Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 25-28, find each value without using a calculator (see Example 4). $$ \cos \left[\cos ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{12}{13}\right)\right] $$

Step-by-Step Solution

Answer

The value is \( \frac{-16}{65} \).

Step 1: Understanding the Problem

The expression to simplify is \( \cos \left[\cos ^{-1}\left(\frac{4}{5}\right)+\sin^{-1}\left(\frac{12}{13}\right)\right] \). We need to determine the cosine of the sum of two inverse trigonometric expressions.

Step 2: Define the Angles

Let \( \alpha = \cos^{-1}\left(\frac{4}{5}\right) \) and \( \beta = \sin^{-1}\left(\frac{12}{13}\right) \). This means \( \cos \alpha = \frac{4}{5} \) and \( \sin \beta = \frac{12}{13} \). We need to find \( \cos(\alpha + \beta) \).

Step 3: Recall the Cosine Addition Formula

The cosine of a sum, \( \cos(\alpha + \beta) \), can be calculated using the formula: \[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \].

Step 4: Find \( \cos \beta \) and \( \sin \alpha \)

Since \( \sin \beta = \frac{12}{13} \), we use the Pythagorean Identity to find \( \cos \beta \): \( \cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13} \). Likewise, for \( \cos \alpha = \frac{4}{5} \), find \( \sin \alpha \): \( \sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5} \).

Step 5: Apply the Values to the Formula

Substitute the values into the cosine addition formula: \[ \cos(\alpha + \beta) = \left(\frac{4}{5}\right)\left(\frac{5}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{12}{13}\right) \].

Step 6: Simplify the Expression

Calculate the expression: \[ \cos(\alpha + \beta) = \frac{4 \cdot 5}{5 \cdot 13} - \frac{3 \cdot 12}{5 \cdot 13} = \frac{20}{65} - \frac{36}{65} = \frac{-16}{65} \]. Thus, \( \cos \left(\alpha + \beta \right) = \frac{-16}{65} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse what the standard trigonometric functions (like sine, cosine, and tangent) do. They take a ratio and give you an angle, rather than taking an angle and giving you a ratio.

• The inverse cosine, or \( \cos^{-1} \), gives you an angle \( \alpha \) whose cosine value is a specific ratio. So, \( \cos^{-1} \left(\frac{4}{5}\right) \) provides an angle such that \( \cos \alpha = \frac{4}{5} \).
• The inverse sine, indicated by \( \sin^{-1} \), works similarly. \( \sin^{-1} \left(\frac{12}{13}\right) \) gives you an angle \( \beta \) for which \( \sin \beta = \frac{12}{13} \).

These functions are vital in calculus and geometry because they allow us to work backwards from known ratios, helping to solve complex trigonometric expressions. They are crucial in many real-world applications where angles need to be determined from measured lengths or heights.

Cosine Addition Formula

The cosine addition formula allows us to find the cosine of the sum of two angles. It is expressed as:
\[\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\]This formula is indispensable in trigonometry for simplifying expressions involving the sum of angles.

• When given the angles \( \alpha \) and \( \beta \), we first find their individual cosine and sine values.
• The formula then combines these values to determine the overall cosine of their sum.

In this problem, using the identities for each angle helps us efficiently apply this formula. Understanding this formula is crucial because it links basic trigonometric identities with more complex calculations, playing an essential role in advanced problem-solving in fields like physics, engineering, and computer science.

Pythagorean Identity

The Pythagorean identity is a foundational concept in trigonometry connecting the squares of sine and cosine for an angle. It states that:
\[\sin^2 \theta + \cos^2 \theta = 1\]This identity is incredibly useful for finding the third trigonometric function when one is known. For example:

• If \( \sin \beta = \frac{12}{13} \), we can find \( \cos \beta \) using \( \cos^2 \beta = 1 - \sin^2 \beta \).
• Applying the square root, we calculate \( \cos \beta = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13} \).
• Similarly, for \( \cos \alpha = \frac{4}{5} \), use \( \sin^2 \alpha = 1 - \cos^2 \alpha \) to find \( \sin \alpha \).

By using the Pythagorean identity, we can complete the set of trigonometric functions needed to use more advanced formulas such as the cosine addition formula. This shows its importance not only in theoretical mathematics but also in practical calculations and problem-solving. It helps bridge the relationships between these trigonometric functions efficiently.