Problem 14 - Thomas Calculus

Question

Given that \(\alpha=\tan ^{-1}(4 / 3),\) find \(\sin \alpha, \cos \alpha, \sec \alpha, \csc \alpha,\) and \(\cot \alpha .\)

Step-by-Step Solution

Answer

\( \sin \alpha = \frac{4}{5}, \cos \alpha = \frac{3}{5}, \sec \alpha = \frac{5}{3}, \csc \alpha = \frac{5}{4}, \cot \alpha = \frac{3}{4} \).

Step 1: Understand the Given Information

We are given that \( \alpha = \tan^{-1}(4/3) \). This implies that \( \tan \alpha = 4/3 \). To find other trigonometric functions for \( \alpha \), we'll use the identity \( \tan \alpha = \frac{\text{opposite side}}{\text{adjacent side}} \). Consider a right triangle with opposite side 4 and adjacent side 3.

Step 2: Calculate the Hypotenuse

Using the Pythagorean Theorem: \[ hypotenuse = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]. So the hypotenuse is 5.

Step 3: Find \( \sin \alpha \)

The sine function is defined as \( \sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} \). Thus, \( \sin \alpha = \frac{4}{5} \).

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

The cosine function is defined as \( \cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}} \). Thus, \( \cos \alpha = \frac{3}{5} \).

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

\( \sec \alpha \) is the reciprocal of \( \cos \alpha \). Therefore, \( \sec \alpha = \frac{5}{3} \).

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

\( \csc \alpha \) is the reciprocal of \( \sin \alpha \). Hence, \( \csc \alpha = \frac{5}{4} \).

Step 7: Find \( \cot \alpha \)

\( \cot \alpha \) is the reciprocal of \( \tan \alpha \). Thus, \( \cot \alpha = \frac{3}{4} \).

Key Concepts

Inverse Trigonometric Functions

The concept of inverse trigonometric functions is crucial in solving for angles when the value of a trigonometric ratio is known. In this exercise, we are given that \( \alpha = \tan^{-1}(4/3) \), which implies that \( \tan \alpha = 4/3 \). Here, \( \tan^{-1} \) is the inverse tangent function, which helps us find the angle whose tangent is \( 4/3 \).
Inverse trigonometric functions include \( \sin^{-1}, \cos^{-1}, \tan^{-1}, \csc^{-1}, \sec^{-1}, \) and \( \cot^{-1} \). They are the reverse operations of the trigonometric functions and allow you to find angle measures from known ratio values.
In this case, the inverse tangent helps you determine the angle that leads to the specified trigonometric identity, forming the backbone for further calculations that link to other trigonometric functions.

Pythagorean Theorem

The Pythagorean Theorem is essential in solving problems involving right triangles, particularly when certain side lengths are known. It states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Mathematically, it is expressed as \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse while \( a \) and \( b \) are the other two sides.
In our exercise, after establishing a right triangle with sides 3 and 4 (as derived from the tangent), the Pythagorean Theorem was employed to calculate the hypotenuse, yielding \( \sqrt{3^2 + 4^2} = 5 \).

• This calculated hypotenuse allows us to determine other trigonometric ratios related to the triangle, demonstrating the interconnectedness of trigonometry concepts.

Sine and Cosine

The sine and cosine functions are among the most fundamental trigonometric functions, representing ratios of sides in a right triangle.
In the context of this problem, with the calculated sides of 4 (opposite), 3 (adjacent), and 5 (hypotenuse), these functions can be found as follows:

• Sine Function (\( \sin \alpha \)): Defined as the ratio of the length of the opposite side to the hypotenuse, calculated here as \( \sin \alpha = \frac{4}{5} \).
• Cosine Function (\( \cos \alpha \)): Defined as the ratio of the length of the adjacent side to the hypotenuse, calculated here as \( \cos \alpha = \frac{3}{5} \).

Each of these functional values provides a distinct perspective of the triangle's geometry and is rooted in the foundational setup of the triangle’s relationship in trigonometry.

Reciprocal Trigonometric Functions

Reciprocal trigonometric functions are derived from the principal trigonometric functions. They are essential in broadening the understanding of trigonometry by providing alternative perspectives.
In this exercise, we consider:

• Secant (\( \sec \alpha \)): The reciprocal of cosine, thus \( \sec \alpha = \frac{1}{\cos \alpha} = \frac{5}{3} \).
• Cosecant (\( \csc \alpha \)): The reciprocal of sine, thus \( \csc \alpha = \frac{1}{\sin \alpha} = \frac{5}{4} \).
• Cotangent (\( \cot \alpha \)): The reciprocal of tangent, thus \( \cot \alpha = \frac{1}{\tan \alpha} = \frac{3}{4} \).

Understanding these functions helps develop a comprehensive grasp of trigonometry. They enhance flexibility in approaching problems and often simplify complex trigonometric expressions.