Problem 16 - Thomas Calculus

Question

Given that \(\alpha=\sec ^{-1}(-\sqrt{13} / 2),\) find \(\sin \alpha, \cos \alpha, \tan \alpha, \csc \alpha\) and \(\cot \alpha .\)

Step-by-Step Solution

Answer

\(\sin \alpha = \frac{3}{\sqrt{13}}, \cos \alpha = -\frac{2}{\sqrt{13}}, \tan \alpha = -\frac{3}{2}, \csc \alpha = \frac{\sqrt{13}}{3}, \cot \alpha = -\frac{2}{3}.\)

Step 1: Understanding the Inverse Secant

The value of \( \alpha \) is given as \( \alpha = \sec^{-1} \left(-\frac{\sqrt{13}}{2}\right) \). This means that \( \sec(\alpha) = -\frac{\sqrt{13}}{2} \). The range for \( \sec^{-1} \) is \([0, \pi)\), excluding \( \frac{\pi}{2} \), so \( \alpha \) must be within this range.

Step 2: Finding the Cosine of Alpha

Since \( \sec(\alpha) = -\frac{\sqrt{13}}{2} \), we have that \( \cos(\alpha) = -\frac{2}{\sqrt{13}} \). The secant function is the reciprocal of the cosine function.

Step 3: Calculating Sine Using Pythagorean Identity

We know from the Pythagorean identity that \( \sin^2(\alpha) + \cos^2(\alpha) = 1 \). Substitute \( \cos(\alpha) = -\frac{2}{\sqrt{13}} \) into the identity: \( \sin^2(\alpha) + \left(-\frac{2}{\sqrt{13}}\right)^2 = 1 \). This simplifies to \( \sin^2(\alpha) = 1 - \frac{4}{13} \), which gives \( \sin^2(\alpha) = \frac{9}{13} \). Hence, \( \sin(\alpha) = \pm \frac{3}{\sqrt{13}} \). Since \( \alpha \) is in the second quadrant, \( \sin(\alpha) \) is positive, so \( \sin(\alpha) = \frac{3}{\sqrt{13}} \).

Step 4: Calculating Tangent

Tangent is the ratio of sine to cosine. Thus, \( \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{3}{\sqrt{13}}}{-\frac{2}{\sqrt{13}}} = -\frac{3}{2} \).

Step 5: Finding Cosecant

Cosecant is the reciprocal of sine, thus \( \csc(\alpha) = \frac{1}{\sin(\alpha)} = \frac{\sqrt{13}}{3} \).

Step 6: Calculating Cotangent

Cotangent is the reciprocal of tangent, so \( \cot(\alpha) = \frac{1}{\tan(\alpha)} = -\frac{2}{3} \).

Key Concepts

Secant Function

The secant function, denoted as \( \sec \), is a fundamental concept in trigonometry. It is defined as the reciprocal of the cosine function. This means that for any angle \( \theta \),

• \( \sec(\theta) = \frac{1}{\cos(\theta)} \)

In the context of inverse functions, \( \sec^{-1} \) refers to the angle whose secant is a specific value. In our exercise, given that \( \alpha = \sec^{-1}\left(-\frac{\sqrt{13}}{2}\right) \), it tells us that \( \sec(\alpha) = -\frac{\sqrt{13}}{2} \).
As important properties of the secant:

• The secant is undefined for angles where the cosine is zero, such as 90 degrees or \( \frac{\pi}{2} \), because division by zero is undefined.

The range of the inverse secant function is typically \([0, \pi)\) excluding \( \frac{\pi}{2} \), placing \( \alpha \) in this range.

Cosine Function

The cosine function is one of the primary functions in trigonometry and is denoted by \( \cos \). It defines the adjacent side over the hypotenuse in a right-angled triangle. For angle \( \alpha \) given in our exercise, since \( \sec(\alpha) = -\frac{\sqrt{13}}{2} \), it implies

• \( \cos(\alpha) = -\frac{2}{\sqrt{13}} \)

because cosine is the reciprocal of secant.

The cosine function can take on values between \(-1\) and \(1\), including all real numbers. In our exercise, the negative result indicates that \( \alpha \) is in the second quadrant of the unit circle. In this quadrant, cosine values are always negative. Understanding the relationship between secant and cosine helps immensely in grasping the basics of trigonometric functions.

Pythagorean Identity

The Pythagorean identity is essential in trigonometry. It states that for any angle \( \theta \):

• \( \sin^2(\theta) + \cos^2(\theta) = 1 \)

This identity is particularly helpful in calculating unknown trigonometric quantities. In our problem, by knowing \( \cos(\alpha) = -\frac{2}{\sqrt{13}} \), we can solve for \( \sin(\alpha) \) using this identity:

{% calculation of sin^2 %}

• \( \sin^2(\alpha) = 1 - \left(-\frac{2}{\sqrt{13}}\right)^2 = \frac{9}{13} \)

Taking the square root gives \( \sin(\alpha) = \pm \frac{3}{\sqrt{13}} \).

Whereas, understanding the concept of quadrants allows us to determine the correct sign. Since \( \alpha \) lies in the second quadrant, the sine value must be positive, leading to \( \sin(\alpha) = \frac{3}{\sqrt{13}} \). This highlights the crucial role of the Pythagorean identity in determining other trigonometric values.

Tangent Function

The tangent function, denoted as \( \tan \), is another foundational trigonometric function. It is the ratio of sine to cosine:

• \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)

For our angle \( \alpha \), this relationship gives:

• \( \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{3}{\sqrt{13}}}{-\frac{2}{\sqrt{13}}} = -\frac{3}{2} \)

The negative sign of the tangent in this circumstance confirms \( \alpha \) is located in the second quadrant, where tangent values are negative.

In problems involving inverse trigonometric functions, we often calculate tangent to understand angle orientation better. It's particularly useful to confirm if assumptions about the quadrant and sign of other trigonometric ratios are correct.