Problem 113 - Precalculus

Question

For the following exercises, evaluate the expressions. $$ \sin ^{-1}\left(\frac{\sqrt{2}}{2}\right) $$

Step-by-Step Solution

Answer

\(\frac{\pi}{4}\)

Step 1: Understand the Problem

The task is to find the value of the inverse sine function, \( \sin^{-1} \), for \( \frac{\sqrt{2}}{2} \). This means finding an angle whose sine is \( \frac{\sqrt{2}}{2} \).

Step 2: Recall Sine Values for Special Angles

Recall that the sine of common angles such as \( \frac{\pi}{4} \), \( \frac{\pi}{3} \), and \( \frac{\pi}{6} \) and so on. We know from trigonometry that \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).

Step 3: Determine the Angle

Since \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \), the inverse sine of \( \frac{\sqrt{2}}{2} \) is \( \frac{\pi}{4} \). This means \( \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \).

Key Concepts

Special Angles

Special angles in trigonometry are certain angles where the values of trigonometric functions like sine, cosine, and tangent are easily recognizable and often used. These angles are integral to many mathematical calculations because they simplify working with trigonometric identities and angles. Some of the most common special angles are:

• \( \frac{\pi}{6} \) (30 degrees)
• \( \frac{\pi}{4} \) (45 degrees)
• \( \frac{\pi}{3} \) (60 degrees)
• \( \frac{\pi}{2} \) (90 degrees)
• \( \pi \) (180 degrees)

These angles are crucial because they allow for exact calculations without approximations, making them a powerful tool in both geometry and calculus. It's helpful to memorize the sine, cosine, and tangent values of these angles for quick reference.
Understanding these angles helps simplify complex trigonometric problems. For instance, knowing the sine of \( \frac{\pi}{4} \) is \( \frac{\sqrt{2}}{2} \), directly helps us evaluate inverse functions swiftly.

Sine Function

The sine function is one of the primary trigonometric functions used to model periodic phenomena. It relates an angle of a right triangle to the ratio of the opposite side to the hypotenuse. For an angle \( \theta \) in a triangle:

• \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)

Trigonometric functions like sine are used not only in geometry but also in various fields like physics, engineering, and even music and arts, due to their cyclical nature.
In the unit circle, the sine of an angle corresponds to the y-coordinate of the point where the terminal side of the angle intersects the circle. This provides a simple and effective way to visualize and derive sine values.
Understanding sine at special angles helps solve problems involving inverse trigonometric functions, as seen in the given tutorial. If you know that \( \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \), then finding the angle for \( \sin^{-1}(\frac{\sqrt{2}}{2}) \) becomes straightforward because you are just reversing the known result.

Angle Evaluation

Angle evaluation in trigonometry often involves identifying angles based on given trigonometric values. When dealing with inverse trigonometric functions like \( \sin^{-1} \), the goal is to find the specific angle that produces a given sine value.

• The function \( \sin^{-1} \), also called arcsine, takes a value and returns the angle whose sine is that particular value.

Using inverse functions is the reverse of computing sine, allowing one to find angles rather than side lengths in a triangle.
In the context of inverse sine, the range typically lies between \(-\frac{\pi}{2}\) and \( \frac{\pi}{2} \). This range ensures that the function is one-to-one and effectively allows you to evaluate the angle precisely.
In problems like identifying that \( \sin^{-1}(\frac{\sqrt{2}}{2}) = \frac{\pi}{4} \), you apply the knowledge of special angles and the properties of the sine function to find solutions quickly. Evaluating these angles accurately involves memorizing trigonometric values and inverse relationships.