Solving Trigonometric Equations, Part 1

Loading...

Из курса от партнера University of California, Irvine

Pre-Calculus: Trigonometry

296 оценки

Explore Related Videos

At Coursera, you will find the best lectures in the world. Here are some of our personalized recommendations for you

University of California, Irvine

Pre-Calculus: Trigonometry

296 оценки

This course covers mathematical topics in trigonometry. Trigonometry is the study of triangle angles and lengths, but trigonometric functions have far reaching applications beyond simple studies of triangles. This course is designed to help prepare students to enroll for a first semester course in single variable calculus. Upon completing this course, you will be able to: 1. Evaluate trigonometric functions using the unit circle and right triangle approaches 2. Solve trigonometric equations 3. Verify trigonometric identities 4. Prove and use basic trigonometric identities. 5. Manipulate trigonometric expressions using standard identities 6. Solve right triangles 7. Apply the Law of Sines and the Law of Cosines

Из урока

Trigonometric Equations

In this module, we will focus on solving equations involving trigonometric functions. These are usually equations in which the variable appears inside of a trigonometric function and we must use a combination of algebra skills and trigonometry manipulation to solve.

Solving Trigonometric Equations, Part 1 5:40

Solving Trigonometric Equations, Part 2 9:35

Solving Trigonometric Equations, Part 3 6:13

Solving Trigonometric Equations - Quadratic in Form5:54

Познакомьтесь с преподавателями

Dr. Sarah Eichhorn
Associate Dean, Distance Learning & Tenured Lecturer
Dr. Rachel Cohen Lehman

Let's look at solving Trig Equations.

For example, let's solve this equation here,

but only on the interval zero to two pi.

We can begin by subtracting three from both sides of this equation,

which would give us 2 times sine of theta is equal to negative 1.

And now, dividing both sides by two gives us that sine of

theta is equal to negative one half.

So, we need to find all angles theta in this interval here,

zero to two pi that have sine equal to negative one half.

Well, let's recall our unit circle.

Remember that the sine is the y coordinate of the point

of intersection of the terminal side of the angle and the unit circle.

So, looking down here in Quadrant three,

we see that the Y coordinate of this point is negative one half,

which corresponds to the angle of 7 pi divided by 6,

but also over here in quadrant four.

The Y coordinate of this point is also negative one half,

which corresponds to the angle of 11 pi divided by 6.

Therefore, the sine of both of these angles will equal negative one half,

which means our answer here then is theta is equal to

7 pi divided by 6 or 11 pi divided by 6.

All right. Let's look at another example.

Again, let's solve this equation only on the interval zero to two pi.

We can begin by subtracting two from both sides,

which gives us secant of theta is equal to negative 2.

Now, remember that secant of theta is equal to 1 divided by cosine of theta,

or cosine of theta is equal to 1 divided by secant of theta.

And, therefore, if secant of theta is equal to negative two,

then we're looking for angles theta

such that cosine of theta is one divided by negative 2,

or negative one half.

Again, recalling our units circle,

and remembering that the cosine of theta is the x coordinate

of the point of intersection of the terminal side of the angle and the unit circle.

We see that the x coordinate of this point here in quadrant two is negative one half,

which corresponds to the angle of 2 pi divided by 3.

But, also down here in Quadrant three.

The x coordinate of this point is also negative one half,

which corresponds to the angle of 4 pi divided by 3.

Therefore, our answer here would be theta is equal to,

2 pi divided by 3,

or 4 pi divided by 3.

All right.

Let's look at one more example.

Again, let's find all solutions to this equation but only in the interval zero to two pi.

We'll begin by adding one to both sides,

which gives us cotangent of theta,

is equal to 1.

Now, remember that cotangent of theta,

is equal to cosine of theta,

divided by sine of theta.

But, we just saw that the cosine of theta was the x coordinate of

the point of intersection of the terminal side of the angle and the unit circle,

and the sine of theta was the y coordinate.

So, this would be equal to x divided by y.

So, therefore, we are looking for angles in the interval zero to two pi,

where this ratio of x to y is equal to 1.

And, this will happen at pi over 4,

as well as 5 pi over 4,

because remember our unit circle,

at pi over 4,

the ratio of x to y here will be equal to 1,

but also down here in Quadrant three,

the ratio of x to y will also equal 1,

which corresponds to the angle of 5 pi divided by 4.

And don't forget this angle down here in Quadrant three,

most students will only remember the angle in Quadrant one.

But, the ratio of x to y in Quadrant three will also yield a positive number.

Therefore, our answer here is,

theta is equal to pi divided by 4 or 5 pi divided by 4.

And this is how we solve trigonometric equations on this interval zero to two pi.

Thank you. And, we'll see you next time.

Ознакомьтесь с нашим каталогом

Присоединяйтесь бесплатно и получайте персонализированные рекомендации, обновления и предложения.

Coursera делает лучшее в мире образование доступным каждому, предлагая онлайн-курсы от ведущих университетов и организаций.

© Coursera Inc., 2019 Все права защищены.

Загрузить из App Store

Загрузить в Google Play