Problem 15 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

\( \sin \alpha = \frac{2\sqrt{5}}{5}, \cos \alpha = -\frac{\sqrt{5}}{5}, \tan \alpha = -2, \csc \alpha = \frac{\sqrt{5}}{2}, \cot \alpha = -\frac{1}{2} \).

Step 1: Define the Relationship

Given that \( \alpha = \sec^{-1}(-\sqrt{5}) \), it implies \( \sec \alpha = -\sqrt{5} \). The secant function is defined as the reciprocal of cosine, therefore \( \cos \alpha = \frac{1}{\sec \alpha} = \frac{1}{-\sqrt{5}} = -\frac{\sqrt{5}}{5} \).

Step 2: Find \( \sin \alpha \) using Pythagorean Identity

Use the Pythagorean identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \). Substitute \( \cos \alpha = -\frac{\sqrt{5}}{5} \) into the equation: \( \sin^2 \alpha + \left(-\frac{\sqrt{5}}{5}\right)^2 = 1 \). Simplify to get \( \sin^2 \alpha = 1 - \frac{5}{25} = \frac{20}{25} = \frac{4}{5} \), thus \( \sin \alpha = \pm\sqrt{\frac{4}{5}} = \pm\frac{2}{\sqrt{5}} = \pm\frac{2\sqrt{5}}{5} \). Since the secant is negative, the angle \( \alpha \) is in the second quadrant where sine is positive, so \( \sin \alpha = \frac{2\sqrt{5}}{5} \).

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

\( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{2\sqrt{5}}{5}}{-\frac{\sqrt{5}}{5}} = \frac{2\sqrt{5}}{5} \times \frac{5}{-\sqrt{5}} = -2 \).

Step 4: Determine \( \csc \alpha \)

Cosecant is the reciprocal of sine, hence \( \csc \alpha = \frac{1}{\sin \alpha} = \frac{1}{\frac{2\sqrt{5}}{5}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2} \).

Step 5: Determine \( \cot \alpha \)

Cotangent is the reciprocal of tangent, therefore \( \cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{-2} = -\frac{1}{2} \).

Key Concepts

Secant Function

The secant function, often denoted as \( \sec \alpha \), is closely related to the cosine function. In fact, it is the reciprocal of cosine. This means that if you know the cosine of an angle, you can find its secant by taking the reciprocal. For example, if \( \cos \alpha = x \), then \( \sec \alpha = \frac{1}{x} \).
In the given problem, the secant is \(-\sqrt{5} \). Thus, \( \sec \alpha = -\sqrt{5} \) implies \( \cos \alpha = \frac{1}{-\sqrt{5}} \). This simplification results in \( \cos \alpha = -\frac{\sqrt{5}}{5} \).
Understanding secant as the reciprocal of cosine helps in finding other trigonometric values of an angle, especially when other functions are involved. Keep in mind that like all trigonometric functions, the secant can have different signs depending on the quadrant of the angle.

Pythagorean Identity

The Pythagorean identity is a fundamental equation in trigonometry that relates sine and cosine functions. It states that \( \sin^2 \alpha + \cos^2 \alpha = 1 \). This identity is crucial for solving problems involving trigonometric functions.
In our particular problem, after determining \( \cos \alpha = -\frac{\sqrt{5}}{5} \), the Pythagorean identity helps in solving for \( \sin alpha \). By substituting the known cosine value into the identity, \( \sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left( -\frac{\sqrt{5}}{5} \right)^2 = \frac{4}{5} \).
By solving for \( \sin \alpha \), the Pythagorean identity ensures that all the trigonometric functions stay consistent with each other, which is key to correctly solving trigonometric equations.

Reciprocal Identities

Reciprocal identities are fundamental in trigonometry, allowing for the interconversion of trigonometric functions. They express one trigonometric function as the reciprocal of another.
For instance:

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

In the given problem, these identities help find the values of \( \csc \alpha \) and \( \cot \alpha \) once \( \sin \alpha \) and \( \tan \alpha \) are known:

• Since \( \sin \alpha = \frac{2\sqrt{5}}{5} \), \( \csc \alpha = \frac{1}{\frac{2\sqrt{5}}{5}} = \frac{\sqrt{5}}{2} \).
• Given \( \tan \alpha = -2 \), \( \cot \alpha = \frac{1}{-2} = -\frac{1}{2} \).

This ensures that all the trigonometric functions are consistent within their reciprocal relationships.

Trigonometric Functions in Quadrants

Trigonometric functions change their signs and values depending on the quadrant in which an angle lies. Knowing which quadrant an angle belongs to helps determine the correct sign for the function values.
In this exercise, \( \alpha = \sec^{-1}(-\sqrt{5}) \) tells us that \( \sec \alpha \) is negative. Since secant is the reciprocal of cosine, \( \cos \alpha \) is negative as well.
This information implies \( \alpha \) is in the second quadrant. This is because in the second quadrant, cosine is negative and sine is positive. Therefore, \( \sin \alpha = \frac{2\sqrt{5}}{5} \) is positive and consistent with this quadrant's properties.
Understanding the behavior of trigonometric functions in different quadrants illuminates how these functions take both sign and value: in the first quadrant, all are positive; in the second quadrant, sine is positive; in the third, only tangent is positive; and in the fourth, cosine is positive. This helps in accurately deriving all associated function values for an angle.