Suppose $\sin X = -\frac{18}{82}$ And $x$ Is An Angle In Quadrant 3. Find:$\cos X = $ $\tan X = $

Feb 25, 2025 by ADMIN 101 views

**Solving Trigonometric Equations in Quadrant 3**

Introduction

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. In this article, we will focus on solving trigonometric equations in Quadrant 3, where the angle $x$ is located in the third quadrant of the unit circle. We will use the given information that $\sin x = -\frac{18}{82}$ and $x$ is an angle in Quadrant 3 to find the values of $\cos x$ and $\tan x$ .

Understanding Quadrant 3

Quadrant 3 is the third quadrant of the unit circle, where both the $x$ -coordinate and the $y$ -coordinate are negative. This means that the angle $x$ is measured counterclockwise from the positive $x$ -axis and lies between $180^\circ$ and $270^\circ$ . In this quadrant, the sine function is negative, and the cosine function is also negative.

Recall of Trigonometric Identities

Before we proceed, let's recall some important trigonometric identities that we will use to solve the problem:

$\sin^2 x + \cos^2 x = 1$
$\tan x = \frac{\sin x}{\cos x}$

Finding $\cos x$

We are given that $\sin x = -\frac{18}{82}$ . To find $\cos x$ , we can use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ . Since $\sin x$ is negative in Quadrant 3, we can write:

\cos^2 x = 1 - \sin^2 x = 1 - \left(-\frac{18}{82}\right)^2

Simplifying the expression, we get:

\cos^2 x = 1 - \frac{324}{6724} = \frac{6400}{6724}

Taking the square root of both sides, we get:

\cos x = \pm \sqrt{\frac{6400}{6724}}

Since $\cos x$ is negative in Quadrant 3, we take the negative square root:

\cos x = -\sqrt{\frac{6400}{6724}} = -\frac{80}{82}

Finding $\tan x$

Now that we have found $\cos x$ , we can find $\tan x$ using the identity $\tan x = \frac{\sin x}{\cos x}$ . Plugging in the values, we get:

\tan x = \frac{-\frac{18}{82}}{-\frac{80}{82}} = \frac{18}{80} = \frac{9}{40}

Conclusion

In this article, we have used the given information that $\sin x = -\frac{18}{82}$ and $x$ is an angle in Quadrant 3 to find the values of $\cos x$ and $\tan x$ . We have used the Pythagorean identity and the definition of the tangent function to solve the problem. The final answers are:

$\cos x = -\frac{80}{82}$
$\tan x = \frac{9}{40}$

Additional Examples

Here are a few more examples of solving trigonometric equations in Quadrant 3:

Suppose $\sin x = -\frac{15}{77}$ and $x$ is an angle in Quadrant 3. Find $\cos x$ and $\tan x$ .
Suppose $\sin x = -\frac{22}{85}$ and $x$ is an angle in Quadrant 3. Find $\cos x$ and $\tan x$ .

Solutions to Additional Examples

Suppose $\sin x = -\frac{15}{77}$ $sin x = - \frac{15}{77}$ and $x$ $x$ is an angle in Quadrant 3. Find $\cos x$ $cos x$ and $\tan x$ $tan x$ .
- Using the Pythagorean identity, we get:
  - $\cos^2 x = 1 - \sin^2 x = 1 - \left(-\frac{15}{77}\right)^2$
  - Simplifying the expression, we get:
    - $\cos^2 x = 1 - \frac{225}{5929} = \frac{5704}{5929}$
  - Taking the square root of both sides, we get:
    - $\cos x = \pm \sqrt{\frac{5704}{5929}}$
  - Since $\cos x$ $cos x$ is negative in Quadrant 3, we take the negative square root:
    - $\cos x = -\sqrt{\frac{5704}{5929}} = -\frac{76}{77}$
  - Using the definition of the tangent function, we get:
    - $\tan x = \frac{-\frac{15}{77}}{-\frac{76}{77}} = \frac{15}{76}$
- Therefore, the final answers are:
  - $\cos x = -\frac{76}{77}$
  - $\tan x = \frac{15}{76}$
Suppose $\sin x = -\frac{22}{85}$ $sin x = - \frac{22}{85}$ and $x$ $x$ is an angle in Quadrant 3. Find $\cos x$ $cos x$ and $\tan x$ $tan x$ .
- Using the Pythagorean identity, we get:
  - $\cos^2 x = 1 - \sin^2 x = 1 - \left(-\frac{22}{85}\right)^2$
  - Simplifying the expression, we get:
    - $\cos^2 x = 1 - \frac{484}{7225} = \frac{6741}{7225}$
  - Taking the square root of both sides, we get:
    - $\cos x = \pm \sqrt{\frac{6741}{7225}}$
  - Since $\cos x$ $cos x$ is negative in Quadrant 3, we take the negative square root:
    - $\cos x = -\sqrt{\frac{6741}{7225}} = -\frac{81}{85}$
  - Using the definition of the tangent function, we get:
    - $\tan x = \frac{-\frac{22}{85}}{-\frac{81}{85}} = \frac{22}{81}$
- Therefore, the final answers are:
  - $\cos x = -\frac{81}{85}$
  - $\tan x = \frac{22}{81}$
    Frequently Asked Questions (FAQs) =====================================

Q: What is Quadrant 3 in the unit circle?

A: Quadrant 3 is the third quadrant of the unit circle, where both the $x$ -coordinate and the $y$ -coordinate are negative. This means that the angle $x$ is measured counterclockwise from the positive $x$ -axis and lies between $180^\circ$ and $270^\circ$ .

Q: What is the value of $\sin x$ in Quadrant 3?

A: In Quadrant 3, the value of $\sin x$ is negative.

Q: How do I find $\cos x$ in Quadrant 3?

A: To find $\cos x$ in Quadrant 3, you can use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ . Since $\sin x$ is negative in Quadrant 3, you can write:

\cos^2 x = 1 - \sin^2 x = 1 - \left(-\frac{18}{82}\right)^2

Simplifying the expression, you get:

\cos^2 x = 1 - \frac{324}{6724} = \frac{6400}{6724}

Taking the square root of both sides, you get:

\cos x = \pm \sqrt{\frac{6400}{6724}}

Since $\cos x$ is negative in Quadrant 3, you take the negative square root:

\cos x = -\sqrt{\frac{6400}{6724}} = -\frac{80}{82}

Q: How do I find $\tan x$ in Quadrant 3?

A: To find $\tan x$ in Quadrant 3, you can use the definition of the tangent function:

\tan x = \frac{\sin x}{\cos x}

Plugging in the values, you get:

\tan x = \frac{-\frac{18}{82}}{-\frac{80}{82}} = \frac{18}{80} = \frac{9}{40}

Q: What are some common mistakes to avoid when solving trigonometric equations in Quadrant 3?

A: Some common mistakes to avoid when solving trigonometric equations in Quadrant 3 include:

Not considering the sign of the trigonometric function in Quadrant 3
Not using the correct Pythagorean identity
Not taking the square root of both sides of the equation
Not considering the quadrant in which the angle lies

Q: How can I practice solving trigonometric equations in Quadrant 3?

A: You can practice solving trigonometric equations in Quadrant 3 by:

Using online resources such as Khan Academy or Mathway
Working with a tutor or teacher
Practicing with sample problems and exercises
Using a calculator to check your answers

Q: What are some real-world applications of trigonometric equations in Quadrant 3?

A: Some real-world applications of trigonometric equations in Quadrant 3 include:

Navigation and mapping
Physics and engineering
Computer graphics and animation
Medical imaging and diagnostics

Q: Can I use trigonometric equations in Quadrant 3 to solve problems in other areas of mathematics?

A: Yes, you can use trigonometric equations in Quadrant 3 to solve problems in other areas of mathematics, such as:

Algebra and geometry
Calculus and differential equations
Statistics and probability

Q: How can I use trigonometric equations in Quadrant 3 to solve problems in science and engineering?

A: You can use trigonometric equations in Quadrant 3 to solve problems in science and engineering by:

Using trigonometric functions to model real-world phenomena
Solving trigonometric equations to find unknown values
Using trigonometric identities to simplify complex equations

Q: Can I use trigonometric equations in Quadrant 3 to solve problems in computer science and programming?

A: Yes, you can use trigonometric equations in Quadrant 3 to solve problems in computer science and programming by:

Using trigonometric functions to model real-world phenomena
Solving trigonometric equations to find unknown values
Using trigonometric identities to simplify complex equations