Problem 6 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

(a) \(\frac{\pi}{4}\), (b) 0, (c) \(\frac{3\pi}{4}\)

Step 1: Understanding the Inverse Cosine Function

The inverse cosine function, \(\cos^{-1}(x)\), gives the angle whose cosine is \x\. The range of \cos^{-1}\ is from \[0, \pi\]. This means that the output angle will always be between \0\ and \pi\.

Step 2(a): Evaluate \( \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) \)

We know that \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\. Thus, \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}\. This angle is within the range of \cos^{-1}\ which is \[0, \pi\].

Step 3(b): Evaluate \( \cos^{-1} 1 \)

We know that \cos(0) = 1\. Therefore, \cos^{-1}(1) = 0\. This is because 0 is clearly within the range \[0, \pi\].

Step 4(c): Evaluate \( \cos^{-1}\left(-\frac{\sqrt{2}}{2}\right) \)

We know that \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}\. Therefore, \cos^{-1}\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}\. This angle is also within the range \[0, \pi\].

Key Concepts

Cosine Function

The cosine function is a fundamental concept in trigonometry. It relates the angles and sides of a right-angled triangle. Specifically, for an angle in a right-angled triangle, the cosine is the ratio of the length of the adjacent side to the hypotenuse.
In mathematical terms, for a given angle \( \theta \), the cosine function is defined as:

• \( \cos(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse}} \)

The cosine function is periodic, with a period of \(2\pi\) radians. This means it repeats its values every \(2\pi\) radians. Its values range from -1 to 1.
The graph of the cosine function is a wave that starts at 1 when \(\theta = 0\), decreases to -1, and rises back to 1. Understanding this wave-like pattern helps in solving trigonometric problems, including those involving inverse functions.

Range of Inverse Functions

Inverse functions essentially reverse the roles of inputs and outputs in the original function. For trigonometric functions like cosine, the inverse tries to find the angle that corresponds to a given cosine value.
The range of inverse functions is crucial when solving expressions involving these reverse operations. For the inverse cosine function, \(\cos^{-1}(x)\), it means specifying the valid angle outputs for given input values between -1 and 1.

• The range of \(\cos^{-1}(x)\) is \([0, \pi]\) radians.
• This means any valid angle as an output will be between 0 and \(\pi\), inclusive.

Choosing this range ensures a unique angle is paired with each cosine value, leading to consistent and reliable solutions. This characteristic is essential when determining specific angles corresponding with given cosine outputs in various quadrants of a unit circle.

Arc Cosine

Arc cosine, denoted as \(\cos^{-1}(x)\), is the inverse of the cosine function. It gives the angle whose cosine value equals \(x\). Solving expressions like \(\cos^{-1}(\frac{\sqrt{2}}{2})\) relies on knowing standard angles and their cosine values.
The process involves:

• Identifying an angle \(\theta\) such that \(\cos(\theta) = x\).
• Ensuring the angle \(\theta\) falls within the range of \([0, \pi]\).

For example, recognizing that \(\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\) allows us to determine that \(\cos^{-1}(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}\). This process is similarly applied to other specific reference angles like \(\cos^{-1}(1) = 0\) and \(\cos^{-1}(-\frac{\sqrt{2}}{2}) = \frac{3\pi}{4}\).
Being comfortable with the arc cosine function enables solving various trigonometric equations and understanding their practical applications.