Problem 27 - Precalculus with Calculus Previews

Question

Find the exact value of the given trigonometric expression. Do not use a calculator. $$ \arcsin \left(\sin \frac{\pi}{16}\right) $$

Step-by-Step Solution

Answer

\( \frac{\pi}{16} \)

Step 1: Understand the arcsin function

Remember that the function \( \arcsin(x) \) gives the angle \( \theta \) such that \( \sin(\theta) = x \) and \( \theta \) is in the range \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\). This means \( \arcsin \) returns the principal value of an angle.

Step 2: Simplify the Expression

Since \( \arcsin(\sin(x)) \) will yield \( x \) if \( x \) is within the range of arcsin, check if \( \frac{\pi}{16} \) belongs to this interval. Since \( 0 \lt \frac{\pi}{16} \lt \frac{\pi}{2} \), the expression simplifies directly to \( \frac{\pi}{16} \).

Key Concepts

Understanding the Arcsin Function

The arcsin function, denoted as \( \arcsin(x) \), is a type of inverse trigonometric function. It is specifically the inverse of the sine function. This means it takes a value \( x \) and gives back the angle \( \theta \) for which \( \sin(\theta) = x \). To ensure it is a true inverse, the angle \( \theta \) is restricted to the range of \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \). This range represents the principal value range for the arcsin function, ensuring that each input corresponds to precisely one output.

In practical terms, this means you can use \( \arcsin \) to figure out what angle would have a particular sine value, as long as it falls within this specified range. For example, if the sine of an angle is 0.5, then \( \arcsin(0.5) \) will give \( \frac{\pi}{6} \), since the sine of \( \frac{\pi}{6} \) is indeed 0.5 and the angle itself is within the valid range.

The main feature of the arcsin function is its principal value, which ensures there's a consistent and predictable result every time it's called upon.

Exploring Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables where the functions are defined. These identities are very useful in simplifying expressions and solving trigonometric equations. Some of the basic identities include

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

These identities help break down complex trigonometric expressions into more manageable pieces, and they are crucial in proving other trigonometric properties.

In our specific problem, the identity used is linked to the simplicity of \( \arcsin(\sin(x)) \). Here, assuming \( x \) is within the principal range, \( \arcsin(\sin(x)) \) simply returns \( x \). This is because you're effectively undoing the sine operation when within the range of \( \arcsin \). Understanding and utilizing these identities can transform a problem that looks intimidating at first glance into something quite straightforward.

Principal Value Concept

The principal value is a fundamental concept when dealing with inverse trigonometric functions, such as arcsin, arccos, and arctan. It refers to a specific angle within a designated range that the function consistently returns. This idea helps maintain consistency by ensuring that each function call yields only one result.

For the arcsin function, the principal value is always found within the range of \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\). This is critical because multiple angles could technically satisfy \( \sin(\theta) = x \). The principal value narrows that down to a single angle by imposing the range.

When solving expressions like \( \arcsin(\sin(x)) \), the principal value concept aids in simplifying the problem. If \( x \) is already within the \( \arcsin \) range, the expression simplifies directly to \( x \), as was done in the exercise with \( \frac{\pi}{16} \).

• This ensures that the results are both predictable and easy to compute.
• It also makes certain that consistency is maintained across mathematical solutions.

Understanding principal values reinforces your grasp on inverse functions and guarantees accurate results in trigonometry.