Problem 11 - Precalculus with Calculus Previews

Question

Find the exact value of the given trigonometric expression. Do not use a calculator. $$ \arctan \left(-\frac{\sqrt{3}}{3}\right) $$

Step-by-Step Solution

Answer

The value is \(-\frac{\pi}{6}\).

Step 1: Understand the Problem

We are tasked with finding the angle whose tangent is \(-\frac{\sqrt{3}}{3}\). The angle should lie within the range of the arctangent function, which is \((-\frac{\pi}{2}, \frac{\pi}{2})\).

Step 2: Recall the Tangent Values

Think of the special angles and their tangent values on the unit circle. Recall that \(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\), \(\tan\left(\frac{\pi}{4}\right) = 1\), and \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\).

Step 3: Simplify the Expression

Recognize that \(\tan\left(-\frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}\). Rewrite \(-\frac{1}{\sqrt{3}}\) as \(-\frac{\sqrt{3}}{3}\), which matches the given expression.

Step 4: Identify the Angle

The angle whose tangent is \(-\frac{\sqrt{3}}{3}\) is \(-\frac{\pi}{6}\), since \(-\frac{\pi}{6}\) lies in the range \((-\frac{\pi}{2}, \frac{\pi}{2})\) of the arctangent function.

Key Concepts

Understanding Special Angles

Special angles are particular angles often used in trigonometry due to their well-known sine, cosine, and tangent values. These angles typically include 30° \((\frac{\pi}{6})\), 45° \((\frac{\pi}{4})\), and 60° \((\frac{\pi}{3})\). These angles correspond to common

• right triangles
• equilateral triangles
• isoceles triangles

where the relationships between their sides yield exact trigonometric values.
This makes them extremely useful when evaluating trigonometric functions without a calculator. For instance, the tangent of 30° or \(\frac{\pi}{6}\) is \(\frac{1}{\sqrt{3}}\), essential for solving trigonometric expressions like the one in our problem.
In this exercise, knowing the special angle relationships prompted us to recognize the pattern in the expression \(-\frac{\sqrt{3}}{3}\) and match it to the negative tangent values of a special angle.

Exploring the Unit Circle

The unit circle is a circle with a radius of 1, centered at the origin of a coordinate plane. It is a valuable tool in trigonometry as it visually connects angles with their corresponding sine, cosine, and tangent values. An angle's position on the unit circle is determined by its measure in radians, starting from the positive x-axis and moving counterclockwise.
On the unit circle:

• the x-coordinate of a point gives the cosine of the angle
• the y-coordinate gives the sine
• the tangent is the y-coordinate divided by the x-coordinate

This makes it easy to see why \(\tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}\) works out as it does: \(\tan\theta = \frac{sin\theta}{cos\theta}\), and by using known sine and cosine values for the angle \(\theta = -\frac{\pi}{6}\), we have our solution.

Evaluating Tangent Values

The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side. On the unit circle, tangent values are derived from the coordinates, specifically, they can be calculated as \(\frac{y}{x}\), where \(y\) and \(x\) are the y and x coordinates, respectively. In this context, knowing the tangent values of special angles, such as \(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\), is crucial.
These specific values help identify angles easily without needing complex calculations.

• For \(\frac{\pi}{6}\), it is \(\frac{1}{\sqrt{3}}\)
• For \(\frac{\pi}{4}\), it is 1
• For \(\frac{\pi}{3}\), it is \(\sqrt{3}\)

Adjusting these values to accommodate negative angles, like \(\tan(-\frac{\pi}{6})\), gives \(-\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}\), helping us solve for exact values.

Trigonometric Functions Basics

Trigonometric functions are mathematical functions that relate angles to side ratios in right triangles. They include sine, cosine, tangent, and their reciprocals — cosecant, secant, and cotangent.

• Sine and cosine oscillate between -1 and 1
• Tangent has no upper bound, as it approaches infinity when its respective angles approach \(\frac{\pi}{2}\) plus multiples of \(\pi\)

In trigonometry, these functions convert between angle measures and lengths of triangle sides.
For the given exercise, understanding arctan or \(\tan^{-1}\) is essential. It helps reverse the tangent function to determine the angle when a tangent value is provided. Given \(-\frac{\sqrt{3}}{3}\), using arctan helps find \(-\frac{\pi}{6}\), the angle in the correct range.