Problem 16 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-22: Find the exact value of the expression whenever it is defined. (a) \(\cot \left[\sin ^{-1}\left(-\frac{2}{5}\right)\right]\) (b) \(\sec \left(\tan ^{-1} \frac{7}{4}\right)\) (c) \(\csc \left(\cos ^{-1} \frac{1}{5}\right)\)

Step-by-Step Solution

Answer

(a) \(-\frac{\sqrt{21}}{2}\), (b) \(\frac{\sqrt{65}}{4}\), (c) \(\frac{5\sqrt{6}}{12}\)."}

Step 1: Understanding Inverse Trigonometric Functions

We need to evaluate expressions involving inverse trigonometric functions. In these problems, functions like \( \sin^{-1} \), \( \tan^{-1} \), and \( \cos^{-1} \) are given, and we need to convert these angles into their corresponding trigonometric functions like \( \cot \), \( \sec \), and \( \csc \).

Step 2: Evaluating \( \cot \left[\sin^{-1}\left(-\frac{2}{5}\right)\right] \)

Start by letting \( \theta = \sin^{-1}\left(-\frac{2}{5}\right) \). This means \( \sin \theta = -\frac{2}{5} \). Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), compute \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(-\frac{2}{5}\right)^2} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5} \). Since \( \theta \) corresponds to a negative sine value, \( \theta \) is in the fourth quadrant. Thus, \( \cos \theta = \frac{\sqrt{21}}{5} \) is positive. Now, \( \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{\sqrt{21}}{5}}{-\frac{2}{5}} = -\frac{\sqrt{21}}{2} \).

Step 3: Evaluating \( \sec \left(\tan^{-1} \frac{7}{4}\right) \)

Let \( \phi = \tan^{-1} \left(\frac{7}{4}\right) \). Therefore, \( \tan \phi = \frac{7}{4} \). Using the identity \( 1 + \tan^2 \phi = \sec^2 \phi \), we find \( \sec \phi = \sqrt{1 + \frac{49}{16}} = \sqrt{\frac{65}{16}} = \frac{\sqrt{65}}{4} \).

Step 4: Evaluating \( \csc \left(\cos^{-1} \frac{1}{5}\right) \)

Let \( \psi = \cos^{-1} \left(\frac{1}{5}\right) \). This tells us \( \cos \psi = \frac{1}{5} \). Using \( \sin^2 \psi + \cos^2 \psi = 1 \), find \( \sin \psi = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} \). Hence, \( \csc \psi = \frac{1}{\sin \psi} = \frac{5}{\sqrt{24}} = \frac{5\sqrt{6}}{12} \) after rationalizing the denominator.

Key Concepts

Pythagorean identity

The Pythagorean identity is a fundamental relation in trigonometry. It states that for any angle \( \theta \), the sum of the squares of the sine and cosine of \( \theta \) is equal to 1. Mathematically, it is expressed as:\[\sin^2 \theta + \cos^2 \theta = 1\]This identity is very useful in finding missing values of sine, cosine, or related trigonometric functions when given partial information. For instance, if you know \( \sin \theta = -\frac{2}{5} \), you can find \( \cos \theta \) using the identity:\[\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (-\frac{2}{5})^2} = \frac{\sqrt{21}}{5}\]Understanding this identity can help in many trigonometric problems, such as the one where \( \cot \) is evaluated. It assures that calculations are consistent within the unit circle. When working with negative sine or other functions, remember the quadrant can affect the sign of the answer.

• Useful for computing unknown trigonometric values
• Applies to all angles
• Maintains the relationship between sine and cosine

cotangent

Cotangent, abbreviated as \( \cot \), is one of the six basic trigonometric functions. It is defined as the reciprocal of the tangent function:\[\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}\]When evaluating \( \cot \left[\sin^{-1}\left(-\frac{2}{5}\right)\right] \), knowing the values of sine and cosine is key. Once we determine \( \sin \theta = -\frac{2}{5} \) and \( \cos \theta = \frac{\sqrt{21}}{5} \) using the Pythagorean identity, we find:\[\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{\sqrt{21}}{5}}{-\frac{2}{5}} = -\frac{\sqrt{21}}{2}\]This calculation exemplifies finding cotangent assuming \( \theta \) is in a specific quadrant. Remember, cotangent is undefined at angles where sine is zero since it would lead to division by zero.

• Reciprocal of the tangent function
• Involves both sine and cosine
• Dependent on quadrant for sign

secant

Secant, denoted as \( \sec \), is another basic trigonometric function. It is the reciprocal of the cosine function and is expressed as:\[\sec \theta = \frac{1}{\cos \theta}\]In the exercise involving \( \sec \left(\tan^{-1} \frac{7}{4}\right) \), we first find the angle \( \phi \) whose tangent is \( \frac{7}{4} \). Using the identity \( 1 + \tan^2 \phi = \sec^2 \phi \), we calculate:\[\sec \phi = \sqrt{1 + \left(\frac{7}{4}\right)^2} = \sqrt{\frac{65}{16}} = \frac{\sqrt{65}}{4}\]This showcases how secant calculations can involve various identities to determine its precise value. Understanding \( \sec \) is crucial as it is undefined when cosine is zero.

• Reciprocal of cosine
• Calculation often involves Pythagorean-like identities
• Undefined where cosine equals zero

cosecant

Cosecant, or \( \csc \), is the reciprocal of the sine function. It is defined as:\[\csc \theta = \frac{1}{\sin \theta}\]In the task to find \( \csc(\cos^{-1} \frac{1}{5}) \), it is crucial to determine \( \sin \theta \) when \( \cos \theta = \frac{1}{5} \). Using the Pythagorean identity:\[\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \frac{\sqrt{24}}{5}\]Subsequently, the value of cosecant is:\[\csc \theta = \frac{1}{\sin \theta} = \frac{5}{\sqrt{24}} = \frac{5\sqrt{6}}{12}\]This calculation illustrates how to find cosecant by converting from the cosine function, underlining its role as the sine's reciprocal. Cosecant is undefined where sine equals zero.

• Reciprocal of sine
• Values depend on \( \sin \theta \)
• Critical in defining certain trigonometric limits