Problem 16 - Precalculus: Functions and Graphs

Question

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) \(\frac{\sqrt{6}}{12}\).

Step 1: Understand the Meaning of Inverse Trigonometric Functions

Inverse trigonometric functions like \(\sin^{-1}\), \(\tan^{-1}\), and \(\cos^{-1}\) determine an angle whose trigonometric function gives a specific value. For example, \(\sin^{-1}(-\frac{2}{5})\) points to an angle whose sine is \(-\frac{2}{5}\).

Step 2: Solve Exercise (a): Determine the Angle

For \(a\), the expression is \(\cot \left[\sin ^{-1}\left(-\frac{2}{5}\right)\right]\). First, set \( \theta = \sin^{-1}(-\frac{2}{5}) \). This means \( \sin(\theta) = -\frac{2}{5} \).

Step 3: Use Pythagorean Identity

For this angle \(\theta\), if \(\sin(\theta) = -\frac{2}{5}\), using the Pythagorean identity, \(\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(-\frac{2}{5}\right)^2} = \frac{\sqrt{21}}{5}\).

Step 4: Calculate the Cotangent

We have \( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{\frac{\sqrt{21}}{5}}{-\frac{2}{5}} = -\frac{\sqrt{21}}{2} \). Thus, the exact value is \(-\frac{\sqrt{21}}{2}\).

Step 5: Solve Exercise (b): Determine the Angle

Set \( \phi = \tan^{-1}(\frac{7}{4}) \), meaning \(\tan(\phi) = \frac{7}{4}\).

Step 6: Use Trigonometric Identity

For \(\phi\), \(\sec(\phi) = \frac{1}{\cos(\phi)}\). From the right triangle definition: \(\tan(\phi) = \frac{7}{4}\), hence \(\cos(\phi) = \frac{4}{\sqrt{65}}\).

Step 7: Calculate the Secant

Using \(\cos(\phi)\) from the previous step, \(\sec(\phi) = \frac{\sqrt{65}}{4}\). Thus, the exact value is \(\frac{\sqrt{65}}{4}\).

Step 8: Solve Exercise (c): Determine the Angle

Set \( \psi = \cos^{-1}(\frac{1}{5}) \), which means \(\cos(\psi) = \frac{1}{5}\).

Step 9: Use Pythagorean Theorem

For angle \(\psi\) in \([0, \pi]\), compute the sine: \(\sin(\psi) = \sqrt{1 - (\frac{1}{5})^2} = \frac{2\sqrt{6}}{5}\).

Step 10: Calculate the Cosecant

Since \(\csc(\psi) = \frac{1}{\sin(\psi)}\), we have \(\csc(\psi) = \frac{5}{2\sqrt{6}} = \frac{\sqrt{6}}{12}\) after rationalization.

Key Concepts

Trigonometric Identities

Trigonometric identities are mathematical equations that are true for all values of the variables where they are defined. They form the foundation for simplifying expressions and solving equations in trigonometry.
Some common identities include:

• Pythagorean Identities:
  • \( \sin^2(\theta) + \cos^2(\theta) = 1 \)
  • \( 1 + \tan^2(\theta) = \sec^2(\theta) \)
  • \( 1 + \cot^2(\theta) = \csc^2(\theta) \)
• Reciprocal Identities:
  • \( \sin(\theta) = \frac{1}{\csc(\theta)} \)
  • \( \cos(\theta) = \frac{1}{\sec(\theta)} \)
  • \( \tan(\theta) = \frac{1}{\cot(\theta)} \)
• Quotient Identities:
  • \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)
  • \( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} \)

These identities help us transition between different trigonometric functions and solve complex trigonometry problems by simplifying expressions through substitution and transformation.

Cotangent

Cotangent, denoted as \( \cot \), is one of the six fundamental trigonometric functions and is defined as the reciprocal of the tangent function.
Specifically, \( \cot(\theta) = \frac{1}{\tan(\theta)} \), which also means \( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} \).

Cotangent is particularly useful in trigonometric identities and problems where complementary angles are involved. In practical terms:

• If \( \theta = \sin^{-1}(-\frac{2}{5}) \), then \( \sin(\theta) = -\frac{2}{5} \).
• We use the identity \( \cos^2(\theta) = 1 - \sin^2(\theta) \) to find \( \cos(\theta) \).
• Substitute into \( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} \) to solve for the value of cotangent, as shown in the exercise.

Understanding cotangent in this context helps bridge the connection between circular and triangle-based interpretations of trigonometric functions.

Secant

Secant, represented as \( \sec \), is the reciprocal of the cosine function. Thus, \( \sec(\theta) = \frac{1}{\cos(\theta)} \). This function is useful in various trigonometric transformations and equations.
Understanding secant requires a firm grasp of both its definition and application:

• When given \( \phi = \tan^{-1}(\frac{7}{4}) \), it implies \( \tan(\phi) = \frac{7}{4} \).
• Use the identity \( \tan(\phi) = \frac{\sin(\phi)}{\cos(\phi)} \) to determine the values of \( \sin(\phi) \) and \( \cos(\phi) \).
• Calculate \( \cos(\phi) \) using the formula for the hypotenuse. Here \( \cos(\phi) = \frac{4}{\sqrt{65}} \).
• Find \( \sec(\phi) = \frac{\sqrt{65}}{4} \) as the reciprocal of the calculated cosine value.

Emphasizing the reciprocal nature of secant clarifies its role in trigonometry, particularly in right triangle problems and when working with angles defined through inverse tangent.

Cosecant

Cosecant is the reciprocal of the sine function, denoted as \( \csc \). Mathematically, \( \csc(\theta) = \frac{1}{\sin(\theta)} \). It's particularly useful in converting between different trigonometric forms, especially where sine is small or fractions are involved.
Key points to understand when working with cosecant:

• For \( \psi = \cos^{-1}(\frac{1}{5}) \), it indicates \( \cos(\psi) = \frac{1}{5} \).
• To find \( \sin(\psi) \), employ the Pythagorean Theorem, leading to \( \sin(\psi) = \sqrt{1 - (\frac{1}{5})^2} = \frac{2\sqrt{6}}{5} \).
• Compute \( \csc(\psi) = \frac{1}{\sin(\psi)} \) to obtain \( \frac{5}{2\sqrt{6}} \), and rationalize if necessary to \( \frac{\sqrt{6}}{12} \).

Cosecant is especially meaningful when working with complementary angles or angles where the sine result is part of multiple trigonometric computations. Its reciprocal relationship simplifies many expressions.