In Exercises 61–86, use reference angles to find the exact value of each expression. Do not use a calculator. tan (-17𝜋/6)

Verified step by step guidance

1. Identify the given angle: \(-\frac{17\pi}{6}\). Since it is negative, we will find a positive coterminal angle by adding \(2\pi\) multiples until the angle is between \(0\) and \(2\pi\).
2. Add \(2\pi\) (which is \(\frac{12\pi}{6}\)) to \(-\frac{17\pi}{6}\) to find a positive coterminal angle: \(-\frac{17\pi}{6} + \frac{12\pi}{6} = -\frac{5\pi}{6}\). Since this is still negative, add \(2\pi\) again: \(-\frac{5\pi}{6} + \frac{12\pi}{6} = \frac{7\pi}{6}\).
3. Now, \(\frac{7\pi}{6}\) is between \(0\) and \(2\pi\), so the reference angle is the acute angle between \(\frac{7\pi}{6}\) and the nearest x-axis multiple. Since \(\frac{7\pi}{6}\) is in the third quadrant, the reference angle is \(\frac{7\pi}{6} - \pi = \frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi}{6}\).
4. Recall that \(\tan(\theta)\) is positive in the third quadrant, so \(\tan\left(\frac{7\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right)\) with a positive sign.
5. Use the exact value of \(\tan\left(\frac{\pi}{6}\right)\), which is \(\frac{1}{\sqrt{3}}\), to write the exact value of \(\tan\left(-\frac{17\pi}{6}\right)\) as \(\frac{1}{\sqrt{3}}\).

Key Concepts

Reference Angles

A reference angle is the acute angle formed between the terminal side of an angle and the x-axis. It helps simplify trigonometric calculations by relating any angle to a corresponding angle in the first quadrant, where trigonometric values are well-known.

Angle Coterminality and Reduction

Angles differing by full rotations (multiples of 2π) share the same terminal side and thus the same trigonometric values. Reducing an angle by subtracting or adding 2π simplifies the angle to an equivalent one within the standard interval [0, 2π) for easier evaluation.

Tangent Function Properties

The tangent function is periodic with period π and is defined as tan(θ) = sin(θ)/cos(θ). Its sign depends on the quadrant of the angle, being positive in the first and third quadrants and negative in the second and fourth, which is crucial when determining the exact value using reference angles.