Problem 23 - Precalculus with Calculus Previews

Question

Find the exact value of the given trigonometric expression. Do not use a calculator. $$ \sin \left(\sin ^{-1} \frac{1}{5}\right) $$

Step-by-Step Solution

Answer

\(\frac{1}{5}\)

Step 1: Understand the expression

The expression is \( \sin \left( \sin ^{-1} \frac{1}{5} \right) \). We need to find the value of this entire expression. \( \sin^{-1} \) is the inverse sine function, which gives an angle whose sine is \( \frac{1}{5} \).

Step 2: Interpret the inverse function

The expression \( \sin^{-1} \frac{1}{5} \) represents an angle, let's call it \( \theta \), such that \( \sin(\theta) = \frac{1}{5} \).

Step 3: Evaluate the outer function

The expression then simplifies to \( \sin(\theta) \) because \( \sin^{-1} \) provides an angle \( \theta \) where \( \sin(\theta) = \frac{1}{5} \). Thus, \( \sin \left( \sin^{-1} \frac{1}{5} \right) = \sin(\theta) \).

Step 4: Conclusion

Since \( \sin(\theta) = \frac{1}{5} \), the value we wanted is already given by the inverse sine as \( \frac{1}{5} \).

Key Concepts

Inverse Sine Function

The inverse sine function, denoted as \( \sin^{-1} \) or arcsin, is a fundamental concept in trigonometry. It is designed to find an angle whose sine has a specified value. Given a number \( y \) within the range of \(-1 \leq y \leq 1\), the inverse sine function returns an angle \( \theta \) such that \( \sin(\theta) = y \). This angle \( \theta \) falls typically within the principal range of \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) radians, or \(-90^\circ \leq \theta \leq 90^\circ \) degrees.It's crucial to note that the inverse sine function only captures one of the infinite number of possible angles due to periodicity of sine function. It selects the angle within the principal range. For example, when you find \( \sin^{-1} \frac{1}{5} \), it means we are searching for an angle \( \theta \) such that \( \sin(\theta) = \frac{1}{5} \), specifically where \( \theta \) is between \(-90^\circ\) and \(90^\circ\).

Evaluation of Trigonometric Functions

Evaluating trigonometric functions involves determining the value associated with a specific angle or expression involving angles. When dealing with expressions such as \( \sin(\sin^{-1}y) \), understanding the processes of inverse functions simplifies the tasks considerably.In the context of our exercise, we aim to evaluate \( \sin(\sin^{-1}\frac{1}{5}) \). We first determine what angle the inverse sine provides us; this angle is \( \theta \). By the definition of inverse functions, \( \sin^{-1}\frac{1}{5} \) returns \( \theta \) such that \( \sin(\theta) = \frac{1}{5} \). After finding \( \theta \), we evaluate the outer function-- here it simplifies back to \( \sin(\theta) \).Because the outcome of \( \sin^{-1}\frac{1}{5} \) is constructed to precisely be an angle where \( \sin(\theta) = \frac{1}{5} \), this makes evaluating the expression straightforward, reinforcing our understanding of function operations.

Sine Function

The sine function, usually denoted as \( \sin \), is one of the primary trigonometric functions. It relates the angle of a right triangle to the ratio of the length of the opposite side to the hypotenuse. In the unit circle framework, the sine of an angle \( \theta \) is the y-coordinate of the point where the terminal side of the angle intersects the unit circle.Key properties of the sine function include its range and periodicity:

• The range of the sine function is between -1 and 1, inclusive.
• The function is periodic, with a period of \(2\pi\). This means \( \sin(\theta + 2\pi) = \sin(\theta) \).

In our exercise, when computing \( \sin\left( \sin^{-1}\frac{1}{5} \right) \), we are actually leveraging these properties. The original equation seeks the sine of an angle that we already know yields \( \sin(\theta) = \frac{1}{5} \). This highlights both the function's reliability and its predictive capability in circular motion and harmonic analysis.