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Chapter 4: Problem 83

In Exercises \(61-86,\) use reference angles to find the exact value of each expression. Do not use a calculator. $$\tan \left(-\frac{17 \pi}{6}\right)$$

Short Answer

Expert verified
The exact value of \(\tan \left(-\frac{17 \pi}{6}\right)\) is \(\frac{\sqrt{3}}{3}\)

Step by step solution

01

Angle Conversion

Start by converting the negative angle into a positive angle. Because every trigonometric function has a period of \(2\pi\), one can add or subtract \(2\pi\) to any angle, and the function鈥檚 value will not change. Therefore, convert \(鈥揬frac{17\pi}{6}\) into a positive angle by adding \(2\pi\) (or \(\frac{12\pi}{6}\)) in terms of sixth multiple to match the denominator, until you land in the interval \(0\) to \(2\pi\). This will give us \(\frac{11\pi}{6}\)
02

Reference Angle Calculation

Next, we need to find the reference angle. The reference angle is always the positive acute angle that a terminal side makes with the x-axis. The negative angle lies in the Fourth Quadrant and its reference angle is given by subtracting the angle from \(2\pi\) or \(\frac{12\pi}{6}\). Thus, the reference angle corresponding to \(\frac{11\pi}{6}\) is \(2\pi - \frac{11\pi}{6} = \frac{1\pi}{6}\)
03

Apply Tangent Function using Reference Angle

Finally, knowing that in the Fourth Quadrant, the tangent function is positive, apply the tangent function to the reference angle to get final result. Hence, \(\tan \left(-\frac{17 \pi}{6}\right) = \tan(\frac{1\pi}{6})\). The standard value of \(\tan(\frac{\pi}{6})\) is known to be \(\frac{\sqrt{3}}{3}\).

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