Problem 5 - Precalculus Mathematics for Calculus

Question

Find the exact value of each expression, if it is defined. (a) \(\sin ^{-1}\left(-\frac{1}{2}\right)\) (b) \(\cos ^{-1} \frac{1}{2}\) (c) \(\tan ^{-1}\left(\frac{\sqrt{3}}{3}\right)\)

Step-by-Step Solution

Answer

(a) \( -\frac{\pi}{6} \); (b) \( \frac{\pi}{3} \); (c) \( \frac{\pi}{6} \).

Step 1: Understanding Inverse Sine Function

We need to find \( \sin^{-1}\left(-\frac{1}{2}\right) \). The inverse sine function, or arcsine, finds an angle whose sine is the given value. The range of \( \sin^{-1}(x) \) is \(-\frac{\pi}{2} \) to \( \frac{\pi}{2}\). For \( \sin(\theta) = -\frac{1}{2} \), \( \theta \) is in the fourth quadrant, so \( \theta = -\frac{\pi}{6}\).

Step 2: Calculating Inverse Cosine Function

For \( \cos^{-1}\left(\frac{1}{2}\right) \), we look for the angle whose cosine is \( \frac{1}{2} \). The range of \( \cos^{-1}(x) \) is from \( 0 \) to \( \pi \). The angle \( \cos(\theta) = \frac{1}{2} \) is \( \frac{\pi}{3} \).

Step 3: Evaluating Inverse Tangent Function

We need to find \( \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) \). The inverse tangent function, or arctangent, returns angles between \(-\frac{\pi}{2} \) and \( \frac{\pi}{2}\). The angle for \( \tan(\theta) = \frac{\sqrt{3}}{3} \) is \( \frac{\pi}{6} \).

Step 4: Summary of Results

We calculated each inverse trigonometric function separately, finding: \( \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \), \( \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \), and \( \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6} \).

Key Concepts

Understanding the Arcsine Function

The arcsine function is the inverse of the sine function. Its purpose is to find an angle whose sine is a given value. Specifically, the range of arcsine, denoted as \( \sin^{-1}(x) \), is from \(-\frac{\pi}{2} \) to \( \frac{\pi}{2}\). This means that the arcsine function will only return angles that lie within these bounds, which corresponds to angles in the first and fourth quadrants.
A practical example for arcsine is when you solve \( \sin^{-1}\left(-\frac{1}{2}\right) \). The goal here is to find an angle \( \theta \) such that \( \sin(\theta) = -\frac{1}{2} \).

• Within its range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), only \( \theta = -\frac{\pi}{6} \) has a sine value of \(-\frac{1}{2}\).
• This angle is in the fourth quadrant.

Thus, the exact value of \( \sin^{-1}\left(-\frac{1}{2}\right) \) is \( -\frac{\pi}{6} \). Understanding the restrictions on the range helps ensure that you identify the correct angle.

Exploring the Arccosine Function

The arccosine function serves as the inverse of the cosine function. It's written as \( \cos^{-1}(x) \), and it gives an angle with a specific cosine value. One key feature of arccosine is that its range stretches from \(0\) to \(\pi\). This range means we're looking at angles in the first and second quadrants.
When tasked with finding \( \cos^{-1}\left(\frac{1}{2}\right) \), we seek an angle \( \theta \) such that \( \cos(\theta) = \frac{1}{2} \).

• The appropriate angle within \([0, \pi]\) is \( \frac{\pi}{3} \).
• This angle resides in the first quadrant.

Here, the value of \( \cos^{-1}\left(\frac{1}{2}\right) \) comes out to be \( \frac{\pi}{3} \). Knowing the correct range of arccosine ensures we identify the right angle corresponding to the given cosine value.

Demystifying the Arctangent Function

The arctangent function, denoted by \( \tan^{-1}(x) \), is the inverse of the tangent function. It aims to find angles associated with a specific tangent value, with its range spanning from \(-\frac{\pi}{2} \) to \( \frac{\pi}{2} \). Because of this range, arctangent values include angles from the first and fourth quadrants.
To solve \( \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) \), the task is to discover an angle \( \theta \) where \( \tan(\theta) = \frac{\sqrt{3}}{3} \).

• The appropriate angle that fits this in its range is \( \frac{\pi}{6} \).
• This angle is part of the first quadrant.

Thus, the exact value of \( \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) \) is \( \frac{\pi}{6} \). Understanding the set range of arctangent allows us to find the correct solution with ease, highlighting the importance of knowing inverse trigonometric functions' definitions and ranges.