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Chapter 10: Problem 141

In Exercises \(131-154\), find the exact value or state that it is undefined. $$ \tan \left(\arccos \left(-\frac{1}{2}\right)\right) $$

Short Answer

Expert verified
The exact value is \(-\sqrt{3}\).

Step by step solution

01

Understand the Problem

We need to find the exact value of \( \tan \left(\arccos \left(-\frac{1}{2}\right)\right) \). This involves finding the tangent of an angle whose cosine is given.
02

Analyze the Inverse Cosine

Recognize that \( \arccos \left(-\frac{1}{2}\right) \) represents an angle \( \theta \) such that \( \cos(\theta) = -\frac{1}{2} \). This is found in the second quadrant, where the angle corresponds to \( \theta = \frac{2\pi}{3} \).
03

Determine the Related Sine Value

Consider that for any angle \( \theta \), the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) holds. With \( \cos(\theta) = -\frac{1}{2} \), calculate \( \sin(\theta) \) as follows: \[ \sin^2(\theta) = 1 - \left(-\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4} \] Therefore, \( \sin(\theta) = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \) in the second quadrant, where sine is positive.
04

Find the Tangent Value

Using the identity for tangent, \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), substitute the known values: \[ \tan \left(\frac{2\pi}{3}\right) = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3} \]
05

Conclusion

We have determined that \( \tan \left(\arccos \left(-\frac{1}{2}\right)\right) = -\sqrt{3} \). The expression is well-defined since the input angle is valid and within the domain of arcsin.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Trigonometric Functions
Inverse trigonometric functions are crucial when working backwards from a trigonometric value to its corresponding angle. These functions allow us to determine the angle that results in a given trigonometric ratio.
  • Each trigonometric function, such as sine, cosine, and tangent, has an inverse: arcsin, arccos, and arctan, respectively.
  • The range of these functions is restricted to provide a unique output. For example, the range of \(\arccos(x)\) is from 0 to \(\pi\), covering angles from the first and second quadrants.
When solving problems like \( \tan(\arccos(-\frac{1}{2})) \), understanding this restricted range allows us to determine the specific angle (i.e., the second quadrant here). This helps deduce the sine and tangent of the angle, as needed for calculating exact trigonometric values.
Tangent Function
The tangent function, a fundamental trigonometric function, is defined as the ratio of the sine to the cosine of an angle.
  • Mathematically, this is expressed as \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).
  • The values of the tangent function can range from \(-\infty\) to \(+\infty\).
When evaluating \( \tan(\arccos(-\frac{1}{2})) \), we use the known values of sine and cosine for the angle derived from the arcsine function. Knowing \( \tan(\theta) \)'s formula helps us find the precise value of tangent once sine and cosine are determined.
For instance, with \( \cos(\theta) = -\frac{1}{2} \) and \( \sin(\theta) = \frac{\sqrt{3}}{2} \), tangent becomes \( \tan(\theta) = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3} \).
Unit Circle
The unit circle is an invaluable tool in trigonometry, providing a simple way to understand the relationships among different trigonometric functions and their corresponding angles.
  • It is a circle with a radius of 1, centered at the origin of a coordinate system.
  • Angles on the unit circle are measured in radians, and important points can easily be associated with angles of 0, \(\frac{\pi}{2}\), \(\pi\), \(\frac{3\pi}{2}\), and so on.
On this circle, the x-coordinate represents the cosine value, while the y-coordinate represents the sine value of an angle. For example, an angle in the second quadrant with \(\cos(\theta) = -\frac{1}{2}\) corresponds to the point \((-\frac{1}{2}, \frac{\sqrt{3}}{2})\) on the unit circle, reflecting the fact that \(\sin(\theta)\) for this angle is \(\frac{\sqrt{3}}{2}\). Working with the unit circle simplifies the process of finding and verifying trigonometric values.
Quadrants
In trigonometry, understanding the four quadrants of the coordinate system is essential for determining the signs and values of trigonometric functions.
  • The first quadrant includes angles from 0 to \(\frac{\pi}{2}\), where both sine and cosine are positive.
  • The second quadrant spans from \(\frac{\pi}{2}\) to \(\pi\), where sine is positive, but cosine is negative, as used in this exercise.
  • The third quadrant (\(\pi\) to \(\frac{3\pi}{2}\)) features both sine and cosine as negative.
  • In the fourth quadrant (\(\frac{3\pi}{2}\) to 2\(\pi\)), sine is negative while cosine is positive.
For \(\arccos(-\frac{1}{2})\), the angle lies in the second quadrant, indicating that the cosine is negative and sine is positive, ultimately affecting the sign of the tangent value as well. This knowledge is fundamental in applying trigonometric identities correctly.

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Most popular questions from this chapter

Discuss with your classmates why \(\arcsin \left(\frac{1}{2}\right) \neq 30^{\circ}\).

A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it touches level ground 360 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place.

In Exercises \(107-118\), assume that the range of arcsecant is \(\left[0, \frac{\pi}{2}\right) \cup\left[\pi, \frac{3 \pi}{2}\right)\) and that the range of arccosecant is \(\left(0, \frac{\pi}{2}\right] \cup\left(\pi, \frac{3 \pi}{2}\right]\) when finding the exact value. $$ \operatorname{arccsc}\left(\csc \left(-\frac{\pi}{2}\right)\right) $$

In Exercises \(165-184\), rewrite the quantity as algebraic expressions of \(x\) and state the domain on which the equivalence is valid. $$ \csc (\arccos (x)) $$

In Exercises \(165-184\), rewrite the quantity as algebraic expressions of \(x\) and state the domain on which the equivalence is valid. $$ \sin (\arccos (2 x)) $$
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