Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 25-28, find each value without using a calculator (see Example 4). $$ \cos \left[2 \sin ^{-1}\left(-\frac{2}{3}\right)\right] $$

Step-by-Step Solution

Answer

\( \cos \left[ 2 \sin^{-1} \left( -\frac{2}{3} \right) \right] = \frac{1}{9} \).

Step 1: Identify the Problem

We need to evaluate \( \cos \left[ 2 \sin^{-1} \left( -\frac{2}{3} \right) \right] \). This is a composite trigonometric expression that involves finding the cosine of twice the inverse sine.

Step 2: Define \( \sin^{-1}(-\frac{2}{3}) \)

Let \( \theta = \sin^{-1} \left( -\frac{2}{3} \right) \). The angle \( \theta \) has a sine value of \(-\frac{2}{3}\), meaning \( \sin(\theta) = -\frac{2}{3} \). We also know by properties of inverse functions that \( \theta \) must be in the range \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\).

Step 3: Use Trigonometric Identity

The expression we need is \( \cos(2\theta) \). Use the double angle identity for cosine: \( \cos(2\theta) = 1 - 2 \sin^2(\theta) \).

Step 4: Calculate \( \sin^2(\theta) \)

Since \( \sin(\theta) = -\frac{2}{3} \), we have \( \sin^2(\theta) = \left(-\frac{2}{3}\right)^2 = \frac{4}{9} \).

Step 5: Use the Identity to Find \( \cos(2\theta) \)

Substitute \( \sin^2(\theta) = \frac{4}{9} \) into the identity: \( \cos(2\theta) = 1 - 2 \cdot \frac{4}{9} = 1 - \frac{8}{9} = \frac{1}{9} \).

Step 6: Verify and Conclude

The value of \( \cos \left[ 2 \sin^{-1} \left( -\frac{2}{3} \right) \right] \) is \( \frac{1}{9} \), which confirms our calculations are correct.

Key Concepts

Trigonometric Identities

Trigonometric identities are equations that hold true for all values of the variables involved. One such essential identity is used in this exercise, namely the double angle identity for cosine. This particular identity is as follows:

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

This identity allows us to express \(\cos(2\theta)\) in terms of \(\sin^2(\theta)\), which can be incredibly useful.
Understanding these identities helps in simplifying more complex trigonometric expressions and solving problems without the need for a calculator.
By breaking down the expression using trigonometric identities, you can easily evaluate things by hand as long as you know basic sine or cosine values.
They form the backbone of more advanced concepts in trigonometry, enabling you to transform and manipulate expressions to suit your problem-solving needs.

Double Angle Formula

The double angle formula is a vital tool in trigonometry, especially relevant when dealing with expressions involving \(2\theta\). In the given exercise, we used the double angle formula for cosine:

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

This formula is particularly useful because it expresses \(\cos(2\theta)\) using the simpler term \(\sin(\theta)\). This transformation helps reduce complexity when working with inverse trigonometric functions, as seen in this exercise.
By substituting specific known values of \(\sin(\theta)\) into the formula, we can seamlessly calculate \(\cos(2\theta)\) without needing a calculator.
The double angle formulas also exist for sine and tangent, adding to the arsenal of tools you can leverage:

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

Understanding when and how to apply these formulas can significantly streamline calculations in trigonometric problems.

Angles and Ranges

Inverse trigonometric functions, such as \(\sin^{-1}(x)\), have specific ranges that dictate the possible values of \(\theta\). For \(\sin^{-1}(x)\), \(\theta\) is constrained to lie between

• \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\)

This range ensures that each value of \(y\) within the interval maps back to a unique angle \(\theta\). In our problem, we found \(\theta = \sin^{-1}(-\frac{2}{3})\), setting \(\theta\) firmly within this specified range.
This restriction is crucial when solving trigonometric problems, as it helps avoid multiple solutions or ambiguity, ensuring that calculations using inverse functions provide one correct outcome.
By understanding these ranges, you can confidently tackle problems involving inverse trigonometric expressions, knowing exactly where your solutions will lie on the unit circle.
This knowledge is especially valuable when evaluating composite expressions, where understanding the permissible range of values is crucial to achieving the correct results.