Problem 24 - Precalculus Mathematics for Calculus

Question

Find the exact value of the trigonometric function. $$\sin \frac{5 \pi}{3}$$

Step-by-Step Solution

Answer

\(-\frac{\sqrt{3}}{2}\)

Step 1: Determine the Reference Angle

The angle \(\frac{5\pi}{3}\) is in radians. We first convert it to degrees to understand its position on the unit circle. Since \(\pi \) radians is equal to \(180^\circ \), \(\frac{5\pi}{3} = \frac{5 \times 180}{3} = 300^\circ\). The reference angle for \(300^\circ\) is \(360^\circ - 300^\circ = 60^\circ\). The reference angle is \(60^\circ\).

Step 2: Identify the Quadrant

The angle \(300^\circ\) lies in the fourth quadrant because it is between \(270^\circ\) and \(360^\circ\). In the fourth quadrant, the sine of an angle is negative.

Step 3: Calculate the Sine of the Reference Angle

The reference angle is \(60^\circ\), which is a standard angle. The sine of \(60^\circ\) is \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\).

Step 4: Apply the Sign Based on the Quadrant

Since \(300^\circ\) is in the fourth quadrant and sine is negative in the fourth quadrant, the sine of \(\frac{5\pi}{3}\) is \[\sin \frac{5\pi}{3} = -\sin 60^\circ = -\frac{\sqrt{3}}{2} \].

Key Concepts

Unit Circle

The unit circle is a fundamental tool in trigonometry. It represents all the angles with a radius of 1. This circle helps us find trigonometric functions for different angles. The circle has a center at the origin (0,0) on the Cartesian plane. An important feature is that any point on the unit circle has coordinates

• (\(\cos\theta\), \(\sin\theta\))

where \(\theta\) is the angle formed with the positive x-axis.
The entire circle is divided into 360 degrees or 2\(\pi\) radians. Understanding angles and their corresponding coordinates on this circle is key to using trigonometric functions. The unit circle also helps to easily determine the sign of trigonometric values, thanks to its division into quadrants.
Knowing the unit circle means you can quickly find sine, cosine, and tangent function values at various angles without needing a calculator.

Reference Angle

A reference angle is the acute angle formed by the terminal side of the given angle

• in standard position and the x-axis

. It is always between 0 and 90 degrees. To determine a reference angle for an angle given in radians or degrees:

• If the angle is in the first quadrant, the reference angle is itself.
• For angles in the second quadrant, subtract the angle from \(180^\circ\).
• In the third quadrant, subtract \(180^\circ\) from the angle.
• For the fourth quadrant, subtract the angle from \(360^\circ\).

As an example, for an angle of \(300^\circ\), the reference angle is \(60^\circ\). Knowing reference angles helps in simplifying the process of finding trigonometric values for various angles.

Quadrant Analysis

Quadrant analysis is essential to determine the sign of trigonometric functions like sine and cosine. The coordinate plane is divided into four quadrants:

• First Quadrant: angles between \(0^\circ\) and \(90^\circ\). All trigonometric functions are positive.
• Second Quadrant: angles between \(90^\circ\) and \(180^\circ\). Sine is positive, cosine is negative.
• Third Quadrant: angles between \(180^\circ\) and \(270^\circ\). Tangent is positive, sine and cosine are negative.
• Fourth Quadrant: angles between \(270^\circ\) and \(360^\circ\). Cosine is positive, sine is negative.

Knowing which quadrant an angle lies in helps with understanding where a trigonometric function will be positive or negative. In our specific case, \(300^\circ\) lies in the fourth quadrant, which means the sine of the angle will be negative.

Sine Function

The sine function is one of the basic trigonometric functions, defining the y-coordinate of a point on the unit circle. For an angle \(\theta\),

• sine is denoted as \(\sin(\theta)\)

. The sine function cycles between -1 and 1 and is periodic, repeating every \(360^\circ\) or \(2\pi\) radians. It is important to know the values of sine for standard angles, such as \(30^\circ\), \(45^\circ\), and \(60^\circ\).
In our problem, we calculated that the \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\). However, since \(\frac{5\pi}{3}\) is in the fourth quadrant, the sine becomes negative, i.e., \(-\frac{\sqrt{3}}{2}\). Understanding the sine function and its properties allows for quick and easy solutions to many trigonometric problems.