Problem 82 - A Graphical Approach to Precalculus with Limits

Question

Give the exact real number value of each expression. Do not use a calculator. $$\sec \left(\sin ^{-1}\left(-\frac{1}{5}\right)\right)$$

Step-by-Step Solution

Answer

The exact value is \( \frac{5\sqrt{6}}{12} \).

Step 1: Understand the problem

We need to find the value of the expression \( \sec(\sin^{-1}(-\frac{1}{5})) \). This involves finding the secant of the angle whose sine is \(-\frac{1}{5}\).

Step 2: Find the angle from the inverse sine

Let \( \theta = \sin^{-1} \left( -\frac{1}{5} \right) \). This means \( \sin \theta = -\frac{1}{5} \). The angle \( \theta \) is in the range \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\).

Step 3: Solve for \( \cos \theta \) using the Pythagorean identity

We use the Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1 \). Substituting \( \sin \theta = -\frac{1}{5} \), \[\left(-\frac{1}{5}\right)^2 + \cos^2 \theta = 1\] \[\frac{1}{25} + \cos^2 \theta = 1\] \[\cos^2 \theta = \frac{24}{25}\]. Since \( \theta \) is in the first or fourth quadrant, \( \cos \theta \) should be positive, thus \( \cos \theta = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5} \).

Step 4: Find \( \sec \theta \)

Secant is the reciprocal of cosine, so \( \sec \theta = \frac{1}{\cos \theta} \). Substitute for \( \cos \theta \): \[\sec \theta = \frac{1}{\frac{2\sqrt{6}}{5}} = \frac{5}{2\sqrt{6}}\]. To rationalize the denominator: \[\sec \theta = \frac{5}{2\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{5 \sqrt{6}}{12}\].

Step 5: Write down the final result

The exact value of \( \sec(\sin^{-1}(-\frac{1}{5})) \) is \( \frac{5\sqrt{6}}{12} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to determine an angle when given a trigonometric ratio. For instance, if you know the sine of an angle, you can use the inverse sine function, denoted as \( \sin^{-1} \), to find the measure of that angle.
This concept is particularly useful in solving trigonometric equations where we need to "invert" the function to obtain the angle in question.

• In the provided problem, \( \sin^{-1}\left(-\frac{1}{5}\right) \) helps us find the angle whose sine is \(-\frac{1}{5}\).
• The range of the inverse sine function is \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\), ensuring that the angle \(\theta\) is in the correct range for its calculations.

By understanding inverse trigonometric functions, you can effectively transition between an angle and its sine, cosine, or tangent values.

Pythagorean Identity

The Pythagorean Identity is key in simplifying trigonometric problems. It states that for any angle \( \theta \), \[ \sin^2 \theta + \cos^2 \theta = 1 \] This relationship allows us to find one trigonometric function if another is known. In the context of our exercise, knowing the sine led us directly to the cosine. When you have \( \sin \theta = -\frac{1}{5} \):

• Substitute it into the Pythagorean identity to solve for \( \cos \theta \).
• \( \cos^2 \theta = 1 - \left(-\frac{1}{5}\right)^2 = \frac{24}{25} \) computed from the identity.
• We take the positive root for \( \cos \theta = \frac{2\sqrt{6}}{5} \) since the angle \( \theta \) is in the range where cosine is positive.

Learning this identity is crucial because it provides a direct connection between the sine and cosine of an angle.

Rationalizing Denominators

Rationalizing the denominator is a technique used in mathematics to eliminate radicals (square roots) from the denominator of a fraction.
It's often required because it usually results in simpler, cleaner expressions that are easier to interpret and compare.When we encountered \( \sec \theta = \frac{5}{2\sqrt{6}} \), the denominator had a radical:

• To rationalize, multiply both the numerator and the denominator by \( \sqrt{6} \).
• This results in \( \frac{5\sqrt{6}}{12} \), a fully rationalized expression.

By practicing rationalization, you become skilled at transforming complex expressions into ones that are more manageable and standardized.