taken simultaneously.

This course is an introduction to graph theory and combinatorics, with a strong emphasis on creative problem-solving techniques and connections with other branches of mathematics. Topics will center around the following: enumeration, Hamiltonian and Eulerian cycles, extremal graph theory, planarity, matching, colorability, Ramsey theory, hypergraphs, combinatorial geometry, and applications of linear algebra, probability, polynomials, and topology to combinatorics.

MATH 4450 Advanced Linear Algebra (Offered Occasionally: 3) Prerequisites: MATH 2210 Linear Algebra and MATH 3310 Introduction to Abstract Algebra.

This proof-based course presents a more rigorous approach to Linear Algebra and covers many topics beyond those in MATH 2210. Topics will include Abstract Vector Spaces and Linear Maps over any field, Modules, Canonical Forms and the Geometry of Bilinear Forms. Additional topics, if time permits, could include the basic theorems of Galois Theory, Matrix Factorization, and applications such as Coding Theory, Factor Analysis and Linear Difference Equations.

MATH 4451 Euclidean and Non-Euclidean Geometry (Fall: 3) Prerequisite: MATH 2216 Introduction to Abstract Mathematics.

This course is an introduction to geometric structure, broadly construed.

Topics may include: Euclidean geometry, hyperbolic and spherical geometry, platonic solids, tilings and wallpaper groups, graph theory, finite geometries, projective geometry, equidecomposition, the isoperimetric problem, surfaces and 3-dimensional manifolds.

MATH 4453 Euclid's Elements (Spring: 3) Prerequisites: None

We will also emphasize clear and creative communication on mathematical ideas, with some attention to the cultural background of the Elements and its place in a modern education.

MATH 4455 Mathematical Problem Solving (Spring: 3) Prerequisites: MATH 2202 Multivariable Calculus, MATH 2210 Linear Algebra, and MATH 2216 Introduction to Abstract Mathematics (or equivalent mathematical background). Permission of the instructor required for students outside the LSOE.

MATH 4460 Complex Variables (Fall/Spring: 3) Prerequisite: MATH 2202 Multivariable Calculus, and at least one of MATH 2210 Linear Algebra or MATH 2216 Introduction to Abstract Mathematics. Not open to MA students.

This course gives an introduction to the theory of functions of a complex variable, a fundamental and central area of mathematics. It is intended for mathematics majors and minors, and science majors.

Topics covered include: complex numbers and their properties, analytic functions and the Cauchy-Riemann equations, the logarithm and other elementary functions of a complex variable, integration of complex functions, the Cauchy integral theorem and its consequences, power series representation of analytic functions, the residue theorem and applications to definite integrals.

MATH 4461 Stochastic Processes (Spring: 3) Prerequisites : MATH 2216 and MATH 4426

A stochastic process describes the evolution of a system that changes over time in a random manner. This course introduces and studies various properties of some fundamental stochastic processes, including Markov chains in discrete and continuous time, renewal processes, and Brownian motion.

MATH 4470 Mathematical Modeling (Fall: 3) Prerequisites: MATH 2202 Multivariable Calculus and MATH 2210 Linear Algebra.

This is a course primarily for mathematics majors with the purpose of introducing the student to the creation, use and analysis of a variety of mathematical models and to reinforce and deepen the mathematical and logical skills required of modelers.

A secondary purpose is to develop a sense of the existing and potential roles of both small and large scale models in our scientific civilization. It proceeds through the study of the model-building process, examination of exemplary models, and individual and group efforts to build or refine models through a succession of problem sets, laboratory exercises, and field work.

MATH 4475 The History of Mathematics (Alternate Fall semesters: 3) Prerequisites: MATH 3310 and MATH 3320, one of which may be taken concurrently. Students must be familiar with abstract algebra (groups, rings, fields...) and rigorous analysis (differentiation and integration of real valued functions, sequences and series of functions...).

This course studies the development of mathematical thought, from ancient times to the twentieth century. Naturally, the subject is much too large for a single semester, so we will concentrate on the major themes and on the contributions of the greatest mathematicians. The emphasis in the course will be on the mathematics. Students will follow the historical arguments and work with the tools and techniques of the period being studied.

MATH 4480 Topics in Mathematics (Offered Occasionally: 3) Topics for this one-semester course vary from year to year according to the interests of faculty and students. With department permission it may be repeated.

MATH 4499 Readings and Research (Fall/Spring: 3) Department permission is required.

This is an independent study course, taken by arrangement with and under the supervision of a Mathematics Department faculty member.

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