Problem 70 - A Graphical Approach to Precalculus with Limits

Question

Give the exact real number value of each expression. Do not use a calculator. $$\cos ^{-1}\left(\cos \left(-\frac{\pi}{6}\right)\right)$$

Step-by-Step Solution

Answer

The exact value is \( \frac{\pi}{6} \).

Step 1: Understanding the Problem

The problem involves finding the inverse cosine, \( \cos^{-1} \), of a cosine \( \cos \) value. Specifically, \( \cos^{-1}(\cos(-\frac{\pi}{6})) \) needs to be evaluated. It is important to note that the inverse cosine function, \( \cos^{-1} \), returns the angle whose cosine is a given number, with the principal range of \( [0, \pi] \).

Step 2: Determine the Cosine Value

First, we find the value of \( \cos(-\frac{\pi}{6}) \). Remember that the cosine function is even, meaning \( \cos(-x) = \cos(x) \). Therefore, \( \cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \).

Step 3: Apply the Inverse Cosine Function

Now apply the inverse cosine function: \( \cos^{-1}(\cos(-\frac{\pi}{6})) = \cos^{-1}(\frac{\sqrt{3}}{2}) \). Since \( \cos^{-1} \) returns angles in the principal range \( [0, \pi] \), we need to find an angle \( x \) such that \( 0 \leq x \leq \pi \) and \( \cos(x) = \frac{\sqrt{3}}{2} \).

Step 4: Identify the Correct Angle

While \(-\frac{\pi}{6} \) does not lie in \([0, \pi] \), its equivalent angle in this range is \( \frac{\pi}{6} \) because both have the same cosine value. Therefore, \( \cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{6} \).

Key Concepts

Understanding Trigonometric Functions

Trigonometric functions are fundamental in mathematics, particularly in geometry and calculus. They include sine, cosine, and tangent, among others. These functions relate angles to ratios of sides in right-angled triangles, providing a way to describe circular motion and oscillations.
- **Sine ( \( \sin \))**: Represents the ratio of the length of the opposite side to the hypotenuse. - **Cosine ( \( \cos \))**: Represents the ratio of the adjoining side to the hypotenuse. - **Tangent ( \( \tan \))**: Represents the ratio of the opposite side to the adjoining side. The cosine function is particularly important here as it connects directly to the concept of inverse cosine. In this problem, we are manipulating the cosine function, both directly and through its inverse, to evaluate an expression.

Exploring the Principal Range

The principal range of a trigonometric function is crucial when dealing with inverse functions. For the cosine function, the principal range is \([0, \pi]\). This means that when we take the inverse cosine, denoted as \( \cos^{-1} \), it returns an angle within this specific range.
The reason for this restriction is to ensure the function remains consistent and gives a single, unique output for every input. Without the principal range, inverse trigonometric functions like \(\cos^{-1}\) could yield multiple results, creating ambiguities.
Thus, even if we start with an angle outside this range, such as the negative angle in our problem, the solution is adjusted to fall within \([0, \pi]\). This is important for arriving at a consistent and correct answer.

The Nature of Even Functions

An even function is defined based on its symmetry. Mathematically, a function \(f(x)\) is considered even if it satisfies the condition \(f(-x) = f(x)\) for all values in its domain.
The cosine function is an excellent example of an even function. This means that for any angle \(x\), \(\cos(-x) = \cos(x)\). This property significantly simplifies problems involving cosine, like our original exercise.
In our case, knowing that \(\cos(-\frac{\pi}{6})\) is the same as \(\cos(\frac{\pi}{6})\), allows us to confidently substitute and solve for the inverse cosine, even if the original angle was negative. This symmetry about the y-axis used by even functions ensures calculations are straightforward, making solutions more accessible and consistent.