Problem 36 - Precalculus Essentials

Question

Involve equations with multiple angles. Solve each equation on the interval \([0,2 \pi)\) $$\cot \frac{3 \theta}{2}=-\sqrt{3}$$

Step-by-Step Solution

Answer

The solutions for \(\theta\) within the interval \([0,2 \pi)\) are \(\theta_1 = \frac{4}{9}\pi\) and \(\theta_2 = \frac{4}{3}\pi\).

Step 1: Identify the Knowns

We know that cotangent is the reciprocal of the tangent, and \(\tan 60^{\circ} = \sqrt{3}\), so \(\cot 60^{\circ} = \frac{1}{\sqrt{3}}\), thus making \(\cot 120^{\circ} = -\frac{1}{\sqrt{3}}\).

Step 2: Solve for the Angle

The equation provided can be equivalently expressed as \(\frac{3\theta}{2} = 120^{\circ} + k \cdot 180^{\circ}\), where \(k\) is an integer. This is because the cotangent function has a period of \(180^{\circ}\), so adding or subtracting multiples of \(180^{\circ}\) will not change its value. In radians, this translates into \(\frac{3\theta}{2} = \frac{2}{3}\pi + k \pi\), where \(\pi\) is approximately 3.142.

Step 3: Solve for theta

We can solve the previous equation for \(\theta\), which gives us \(\theta = \frac{4}{9}\pi + \frac{2}{3}k \pi\).

Step 4: Determine Angles within the Interval

In the interval \([0,2 \pi)\), k can take values of 0, 1, and 2. This leads to the solutions \(\theta_1 = \frac{4}{9}\pi\), \(\theta_2 = \frac{4}{3}\pi\), and \(\theta_3 = \frac{20}{9}\pi\). However, \(\theta_3\) is not within the stated interval, so the valid solutions are \(\theta_1\) and \(\theta_2\).

Key Concepts

Cotangent Function

In trigonometry, the cotangent function, often denoted as \( \cot \theta \), is one of the basic trigonometric functions. It is defined as the reciprocal of the tangent function. In mathematical terms, this means \( \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \).
The cotangent function is particularly useful in various mathematical problems, especially those involving right triangles or trigonometric identities.
Some important characteristics of the cotangent function include:

• It is undefined when the sine of \( \theta \) is zero because division by zero is not allowed. This occurs at angles such as \( \theta = 0, \pi, 2\pi, \) etc.
• The cotangent function has a period of \( \pi \), meaning that it repeats its values every \( \pi \) radians.
• It decreases as \( \theta \) increases, giving it a downward slope when you graph it over a period.

Understanding the properties of the cotangent function helps solve trigonometric equations involving cot such as the one given.

Angle Solutions

Solving trigonometric equations often involves finding angles that satisfy the given conditions. In our exercise, the task is to solve \( \cot \frac{3 \theta}{2} = -\sqrt{3} \) within the interval \([0, 2\pi)\).
To find these angle solutions, we first equate the cotangent to known angle values.
This exercise identifies that \( \cot 120^{\circ} = -\sqrt{3} \), which suggests using angles such as \( 120^{\circ} \) and adding multiples of the angle's period to find additional valid solutions.
Once we solve the equation \( \frac{3\theta}{2} = \frac{2}{3}\pi + k\pi \), converting into \( \theta = \frac{4}{9}\pi + \frac{2}{3}k \pi \), where \( k \) is an integer, helps us to find specific solutions that fit our required interval:

• For \( k = 0 \), \( \theta_1 = \frac{4}{9}\pi \)
• For \( k = 1 \), \( \theta_2 = \frac{4}{3}\pi \)
• For \( k = 2 \), \( \theta_3 = \frac{20}{9}\pi \)

It is crucial to verify each potential solution to ensure it lies within the specified interval of \([0, 2\pi)\). Here both \( \theta_1 \) and \( \theta_2 \) are valid solutions while \( \theta_3 \) exceeds the interval bounds.

Periodic Functions

In the realm of mathematics, periodic functions are those which repeat their values at regular intervals. Trigonometric functions are classical examples of periodic functions.
The cotangent function specifically, repeats every \( \pi \) radians.
This property of periodicity is essential because it allows us to deduce all possible solutions of trigonometric equations over different intervals. If a function is periodic with a certain period, such as \( \pi \) for \( \cot \theta \), this tells us that:

• The function takes the same value at each integer multiple of the period added to a known point on the function.
• This allows compact representation of infinite angle solutions.
• Understanding this concept helps to simplify problem solving by recognizing that it suffices to find solutions within just one period and then replicate them by adding appropriate multiples of the period.

For the given equation \( \cot \frac{3 \theta}{2} = -\sqrt{3} \), understanding the periodicity of the cotangent helps us determine appropriate values for the integer \( k \) to find all valid solutions within the given interval.