Problem 103 - Algebra and Trigonometry

Question

find two values of \(\theta, 0 \leq \theta<2 \pi,\) that satisfy each equation. $$ \tan \theta=-\sqrt{3} $$

Step-by-Step Solution

Answer

The two values of \(\theta\) that satisfy the equation \(\tan \theta = -\sqrt{3}\) in the interval \(0 \leq \theta<2 \pi\) are \(4\pi/3\) and \(5\pi/3\).

Step 1: Determine the related acute angle

First, let's recall that the tangent of \(\pi/3\) or \(60^\circ\) is \(\sqrt{3}\). Since our equation involves a negative sign, our related acute angle is \(\pi/3\) or \(60^\circ\). So \(\theta_a = \pi/3\).

Step 2: Find theta in the third quadrant

In the third quadrant, the tangent is positive, so to make the tangent negative, we have to take the supplementary angle of the related acute angle. The supplementary angle from the third quadrant is \(\pi + \pi/3 = 4\pi/3\). So, the possible value of \(\theta\) in the third quadrant is \(\theta_1 = 4\pi/3\).

Step 3: Find theta in the fourth quadrant

In the fourth quadrant, the tan value is also positive. So to make it negative, we need to subtract our related acute angle from \(\pi \times 2\). This gives us \(\theta = 2\pi - \pi/3 = 5\pi/3\). So, the possible value of \(\theta_2\) is \(5\pi/3\).

Key Concepts

Tangent Function

The tangent function is an essential concept in trigonometry and calculus that relates the angles of a right triangle to the ratio of its opposite and adjacent sides. More specifically, for an angle \( \theta \) in a right triangle, the tangent function is defined as \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). It is also represented as the slope of the line generated by the angle on the unit circle.

Unlike sine and cosine, the tangent function can take any real number value, meaning it can be positive or negative and is not restricted to a range of -1 to 1. Tangent values are positive in the first and third quadrants of the unit circle, where both sine and cosine have the same sign, and negative in the second and fourth quadrants, where sine and cosine have opposite signs.

Understanding the sign of the tangent function in different quadrants helps solve equations such as \( \tan \theta = -\sqrt{3} \), as seen in the exercise. In this scenario, finding the angle \( \theta \) that results in a negative tangent value involves knowledge of the unit circle and the behavior of the tangent function in its quadrants.

Quadrants of the Unit Circle

The quadrants of the unit circle are a fundamental aspect when solving trigonometric equations because each quadrant provides important information about the signs of trigonometric functions. The unit circle is divided into four quadrants by the x-axis and y-axis:

First Quadrant (Q1):

• All trigonometric functions (sine, cosine, and tangent) are positive in this quadrant.

Second Quadrant (Q2):

• Sine is positive, while cosine and tangent are negative.

Third Quadrant (Q3):

• Tangent and sine are positive, while cosine is negative.

Fourth Quadrant (Q4):

• Cosine is positive, while sine and tangent are negative.

For the given problem, the negative tangent value indicates that the solutions must be found in either the second or fourth quadrant where the tangent function is negative. However, since the related acute angle is in the first quadrant, we must adjust the angle measurements according to the proper quadrant to find the solutions for \( \theta \).

Related Acute Angle

The concept of a 'related acute angle' is key to solving trigonometric equations. A related acute angle is an angle less than \( 90^\circ \) (or \( \pi/2 \) radians) that can provide insight into the trigonometric function values of larger angles.

In the context of the given problem, the angle whose tangent is \( \sqrt{3} \) is known to be \( \pi/3 \) or \( 60^\circ \) – this is our related acute angle. Whenever we encounter the tangent function with a negative value, we can use the related acute angle to help us find the angle \( \theta \) in the other quadrants. By adding \( \pi \) to the related acute angle, we can find a corresponding angle in the third quadrant where tangent is positive. Conversely, subtracting the acute angle from \( 2\pi \) leads us to the appropriate angle in the fourth quadrant.

Identifying the related acute angle is the first step in solving such trigonometric equations. Then through symmetry and understanding the properties of the tangent values in different quadrants, we can determine the possible values of \( \theta \).