Problem 13 - Thomas Calculus

Question

Given that \(\alpha=\sin ^{-1}(5 / 13),\) find \(\cos \alpha, \tan \alpha, \sec \alpha, \csc \alpha\) and \(\cot \alpha .\)

Step-by-Step Solution

Answer

\(\cos \alpha = \frac{12}{13}, \tan \alpha = \frac{5}{12}, \sec \alpha = \frac{13}{12}, \csc \alpha = \frac{13}{5}, \cot \alpha = \frac{12}{5}.\)

Step 1: Understanding the Given Information

We are given that \( \alpha = \sin^{-1} \left( \frac{5}{13} \right) \). This means that \( \sin \alpha = \frac{5}{13} \). Our task is to find the other trigonometric functions: \( \cos \alpha, \tan \alpha, \sec \alpha, \csc \alpha, \text{and } \cot \alpha \).

Step 2: Use the Pythagorean Identity to Find \(\cos \alpha\)

From the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \), substitute \( \sin \alpha = \frac{5}{13} \):\[ \left( \frac{5}{13} \right)^2 + \cos^2 \alpha = 1 \]Calculate the square of \( \sin \alpha \):\[ \frac{25}{169} + \cos^2 \alpha = 1 \]Rearrange the equation to solve for \( \cos^2 \alpha \):\[ \cos^2 \alpha = 1 - \frac{25}{169} = \frac{144}{169} \]Take the square root to find \( \cos \alpha \):\[ \cos \alpha = \pm \frac{12}{13} \]Given the range of the inverse sine function \([-\frac{\pi}{2}, \frac{\pi}{2}]\), \( \cos \alpha \) is positive, so \( \cos \alpha = \frac{12}{13} \).

Step 3: Calculate \(\tan \alpha\)

The tangent of \( \alpha \) can be found using \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \).Substitute the known values:\[ \tan \alpha = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \]

Step 4: Find \(\sec \alpha\)

The secant is the reciprocal of the cosine:\[ \sec \alpha = \frac{1}{\cos \alpha} = \frac{1}{\frac{12}{13}} = \frac{13}{12} \]

Step 5: Find \(\csc \alpha\)

The cosecant is the reciprocal of the sine:\[ \csc \alpha = \frac{1}{\sin \alpha} = \frac{1}{\frac{5}{13}} = \frac{13}{5} \]

Step 6: Find \(\cot \alpha\)

The cotangent is the reciprocal of the tangent:\[ \cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{\frac{5}{12}} = \frac{12}{5} \]

Key Concepts

Inverse Sine Function

The inverse sine function, often denoted as \( \sin^{-1} \) or arcsin, is a crucial concept in trigonometry. It helps us find the angle when the sine value is known. For a value \( y \) where \( -1 \leq y \leq 1 \), the inverse sine function returns an angle \( \alpha \) such that \( \sin \alpha = y \). This angle is always within the range of \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), ensuring that only one principal value is assigned to every sine value.

In our exercise, we know \( \alpha = \sin^{-1}\left(\frac{5}{13}\right) \). This means that the sine of angle \( \alpha \) is \( \frac{5}{13} \). Understanding the inverse sine is key to finding other trigonometric values associated with \( \alpha \) by leveraging the known sine value.

Pythagorean Identity

The Pythagorean identity is a core trigonometric principle that relates the squares of sine and cosine. Specifically, it states \( \sin^2 \alpha + \cos^2 \alpha = 1 \). This identity serves as a building block for finding unknown trigonometric functions when at least one is known.

In our problem, we start with \( \sin \alpha = \frac{5}{13} \). We substitute this into the identity: \( \left(\frac{5}{13}\right)^2 + \cos^2 \alpha = 1 \). By solving for \( \cos^2 \alpha \), we find \( \cos^2 \alpha = \frac{144}{169} \) and then determine \( \cos \alpha \). The positive value \( \frac{12}{13} \) is chosen, adhering to the range permitted by the inverse sine function.

Reciprocal Trigonometric Functions

Reciprocal trigonometric functions are crucial in fully expressing the relationships between different trigonometric ratios. They include secant (sec), cosecant (csc), and cotangent (cot). Each of these is the reciprocal of a primary trigonometric function:

• Secant is the reciprocal of cosine: \( \sec \alpha = \frac{1}{\cos \alpha} \).
• Cosecant is the reciprocal of sine: \( \csc \alpha = \frac{1}{\sin \alpha} \).
• Cotangent is the reciprocal of tangent: \( \cot \alpha = \frac{1}{\tan \alpha} \).

For example, knowing \( \cos \alpha = \frac{12}{13} \), we can determine \( \sec \alpha = \frac{13}{12} \). Similarly, with \( \sin \alpha = \frac{5}{13} \), we find \( \csc \alpha = \frac{13}{5} \). Understanding these reciprocals helps in computing trigonometric functions quickly.

Trigonometric Ratios

Trigonometric ratios are foundational in understanding angles and lengths in right triangles. They consist of sine, cosine, and tangent:

• Sine \( \sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} \).
• Cosine \( \cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}} \).
• Tangent \( \tan \alpha = \frac{\text{opposite}}{\text{adjacent}} \).

These ratios help interpret the relationship between the sides of a right triangle and an angle inside it. Using \( \sin \alpha = \frac{5}{13} \) as the opposite over the hypotenuse, with a known cosine, the tangent becomes \( \tan \alpha = \frac{5}{12} \). Familiarity with these ratios allows for quick calculations of related trigonometric functions.