Problem 6 - Essential Calculus Early Transcendentals

Question

Find the exact value of each expression. (a) \(\tan \left(\sec ^{-1} 4\right)\) (b) \(\sin \left(2 \sin ^{-1}\left(\frac{3}{5}\right)\right)\)

Step-by-Step Solution

Answer

(a) \( \sqrt{15} \); (b) \( \frac{24}{25} \)

Step 1: Understand the Expression

The expression \( \tan(\sec^{-1} 4) \) implies finding the tangent of the angle whose secant is 4. Remember, \( \sec(\theta) = \frac{1}{\cos(\theta)} \).

Step 2: Determine the Angle

To find the angle \( \theta \) where \( \sec(\theta) = 4 \), we know \( \cos(\theta) = \frac{1}{4} \) because the secant is the reciprocal of the cosine.

Step 3: Find the Sine of the Angle

Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we can find \( \sin(\theta) \). Thus, \( \sin^2(\theta) = 1 - \left(\frac{1}{4}\right)^2 = \frac{15}{16} \). So, \( \sin(\theta) = \frac{\sqrt{15}}{4} \).

Step 4: Calculate the Tangent

Now, \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{\sqrt{15}}{4}}{\frac{1}{4}} = \sqrt{15} \).

Step 5: Address Part (b)

For \( \sin(2 \sin^{-1}(\frac{3}{5})) \), start by letting \( \theta = \sin^{-1}\left(\frac{3}{5}\right) \), hence \( \sin(\theta) = \frac{3}{5} \).

Step 6: Use the Double Angle Identity

The expression \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \). Here, \( \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5} \).

Step 7: Compute \( \sin(2\theta) \)

\( \sin(2\theta) = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25} \).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in mathematics that help simplify and solve various trigonometric problems. They are equations that are true for all angles and relate different trigonometric functions to one another. In the exercise, the identity \( \sec(\theta) = \frac{1}{\cos(\theta)} \) is crucial.
Here, the inverse secant function \( \sec^{-1} \) helps us find the angle whose secant is given as 4. By using the identity, we understand this situation's reciprocal relationship, allowing us to convert it to a more familiar cosine problem. Other key identities that are often used include:

• \( \sin^2(\theta) + \cos^2(\theta) = 1 \)
• \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)

These identities assist us in expressing trigonometric functions in terms of each other, making the calculations more straightforward and manageable.

Angle Calculation

Calculating angles involves determining specific angle measures based on given trigonometric values or relations. This process often uses inverse trigonometric functions, allowing us to work backward from known ratios to find the corresponding angle.
In part (a) of the original exercise, we start by finding an angle for which the secant is 4. Since the secant is the reciprocal of the cosine, we deduce that \( \cos(\theta) = \frac{1}{4} \). By working through these calculations step by step, we can then easily find other related trigonometric values for the same angle.
In part (b), the calculation involves finding an angle \( \theta \) where \( \sin(\theta) = \frac{3}{5} \). Recognizing this angle allows us to use it in the double angle formula, demonstrating a practical application of angle calculation.

Double Angle Formulas

Double angle formulas are trigonometric identities used to find the trigonometric function values of double angles, such as \( 2\theta \). They help in simplifying expressions where angles are doubled, a common occurrence in trigonometry problems.
The key formula used in part (b) is the sine double angle formula: \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \). This formula cleverly combines sine and cosine of a single angle, showing how these functions interact when the angle is doubled.
For \( \sin(2\sin^{-1}(\frac{3}{5})) \), we first determine \( \cos(\theta) \) using \( \cos(\theta) = \sqrt{1 - \sin^2(\theta)} \). Substituting into the formula gives us the exact value of the expression. Such transformations allow for straightforward calculations and problem-solving by relating complex expressions back to foundational identities.