Problem 24 - Algebra and Trigonometry

Question

\(17-34\) . An equation is given. (a) Find all solutions of the equation. (b) Find the solutions in the interval \([0,2 \pi) .\) $$ \tan \frac{\theta}{4}+\sqrt{3}=0 $$

Step-by-Step Solution

Answer

The solution is \( \theta = \frac{4\pi}{3} \).

Step 1: Isolate the Tangent Expression

Start with the equation \( \tan \frac{\theta}{4} + \sqrt{3} = 0 \). Subtract \( \sqrt{3} \) from both sides to isolate the tangent expression: \( \tan \frac{\theta}{4} = -\sqrt{3} \).

Step 2: Identify the General Solution for Tangent

Recall that the tangent function gives \( \tan x = -\sqrt{3} \) at angles corresponding to \( x = \arctan(-\sqrt{3}) + k\pi \), where \( k \) is any integer. For the unit circle, \( \tan(x) = -\sqrt{3} \) at angles \( x = -\frac{\pi}{3} \) and \( x = \frac{2\pi}{3} \), plus all periodic additions of \( \pi \).

Step 3: Solve for Theta

From the tangent function solution, \( \frac{\theta}{4} = -\frac{\pi}{3} + k\pi \) and \( \frac{\theta}{4} = \frac{2\pi}{3} + k\pi \). Multiply each equation by 4 to solve for \( \theta \):1. \( \theta = -\frac{4\pi}{3} + 4k\pi \)2. \( \theta = \frac{8\pi}{3} + 4k\pi \)

Step 4: Apply the Interval Restriction

We need to find solutions for \( \theta \) within the interval \([0, 2\pi)\). - For \( \theta = -\frac{4\pi}{3} + 4k\pi \), plug in \( k = 1 \) to get \( \theta = 8\pi/3 - 4\pi/3 = 4\pi/3 \), which is in the interval. - For \( \theta = \frac{8\pi}{3} + 4k\pi \), \( \theta \) already simplifies to \( \frac{8\pi}{3} \) which is outside the interval. Instead, use \( k = -1 \) to get \( \theta = \frac{8\pi}{3} - 4\pi \ =4\pi/3 -2\pi =-2\pi/3+2 \pi=4\pi/3\).

Step 5: Final Solution in the Interval

Both sub-expressions lead to \( \theta = \frac{4\pi}{3} \) in the interval \([0, 2\pi)\). So the final solution is \( \theta = 4\pi/3 \).

Key Concepts

Tangent Function

The tangent function, typically denoted as \( \tan(x) \), is one of the fundamental trigonometric functions. It relates to the sine and cosine functions as follows: \( \tan(x) = \frac{\sin(x)}{\cos(x)} \). This function can be derived from the unit circle and is known for its periodic and unbounded nature.

• The tangent function has a period of \( \pi \), meaning it repeats every \( \pi \) units.
• It is undefined where \( \cos(x) = 0 \), which is at odd multiples of \( \frac{\pi}{2} \).

The tangent function is usually used to solve right-angle triangles or to convert an angle measure into a ratio. In equations, it often leads to a recurring pattern, assisting us with determining general solutions that include a term \( k\pi \), where \( k \) is an integer. Understanding the behavior of the tangent function on the unit circle gives us valuable insights for solving trigonometric equations.

Unit Circle

The unit circle is a crucial concept in trigonometry, a perfect circle with a radius of one centered at the origin of a coordinate plane. It provides an intuitive way to define and understand trigonometric functions.

• Each point on the unit circle has coordinates that correspond to \( (\cos(\theta), \sin(\theta)) \).
• The tangent function is represented by the slope of the line from the origin to a point on the circle, which can be expressed as \( \frac{\sin(\theta)}{\cos(\theta)} \).

On the unit circle, the function \( \tan(\theta) \) will equal \(-\sqrt{3}\) at angles \( -\frac{\pi}{3} \) and \( \frac{2\pi}{3} \). These angles each correspond to specific points on the circle, reflecting the repetitive and symmetric properties of trigonometric functions. Using the unit circle helps visualize when and why certain angles solve our trigonometric equations efficiently.

Arctan Function

The arctan function, also known as the inverse tangent function, is denoted by \( \arctan(x) \) or sometimes \( \tan^{-1}(x) \). It is the inverse process of the tangent function, taking a ratio and outputting an angle.

• \( \arctan(x) \) returns an angle whose tangent is \( x \).
• The basic range of \( \arctan \) is \((-\frac{\pi}{2}, \frac{\pi}{2})\), meaning it only yields angles in this range.

Using the arctan function can help us work backwards from a tangent value to find the angle that produced it. In our exercise, \( \tan(x) = -\sqrt{3} \) leads us to use the arctan function to decide which angles are solutions. Arctan is particularly helpful in calculations that need an exact angle measure, as it directly helps in determining the principal value of the angle in the standard trigonometric circles.