Problem 11 - Algebra and Trigonometry with Analytic Geometry

Question

Find the exact value. (a) \(\tan (5 \pi / 6)\) (b) \(\tan (-\pi / 3)\)

Step-by-Step Solution

Answer

(a) \( -\frac{\sqrt{3}}{3} \), (b) \( -\sqrt{3} \).

Step 1: Understanding the angle for \( \tan (\frac{5\pi}{6}) \)

The angle \( \frac{5\pi}{6} \) is in the second quadrant. In the second quadrant, the tangent of an angle is negative. The reference angle is \( \pi - \frac{5\pi}{6} = \frac{\pi}{6} \).

Step 2: Finding \( \tan (\frac{\pi}{6}) \)

The tangent of \( \frac{\pi}{6} \) is known from trigonometric identities: \( \tan (\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \).

Step 3: Applying the sign for \( \tan (\frac{5\pi}{6}) \)

Since \( \frac{5\pi}{6} \) is in the second quadrant, \( \tan (\frac{5\pi}{6}) \) will be negative. Therefore, \( \tan (\frac{5\pi}{6}) = -\frac{1}{\sqrt{3}} \).

Step 4: Simplifying \( -\frac{1}{\sqrt{3}} \)

To rationalize the denominator, multiply by \( \frac{\sqrt{3}}{\sqrt{3}} \):\[ -\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \]. So, \( \tan(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{3} \).

Step 5: Understanding the angle for \( \tan (-\frac{\pi}{3}) \)

The angle \( -\frac{\pi}{3} \) can be interpreted as moving \( \frac{\pi}{3} \) downwards from 0, placing the angle in the fourth quadrant. In the fourth quadrant, the tangent is negative.

Step 6: Finding \( \tan (\frac{\pi}{3}) \)

The tangent of \( \frac{\pi}{3} \) is \( \sqrt{3} \) from trigonometric identities.

Step 7: Applying the sign for \( \tan (-\frac{\pi}{3}) \)

Since \( -\frac{\pi}{3} \) is in the fourth quadrant where tangent is negative, \( \tan(-\frac{\pi}{3}) = -\sqrt{3} \).

Key Concepts

Tangent Function

The tangent function, often denoted as \( \tan \theta \), is one of the primary trigonometric functions. It relates the angles of a right triangle to the ratios of two of its sides. Specifically, it is the ratio of the length of the opposite side to the length of the adjacent side in a right-angled triangle. Mathematically, it is represented as:- \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \).In terms of the unit circle, the tangent function can also be defined using sine and cosine:- \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).This function has a periodic nature, repeating every \( \pi \) radians, or 180 degrees. It's important to note that the tangent function is undefined at angles where the cosine is zero, such as \( \theta = \frac{\pi}{2} \) or multiples of it, because division by zero is undefined.

Quadrants of the Unit Circle

The unit circle is a fundamental concept in trigonometry, providing a way to understand angle measures that go beyond right triangles. It is a circle with radius 1 centered at the origin of a coordinate plane.The circle is divided into four quadrants:- **First Quadrant (0 to \( \frac{\pi}{2} \)):** Both sine and cosine are positive, so tangent is also positive.- **Second Quadrant (\( \frac{\pi}{2} \) to \( \pi \)):** Sine is positive, cosine is negative, making tangent negative.- **Third Quadrant (\( \pi \) to \( \frac{3\pi}{2} \)):** Both sine and cosine are negative, so tangent is positive.- **Fourth Quadrant (\( \frac{3\pi}{2} \) to \( 2\pi \)):** Sine is negative and cosine is positive, making tangent negative.Understanding which quadrant an angle falls into allows us to determine the sign of the tangent function. For angles like \( \frac{5\pi}{6} \) and \( -\frac{\pi}{3} \) mentioned in the problem, knowledge of quadrants helps us immediately know that tangent will be negative.

Reference Angles

Reference angles are a helpful tool in trigonometry to simplify the evaluation of trigonometric functions for angles outside the typical range of 0 to \( \frac{\pi}{2} \) radians. A reference angle is the smallest angle between the terminal side of the given angle and the x-axis. They are always positive and between 0 and \( \frac{\pi}{2} \).To find the reference angle if the given angle is in:- **Second Quadrant:** Use \( \text{Reference Angle} = \pi - \theta \).- **Third Quadrant:** Use \( \text{Reference Angle} = \theta - \pi \).- **Fourth Quadrant:** Use \( \text{Reference Angle} = 2\pi - \theta \).In the context of our problem, the reference angle for \( \frac{5\pi}{6} \) is \( \frac{\pi}{6} \), giving a simpler way to compute \( \tan \theta \); knowing that the tangent of \( \frac{\pi}{6} \) is \( \frac{1}{\sqrt{3}} \), we adapted this with the correct sign based on quadrant location.

Rationalizing Denominators

Rationalizing the denominator is a technique used to eliminate radicals from the bottom of a fraction. This process makes expressions easier to work with and is often required for final answers in trigonometry and mathematics in general.Here’s the general process for rationalizing a denominator:- If you have a fraction like \( \frac{a}{\sqrt{b}} \), you multiply both numerator and denominator by \( \sqrt{b} \) to get \( \frac{a\sqrt{b}}{b} \).For instance, in the exercise solution, we had \( -\frac{1}{\sqrt{3}} \), and multiplying by \( \frac{\sqrt{3}}{\sqrt{3}} \) simplifies this to \( -\frac{\sqrt{3}}{3} \).This process not only gives a cleaner expression but also aligns with many mathematical conventions and standards that prefer radical-free denominators.