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Chapter 5: Problem 149

Suppose that \(-90^{\circ}<\theta<90^{\circ}\). Find the sign of each finction value. $$\sec (-\theta)$$

Short Answer

Expert verified
The sign of \(\text{sec}(-\theta)\) is positive.

Step by step solution

01

Recall the Range of Secant Function

The secant function, \(\text{sec}(\theta)\), is the reciprocal of the cosine function, \(\text{cos}(\theta)\). So, \(\text{sec}(\theta) = \frac{1}{\text{cos}(\theta)}\). We need to understand how the range of the cosine function affects the secant function within the provided domain for \(\theta\).
02

Identify the Quadrant for \(-\theta\)

Given \(-90^{\backslash\circ} < \theta < 90^{\backslash\circ}\), \(-\theta\) would also be in the same range: \(-90^{\backslash\circ} < -\theta < 90^{\backslash\circ}\). This ensures that \(-\theta\) is still in Quadrant I or Quadrant IV.
03

Determine the Sign of \(\text{cos}(-\theta)\)

In both Quadrant I and Quadrant IV, the cosine function is positive. Since \(\text{cos}(-\theta)\) is positive, \(\text{sec}(-\theta)\) being the reciprocal will also be positive.
04

Apply Even-Odd Identities

The secant function is an even function, meaning \(\text{sec}(-\theta) = \text{sec}(\theta)\). Therefore, the positivity derived in the previous step applies equally to \(\theta\).

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Key Concepts

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

Secant Function
The secant function, denoted as \(\text{sec}(\theta)\), is essentially the reciprocal of the cosine function \(\text{cos}(\theta)\). This means that \(\text{sec}(\theta) = \frac{1}{\text{cos}(\theta)}\). By understanding this reciprocal relationship, you get better insights into how secant behaves depending on the value of cosine. For instance, if the cosine of an angle is positive, then the secant of that angle will also be positive. Similarly, if cosine is negative, secant will follow suit and be negative too.
The behavior of secant is largely determined by the angles you鈥檙e examining and in which quadrant those angles lie. Since cosine is positive in the first and fourth quadrants, secant will also be positive in those quadrants.
This reciprocal property is a key feature when solving trigonometric problems involving the secant function.
Cosine Function
The cosine function, denoted as \(\text{cos}(\theta)\), is one of the fundamental trigonometric functions. It represents the ratio of the adjacent side to the hypotenuse in a right-angled triangle. This function takes on values between -1 and 1 for any angle \(\theta\). Knowing the sign and range of cosine values is crucial while dealing with trigonometric functions.
For angles between -90掳 and 90掳, the cosine function is always positive because these angles lie within the first and fourth quadrants. Therefore, for any angle \(-\theta \) within that range, \(\text{cos}(-\theta)\) will also be positive.
Even-Odd Identities
Even-odd identities are essential in trigonometry as they help simplify expressions and solve equations more efficiently. An even function satisfies the property \(\text{f}(-x) = \text{f}(x)\). The secant function is an even function because \(\text{sec}(-\theta) = \text{sec}(\theta)\).
This property implies that the secant of an angle is unaffected by the sign of that angle, making it easier to determine the values and signs of trigonometric expressions.
On the other hand, odd functions satisfy \(\text{f}(-x) = -\text{f}(x)\). Knowing whether a function is even or odd can greatly simplify the problem-solving process in trigonometry.

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Most popular questions from this chapter

Work each problem. In these exercises, assume the course of a plane or ship is on the indicated bearing. Two ships leave a port at the same time. The first ship sails on a bearing of \(52^{\circ}\) at 17 knots and the second on a bearing of \(322^{\circ}\) at 22 knots. How far apart are they after 2.5 hr?

Use identities to solve each of the following. Find \(\tan \theta,\) given that \(\sin \theta=0.49268329\) and \(\theta\) is in quadrant II.

When highway curves are designed, the outside of the curve is often slightly elevated or inclined above the inside of the curve. See the figure. This inclination is the superelevation. For safety reasons, it is important that both the curve's radius and superelevation be correct for a given speed limit. If an automobile is traveling at velocity \(V\) (in feet per second), the safe radius \(R\) for a curve with superelevation \(\theta\) is modeled by the formula $$R=\frac{V^{2}}{g(f+\tan \theta)}.$$ where \(f\) and \(g\) are constants. (Source: Mannering, \(\mathbf{F}\), and \(\mathbf{W}\). Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.) (a) A roadway is being designed for automobiles traveling at 45 mph. If \(\theta=3^{\circ}\) \(g=32.2,\) and \(f=0.14,\) calculate \(R\) to the nearest foot. (Hint: \(45 \mathrm{mph}=66 \mathrm{ft}\) per sec). (b) Determine the radius of the curve, to the nearest foot, if the speed in part (a) is increased to 70 mph. (c) How would increasing the angle \(\theta\) affect the results? Verify your answer by repeating parts (a) and (b) with \(\theta=4^{\circ}.\)

Find the five remaining trigonometric finction values for each angle. \(\csc \theta=2,\) and \(\theta\) is in quadrant II.

Use the trigonometric finction values of quadrantal angles given in this section to evaluate each expression. An expression such as cot' \(90^{\circ}\) means (cot \(90^{\circ}\) ) \(^{2}\), which is equal to \(0^{2}=0\). $$-2 \sin ^{4} 0^{\circ}+3 \tan ^{2} 0^{\circ}$$
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