Problem 11 - University Calculus: Early Transcendentals

Question

Find the values. $$\tan \left(\sin ^{-1}\left(-\frac{1}{2}\right)\right)$$

Step-by-Step Solution

Answer

The value is \(-\frac{\sqrt{3}}{3}\).

Step 1: Understand the problem context

We are given the expression \( \tan(\sin^{-1}(-\frac{1}{2})) \) and we need to find its value. The goal is to first find \( \sin^{-1}(-\frac{1}{2}) \), then compute the tangent of that angle.

Step 2: Find \( \theta = \sin^{-1}(-\frac{1}{2}) \)

We know from inverse trigonometry that \( \sin^{-1}(-\frac{1}{2}) \) is the angle \( \theta \) such that \( \sin(\theta) = -\frac{1}{2} \). Since the range of \( \sin^{-1}x \) is \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \), the angle that gives \( \sin(\theta) = -\frac{1}{2} \) is \( \theta = -\frac{\pi}{6} \).

Step 3: Compute \( \tan(-\frac{\pi}{6}) \)

Now that we know \( \theta = -\frac{\pi}{6} \), we need to find \( \tan(-\frac{\pi}{6}) \). Using the identity \( \tan(-x) = -\tan(x) \) and knowing that \( \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \), we calculate \( \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} \).

Step 4: Simplify the result

Simplifying \( -\frac{1}{\sqrt{3}} \) using the rationalization process involves multiplying both the numerator and the denominator by \( \sqrt{3} \). Thus, \( -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \).

Key Concepts

Trigonometric Identities

In the realm of trigonometry, identities are key tools that allow us to manipulate and simplify expressions. One fundamental identity to remember is that for any angle \( x \), the tangent function fulfills the property \( \tan(-x) = -\tan(x) \). This particular identity is highly useful when dealing with negative angles, as seen in tasks involving inverse trigonometric functions.

In addition, knowing that \( \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \) is crucial. Because of the identity \( \tan(-x) = -\tan(x) \), it immediately follows that \( \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} \). This illustrates how identities not only simplify calculations but also provide insights into the symmetries inherent in trigonometric functions.

When we collectively use these identities, we can effortlessly abbreviate and conclude our computations with confidence, resulting in simplified, accurate results.

Inverse Trigonometric Range

Understanding the range of inverse trigonometric functions is crucial. These ranges help determine the possible output angles for given inputs within these functions, which are often limited to ensure they remain functions and pass the vertical line test. Specifically, for \( \sin^{-1}(x) \), the range is \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \). This range corresponds to the right side of the unit circle, covering the first and fourth quadrants.

In the example of finding \( \theta = \sin^{-1}(-\frac{1}{2}) \), the result \( \theta = -\frac{\pi}{6} \) is valid because it fits within the specified range of the inverse sine function. This ensures that any angle derived from \( \sin^{-1}(x) \) will produce a unique, unambiguous output value, making calculations clear and consistent across different applications.

Trigonometric Simplification

Simplifying trigonometric solutions often includes rationalizing denominators. For instance, in the solution \( \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} \), further simplification involves eliminating the square root from the denominator.

Rationalization is achieved by multiplying both the numerator and denominator by the square root present in the denominator. So, \( -\frac{1}{\sqrt{3}} \) becomes \(-\frac{\sqrt{3}}{3} \) after multiplying by \( \sqrt{3}/\sqrt{3} \).

This final expression is often preferred in mathematical practice and education since it avoids square roots in the denominator—making it look cleaner and more conventional. Simplification techniques like these are incredibly helpful in achieving precise and standardized answers in trigonometric computations.