Problem 3 - Precalculus with Calculus Previews

Question

Find the indicated value without the use of a calculator. $$ \cot \frac{13 \pi}{6} $$

Step-by-Step Solution

Answer

\( \sqrt{3} \)

Step 1: Determine Reference Angle

First, recognize that the angle \( \frac{13\pi}{6} \) is larger than \( 2\pi \). To find the reference angle, subtract \( 2\pi \) from \( \frac{13\pi}{6} \).\[ \frac{13\pi}{6} - 2\pi = \frac{13\pi}{6} - \frac{12\pi}{6} = \frac{\pi}{6} \]So, the reference angle is \( \frac{\pi}{6} \).

Step 2: Identify the Quadrant

Since \( \frac{13\pi}{6} \) is equivalent to \( \frac{\pi}{6} \) less one full circle (which is \( 2\pi \)), it lands in the 1st quadrant where all trigonometric functions have the same sign as their reference angles.

Step 3: Evaluate the Cotangent Function

We need \( \cot \frac{\pi}{6} \). Recall that \( \cot \theta = \frac{1}{\tan \theta} \) and \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \). Thus, the cotangent is:\[ \cot \frac{\pi}{6} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \]Thus, \( \cot \frac{13\pi}{6} = \sqrt{3} \).

Key Concepts

Reference Angle

The concept of a reference angle is a crucial component in trigonometry, especially when dealing with angles larger than a full circle (i.e., greater than \(2\pi\) radians or 360 degrees). A reference angle allows us to find equivalent angles within the first cycle of a unit circle. To compute the reference angle for \(\frac{13\pi}{6}\), recognize that this angle exceeds \(2\pi\). Therefore, subtract \(2\pi\) from \(\frac{13\pi}{6}\), the calculation of which is:

• \(\frac{13\pi}{6} - 2\pi = \frac{13\pi}{6} - \frac{12\pi}{6} = \frac{\pi}{6}\).

Thus, the reference angle is \(\frac{\pi}{6}\).
This means that \(\frac{13\pi}{6}\) and \(\frac{\pi}{6}\) form an acute angle with the x-axis, and have the same sine, cosine, and tangent values as if it were within the first cycle, given the quadrant's sign considerations.

Cotangent Function

Understanding the cotangent function is important, particularly when solving for trigonometric ratios. The cotangent is defined as the reciprocal of the tangent function:

• \(\cot \theta = \frac{1}{\tan \theta}\)

In this exercise, we need \(\cot \frac{\pi}{6}\). First, recall that \(\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\). Thus, using the reciprocal relationship:

• \(\cot \frac{\pi}{6} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3}\).

So, if asked to find \(\cot \frac{13\pi}{6}\), we can determine that it is equivalent to \(\cot \frac{\pi}{6}\) because they share the same reference angle and reside in a quadrant where their sign is positive. Hence, the solution is \(\sqrt{3}\).

Trigonometric Quadrants

Each trigonometric function's sign depends on the quadrant in which its terminal side lies. The standard unit circle is divided into four quadrants:

• 1st Quadrant (0 to \(\pi/2\)), where all trigonometric functions are positive.
• 2nd Quadrant (\(\pi/2\) to \(\pi\)), where sine is positive.
• 3rd Quadrant (\(\pi\) to \(3\pi/2\)), where tangent is positive.
• 4th Quadrant (\(3\pi/2\) to \(2\pi\)), where cosine is positive.

Here, \(\frac{13\pi}{6}\) or equivalently \(\frac{\pi}{6}\), lands in the 1st quadrant after one complete rotation. In the 1st quadrant, each trigonometric function, including the cotangent, holds the same positive sign as its reference angle.
Therefore, when a problem asks for the cotangent of \(\frac{13\pi}{6}\), we can confidently apply the same sign and result as \(\cot \frac{\pi}{6}\), confirming it to be \(\sqrt{3}\).