Problem 141 - Precalculus

Question

For the following exercises, find the exact value, if possible, without a calculator. If it is not possible, explain why. $$ \cos \left(\sin ^{-1}\left(\frac{1}{2}\right)\right) $$

Step-by-Step Solution

Answer

The exact value is \( \frac{\sqrt{3}}{2} \).

Step 1: Understand the Problem

We need to find the value of \( \cos \left(\sin ^{-1}\left(\frac{1}{2}\right)\right) \). This involves determining the cosine of an angle whose sine value is \( \frac{1}{2} \).

Step 2: Determine \( \theta \) from \( \sin^{-1}\)

We know that \( \sin \theta = \frac{1}{2} \). By definition of inverse sine, \( \theta = \sin^{-1}\left(\frac{1}{2}\right) \) is the angle whose sine is \( \frac{1}{2} \), and this angle should be in the range of \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\).

Step 3: Recall standard angle values

From trigonometry, we know that \( \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \). Thus, \( \theta = \frac{\pi}{6} \) is a possible solution, as it is in the proper range for \( \sin^{-1} \).

Step 4: Apply the Cosine Function

Since \( \theta = \frac{\pi}{6} \), we need to find \( \cos \left(\frac{\pi}{6}\right) \).

Step 5: Calculate \( \cos \left(\frac{\pi}{6}\right) \)

From standard trigonometric values, \( \cos \left(\frac{\pi}{6}\right) = \sqrt{3}/2 \). Thus, the value of \( \cos \left(\sin^{-1}\left(\frac{1}{2}\right)\right) = \sqrt{3}/2 \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are useful tools in mathematics that allow us to find angles when given trigonometric ratios. In this exercise, we have the expression \( \sin^{-1}\left(\frac{1}{2}\right) \), which is an example of an inverse sine function. This notation can be a bit confusing, as it does not mean dividing sine by 2.

Instead, \( \sin^{-1}\left(\frac{1}{2}\right) \) asks: "What angle has a sine of \( \frac{1}{2} \)?" The inverse sine function is defined to return values within the interval \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\). In this interval, it allows us to pinpoint the exact angle corresponding to the given sine value. This property is key to solving inverse trigonometric problems without a calculator. In our particular case, this leads us to the angle \( \frac{\pi}{6} \).

Trigonometric Functions

Trigonometric functions, like sine and cosine, are fundamental in right triangle geometry and the study of periodic phenomena. They help relate angles to side lengths in right triangles. In this context, we need to understand how cosine relates to the angle we identified from the inverse sine function.

Once we have the angle \( \theta = \frac{\pi}{6} \), which corresponds to \( \sin \theta = \frac{1}{2} \), we can easily apply this to find its cosine value. The cosine of an angle \( \theta \) in a right-angled triangle is defined as the ratio of the adjacent side to the hypotenuse. From trigonometric tables or knowledge, we know \( \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \). This step exemplifies how different trigonometric functions can interrelate to solve problems involving angular measurements.

Exact Values in Trigonometry

Exact values in trigonometry are derived from the properties of special angles and the unit circle. They provide precise ratio values for certain angles without the need for extensive calculations or approximations. Special angles like \( \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3} \) have well-known sine, cosine, and tangent values that can be memorized or quickly referenced.

For instance, \( \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \) and \( \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \). These values are part of the foundational trigonometric identities frequently used in calculus and geometry. Understanding exact values is crucial in quickly solving trigonometric equations and understanding transformations without the constant need for computation aids. It not only boosts calculation efficiency but also deepens the comprehension of trigonometric relationships.