Question

Find the exact value of each expression. Do not use a calculator. $$ \tan \frac{8 \pi}{3} $$

Short answer

The exact value of \( \tan \frac{8 \pi}{3} \) is \( -\sqrt{3} \).

Step-by-step solution

Understand the problem

The goal is to find the exact value of the trigonometric function \( \tan \frac{8 \pi}{3} \). To solve this without a calculator, first express the angle in a more familiar form.

Convert the angle

The angle \( \frac{8 \pi}{3} \) is larger than \( 2\pi \), so subtract \( 2\pi \) to find a coterminal angle. Perform the subtraction: \( \frac{8 \pi}{3} - 2 \pi = \frac{8 \pi}{3} - \frac{6 \pi}{3} = \frac{2 \pi}{3} \). Thus, \( \tan \frac{8 \pi}{3} = \tan \frac{2 \pi}{3} \).

Determine the reference angle

The angle \( \frac{2 \pi}{3} \) is in the second quadrant. The reference angle for \( \frac{2 \pi}{3} \) can be found by subtracting it from \ \pi \ (180 degrees): \( \pi - \frac{2 \pi}{3} = \frac{3 \pi}{3} - \frac{2 \pi}{3} = \frac{ \pi}{3} \).

Evaluate the tangent function

In the second quadrant, the tangent function is negative. Therefore, \( \tan \frac{2 \pi}{3} = -\tan \frac{ \pi}{3}\). Recall that \(\tan \frac{ \pi}{3} = \sqrt{3} \), so \( \tan \frac{2 \pi}{3} = -\tan \frac{ \pi}{3} = -\sqrt{3} \).

Key concepts

trigonometric identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the variable where both sides of the identity are defined. Some of the most common trigonometric identities include the Pythagorean identities, addition and subtraction formulas, and double-angle formulas. For example, one key identity is \(\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}\). These identities are incredibly useful for simplifying expressions and solving trigonometric equations without a calculator.

reference angles

A reference angle is the acute angle formed between the terminal side of an angle and the x-axis. It is always positive and between 0 and 90 degrees (or 0 and \(\frac{\rect{\footnotesize\(\text{π}\)}}{2}\) radians). Finding the reference angle is a crucial step in solving trigonometric problems because it allows you to use the known values of trigonometric functions for standard angles to determine the function values for any angle. For example, the reference angle for \(\frac{\rect{\footnotesize\(2\)\footnotesize\(\text{π}\)}}{3}\) is found by subtracting this angle from \(\text{π}\), making it \(\rect{\footnotesize\(\text{π}\)} - \rect{\footnotesize\(\text{π}\)}{/ 3} = \rect{\footnotesize\(\text{π}\)/ 3}\). This simplifies solving the original problem.

coterminal angles

Coterminal angles are angles that share the same terminal side but may have different rotations around the unit circle. Essentially, they are angles that differ by a multiple of \(\rect{\footnotesize\(2\)\footnotesize\(\text{π}\)}\) radians or 360 degrees. Finding a coterminal angle often involves adding or subtracting \(\rect{\footnotesize \(2\)\footnotesize\(\text{π}\)}\). This is useful because it allows us to work with angles within a standard 0 to \(\rect{\footnotesize\(2\)\footnotesize\(\text{π}\)}\) range. For example, in the given problem, \(\frac{\rect{\footnotesize\(8\)\footnotesize\(\text{π}\)}}{3}\) is coterminal with \(\frac{\rect{\footnotesize\(8\)\footnotesize\(\text{π}\)}}{3} - \rect{\footnotesize\(2\)\footnotesize\(\text{π}\)} = \frac{\rect{\footnotesize\(2\)\footnotesize\(\text{π}\)}}{3}\). This simplifies solving the trigonometric functions by reducing the angle to a more familiar or simpler form.

unit circle

The unit circle is a circle of radius 1 centered at the origin (0,0) in the coordinate plane. It is a fundamental tool in trigonometry because the coordinates of any point on the circle are cosine and sine values of the corresponding angle. Every angle can be represented on the unit circle, making it easier to visualize and solve trigonometric problems. For example, in the given problem, finding \(\frac{\rect{\footnotesize\(2\)\footnotesize\(\text{π}\)}}{3}\) on the unit circle helps determine the coordinates and thus the sine, cosine, and tangent values. The unit circle shows that in the second quadrant where \(\frac{\rect{\footnotesize\(2\)\footnotesize\(\text{π}\)}}{3}\) lies, the x-coordinate (cosine) is negative while the y-coordinate (sine) is positive, making the tangent (sine/cosine) negative, which aligns with \(-\rect{\footnotesize\(\text{√3}\)}\) in this problem.