Problem 4 - Precalculus with Calculus Previews

Question

Find the exact value of the given trigonometric expression. Do not use a calculator. $$ \arcsin \frac{\sqrt{3}}{2} $$

Step-by-Step Solution

Answer

\( \arcsin \frac{\sqrt{3}}{2} = \frac{\pi}{3} \).

Step 1: Understanding the Inverse Sine Function

The function \( \arcsin(x) \) gives the angle \( \theta \) whose sine is \( x \). We need to find the angle \( \theta \) such that \( \sin(\theta) = \frac{\sqrt{3}}{2} \).

Step 2: Recall Common Sine Values

Recall the common angles and their sine values. We know \( \sin(30^\circ) = \frac{1}{2} \), \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \), and \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \).

Step 3: Identify the Angle

From Step 2, we see that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \). Therefore, the angle \( \theta \) we are looking for is \( 60^\circ \). In radians, this angle is \( \frac{\pi}{3} \).

Step 4: Verify the Principal Value Range

The principal value range for \( \arcsin(x) \) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), and since \( \frac{\pi}{3} \approx 1.047 \), it lies within this range. Therefore, \( \arcsin \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3} \) is valid.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \arcsin \), are pivotal in finding angles when you know the sides of a triangle. Essentially, if you know the sine, cosine, or tangent of an angle, these inverse functions will help you find the original angle. In the case of \( \arcsin(x) \), you're looking for an angle \( \theta \) where the sine value equals \( x \).
\( \arcsin(x) \) specifically returns the angle in the range where sine values are defined, ensuring that it gives a consistent result for any valid input. This is why understanding inverse trigonometric functions is essential for converting trigonometric values back into angles.
When solving exercises like the given one, you leverage these functions to quickly determine angles based on known sine values, simplifying problems across various trigonometric applications.

Common Angle Values

Recognizing common angle values and their corresponding trigonometric function results are crucial for solving trigonometric problems efficiently. In our example, knowing that:

• \( \sin(30^\circ) = \frac{1}{2} \)
• \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)
• \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)

helps you quickly identify angles just by looking at a given sine value. This knowledge saves time, as you don't need a calculator to find these angles.
In mathematics, these angles (in degrees: 30°, 45°, 60°) are frequently used, and their corresponding values demonstrate symmetry and consistency in trigonometric functions. Recognizing them allows you not only to navigate exams and exercises swiftly but also deepens your intuition for trigonometry.

Principal Value Range

When dealing with inverse trigonometric functions, the principal value range is an important concept because it defines where the function yields results. For \( \arcsin(x) \), the principal value range is \([ -\frac{\pi}{2}, \frac{\pi}{2} ]\). This range ensures that the \( \arcsin(x) \) function produces a unique, unambiguous angle corresponding to each sine value.
The principal value range is crucial in ensuring that inverse functions give consistent results. For instance, for \( \arcsin \frac{\sqrt{3}}{2} \), the principal range confirms that the result \( \frac{\pi}{3} \) (or 60°) lies comfortably within \([ -\frac{\pi}{2}, \frac{\pi}{2} ]\), making it a valid solution.