Problem 32 - Calculus Single Variable

Question

For each \(\theta\) in \(\\{0, \pi / 6, \pi / 4, \pi / 3, \pi / 2\\},\) find an integer value of \(n\) such that \(\sin (\theta)=\sqrt{n} / 2\) (The pattern found in this exercise is sometimes used as a memory aid.)

Step-by-Step Solution

Answer

The values of \( n \) are 0, 1, 2, 3, and 4 for \( \theta = 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} \), respectively.

Step 1: Understanding the Problem

The problem requires us to find an integer value of \( n \) such that \( \sin(\theta) = \frac{\sqrt{n}}{2} \) for given angles \( \{ 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} \} \). We need to use known values of the sine function for these angles.

Step 1: Calculate \( \sin(0) \)

For \( \theta = 0 \), we have \( \sin(0) = 0 \). Matching this with \( \frac{\sqrt{n}}{2} \), we get \( \frac{\sqrt{n}}{2} = 0 \). Therefore, \( \sqrt{n} = 0 \), so \( n = 0 \).

Step 2: Calculate \( \sin(\pi/6) \)

For \( \theta = \frac{\pi}{6} \), we have \( \sin(\frac{\pi}{6}) = \frac{1}{2} \). Matching this with \( \frac{\sqrt{n}}{2} \), we have \( \frac{\sqrt{n}}{2} = \frac{1}{2} \). Therefore, \( \sqrt{n} = 1 \), so \( n = 1 \).

Step 3: Calculate \( \sin(\pi/4) \)

For \( \theta = \frac{\pi}{4} \), we have \( \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \). Matching this with \( \frac{\sqrt{n}}{2} \), we have \( \frac{\sqrt{n}}{2} = \frac{\sqrt{2}}{2} \). Thus, \( \sqrt{n} = \sqrt{2} \) and \( n = 2 \).

Step 4: Calculate \( \sin(\pi/3) \)

For \( \theta = \frac{\pi}{3} \), we have \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \). Matching this with \( \frac{\sqrt{n}}{2} \), we have \( \frac{\sqrt{n}}{2} = \frac{\sqrt{3}}{2} \). So, \( \sqrt{n} = \sqrt{3} \) and \( n = 3 \).

Step 5: Calculate \( \sin(\pi/2) \)

For \( \theta = \frac{\pi}{2} \), we have \( \sin(\frac{\pi}{2}) = 1 \). Matching this with \( \frac{\sqrt{n}}{2} \), we have \( \frac{\sqrt{n}}{2} = 1 \). Thus, \( \sqrt{n} = 2 \) and \( n = 4 \).

Key Concepts

The Sine Function Explained

The sine function is a fundamental concept in trigonometry that helps us understand the relationship between angles and side lengths in right triangles. The sine of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse in a right-angled triangle. Mathematically, we denote this as:

• \( ext{sin}( heta) = \frac{ ext{Opposite side}}{ ext{Hypotenuse}} \)

This function is periodic and oscillates between -1 and 1. It is important to note that the sine function has specific values at commonly used angles such as 0, \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\), and \(\frac{\pi}{2}\). These values are crucial in solving many trigonometric problems and can often be expressed using square roots and integers for simplicity.

Understanding Angle Values in Radians

In trigonometry, angles are often measured in radians instead of degrees. The radian measure is based on the radius of a circle, making it a natural way to describe angles arising from real-world phenomena. For instance:

• \( 0 \text{ radians} = 0^\circ \)
• \( \frac{\pi}{6} \text{ radians} = 30^\circ \)
• \( \frac{\pi}{4} \text{ radians} = 45^\circ \)
• \( \frac{\pi}{3} \text{ radians} = 60^\circ \)
• \( \frac{\pi}{2} \text{ radians} = 90^\circ \)

These radian values correspond to significant points on the unit circle where the sine function takes on predictable values. Utilizing radians helps in simplifying calculations, especially when working with periodic functions like sine and cosine.

Memory Aid for Trigonometric Values

Memorizing trigonometric values for standard angles can be challenging, but thankfully, there are clever memory aids that simplify this task. The pattern highlighted in the exercise is an excellent example:

• For \( \text{sin}(0) \), the value is \( \frac{0}{2} \). Hence, \( n = 0 \).
• For \( \text{sin}(\frac{\pi}{6}) \), the value is \( \frac{1}{2} \). Thus, \( n = 1 \).
• For \( \text{sin}(\frac{\pi}{4}) \), the value is \( \frac{\sqrt{2}}{2} \). Therefore, \( n = 2 \).
• For \( \text{sin}(\frac{\pi}{3}) \), the value is \( \frac{\sqrt{3}}{2} \). So, \( n = 3 \).
• For \( \text{sin}(\frac{\pi}{2}) \), the value is \( 1 \). Hence, \( n = 4 \).

Try to remember this pattern as using these particular values with their respective square roots makes calculating sine much easier.

Importance of Square Roots in Trigonometry

Square roots frequently appear in trigonometry, especially in problems involving the sine and cosine functions of standard angles. Here's why they are so vital:

• They allow us to express the ratio of side lengths in a triangle in the simplest form.
• Square roots help in identifying neat and simple expressions for trigonometric values, improving calculation accuracy and understanding.

For example, the sine values calculated in the exercise make use of square roots, as it becomes necessary when representing the side ratios for certain angles on the unit circle. Understanding the square root concept, therefore, broadens our ability to tackle a wide range of trigonometric problems efficiently.