Problem 11 - Calculus Single Variable

Question

Calculate the value of the given inverse trigonometric function at the given point. $$ \operatorname{arcsec}(\sqrt{2}) $$

Step-by-Step Solution

Answer

\(\text{arcsec}(\sqrt{2}) = \frac{\pi}{4}\).

Step 1: Understanding the Problem

We need to calculate the value of the inverse secant function, \( ext{arcsec}(\sqrt{2})\). This function gives us an angle whose secant is \(\sqrt{2}\). Our goal is to find that angle.

Step 2: Recalling Definitions and Relevant Range

The secant function, \( ext{sec}(\theta) = \frac{1}{\cos(\theta)}\), is defined for angles where \(|\cos(\theta)| \le 1\). For \(\text{arcsec}(x)\), \(x\) must be \(|x| \ge 1\). The principal value of \(\text{arcsec}(x)\) lies in the interval \([0, \pi)\) excluding \(\frac{\pi}{2}\).

Step 3: Calculating the Angle

We know \(\text{sec}(\theta) = \sqrt{2}\), which implies \(\frac{1}{\cos(\theta)} = \sqrt{2}\). Solving \(\cos(\theta) = \frac{1}{\sqrt{2}}\), we recognize that \(\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\). Thus, \(\theta = \frac{\pi}{4}\) is in the principal range.

Step 4: Confirming the Solution

The angle \(\theta = \frac{\pi}{4}\) satisfies the equation \(\text{sec}(\theta) = \sqrt{2}\) and falls within the range \([0, \pi)\). Therefore, \(\text{arcsec}(\sqrt{2}) = \frac{\pi}{4}\).

Key Concepts

Arcsecant

In the world of trigonometry, the arcsecant (often written as \( \text{arcsec} \)) is the inverse function of secant. The secant function itself is represented as \( \text{sec}(\theta) = \frac{1}{\cos(\theta)} \). In simpler terms, arcsecant helps us find the angle \( \theta \) whose secant is a given number. For example, to find \( \text{arcsec}(\sqrt{2}) \), we are looking for an angle for which \( \text{sec}(\theta) = \sqrt{2} \). This can seem complicated at first, but understanding it means remembering some basic function reversal.

It is very important to note that for the arcsecant, the value of \( x \) must be \(|x| \ge 1\). This means that when finding the arcsec of a number, make sure that the number is either greater than or equal to 1, or less than or equal to -1. That helps ensure the function is properly defined.

Trigonometric Identities

To solve problems involving arcsecant, a good grasp of trigonometric identities is critically helpful. The core identity to remember here is \( \text{sec}(\theta) = \frac{1}{\cos(\theta)} \). This simple identity is the foundation for finding inverse values like arcsecant. When you see \( \text{arcsec}(\sqrt{2}) \), it's essentially asking: "For which angle is \( \sec(\theta) = \sqrt{2} \) true?"

Another key identity involved is recognizing that \( \cos(\theta) = \frac{1}{\sqrt{2}} \) for special angles like \( \theta = \frac{\pi}{4} \). Recognizing these special angles and their cosine values is a great aid when working with secant and arcsecant.

• Remember \( \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \)
• Hence, \( \sec(\frac{\pi}{4}) = \sqrt{2} \)

These trigonometric identities simplify the evaluation of inverse trig functions significantly.

Principal Value Range

The principal value range is an essential concept when dealing with inverse trigonometric functions, including arcsecant. The principal value defines the specific outputs that the inverse function can choose from. For arcsecant, the range is typically \([0, \pi)\), excluding \(\frac{\pi}{2}\). This range specifies that when we're calculating \( \text{arcsec}(x) \), we should look for angles \( \theta \) such that \( \theta \) falls into this interval.

In the context of the problem \( \text{arcsec}(\sqrt{2}) \), the calculated angle \( \theta = \frac{\pi}{4} \) lies within the acceptable principal range \([0, \pi)\), making it a valid answer. Understanding this range helps avoid errors, ensuring that the result adheres to the expected behavior of the inverse trigonometric functions.

• Arcsecant produces angles in \([0, \pi)\), while \(\cos(\theta) eq 0\).
• Checking the principal value range is a crucial step in verifying your final answer.