Problem 40 - Algebra and Trigonometry

Question

\(39-56 \approx\) Solve the given equation. $$ (\tan \theta-2)\left(16 \sin ^{2} \theta-1\right)=0 $$

Step-by-Step Solution

Answer

The solutions are \(\theta = \tan^{-1}(2) + n\pi\) and \(\theta = \sin^{-1}(\pm \frac{1}{4}) + k\pi\), where \(n, k\) are integers.

Step 1: Set Each Factor to Zero

The equation is a product set to zero: \[( an \theta - 2)(16 \sin^2 \theta - 1) = 0\]Set each factor to zero separately:1. \(\tan \theta - 2 = 0\)2. \(16 \sin^2 \theta - 1 = 0\)

Step 2: Solve First Factor

Solve \(\tan \theta - 2 = 0\):Add 2 to both sides to get \(\tan \theta = 2\).Since \(\tan \theta = 2\), \(\theta = \tan^{-1}(2)\).

Step 3: Solve Second Factor

Solve \(16 \sin^2 \theta - 1 = 0\):Add 1 to both sides: \(16 \sin^2 \theta = 1\).Divide by 16: \(\sin^2 \theta = \frac{1}{16}\).Take the square root of both sides: \(\sin \theta = \pm \frac{1}{4}\).Thus, \(\theta = \sin^{-1}\left(\frac{1}{4}\right)\) or \(\theta = \sin^{-1}\left(-\frac{1}{4}\right)\).

Step 4: Find General Solutions

For \(\tan \theta = 2\), the general solution for tangent is:\[\theta = \tan^{-1}(2) + n\pi\]where \(n\) is an integer.For \(\sin \theta = \frac{1}{4}\), the general solution is:\[\theta = \sin^{-1}(\frac{1}{4}) + 2k\pi \text{ and } \theta = \pi - \sin^{-1}(\frac{1}{4}) + 2k\pi\] where \(k\) is an integer.Similarly, for \(\sin \theta = -\frac{1}{4}\), the general solution is:\[\theta = -\sin^{-1}(\frac{1}{4}) + 2m\pi \text{ and } \theta = \pi + \sin^{-1}(\frac{1}{4}) + 2m\pi\]where \(m\) is an integer.

Key Concepts

Tangent Function

The tangent function, often denoted as \(\tan(\theta)\), is one of the fundamental trigonometric functions. It relates to the angle \(\theta\) of a right-angled triangle through the ratio of the length of the opposite side to the adjacent side. It's important to remember:

• The tangent function is periodic with a period of \(\pi\).
• It has vertical asymptotes whenever \(\cos(\theta) = 0\).

The solution, \(\tan(\theta) = 2\), involves finding the angle \(\theta\) whose tangent value is 2. The inverse trigonometric function \(\tan^{-1}(2)\) is used to find this angle. Remember, due to the periodic nature of the tangent function, there are infinitely many solutions, differing by multiples of \(\pi\).

Sine Function

The sine function, represented as \(\sin(\theta)\), is another essential trigonometric function. It gives the ratio of the length of the opposite side to the hypotenuse in a right triangle for a given angle \(\theta\). Key aspects include:

• Sine values range between -1 and 1.
• It has a period of \(2\pi\).

In this problem, we focus on solving \(16\sin^2(\theta) = 1\). By simplifying, we find \(\sin(\theta) = \pm \frac{1}{4}\). This requires the use of the inverse sine function to determine the angle values corresponding to these sine values. This step is crucial for solving trigonometric equations involving powers of sine.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find angles when the values of trigonometric functions are given. These functions are denoted as \(\sin^{-1}, \tan^{-1}\), etc. They help us by reversing the trigonometric functions:

• \(\tan^{-1}(x)\) gives the angle whose tangent is \(x\).
• \(\sin^{-1}(x)\) provides the angle with sine \(x\), restricted to the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

In this exercise, \(\tan^{-1}(2)\) yields the angle where \(\tan(\theta) = 2\), and similarly, \(\sin^{-1}(\pm \frac{1}{4})\) gives angles where \(\sin(\theta) = \pm \frac{1}{4}\). These functions are fundamental when deriving exact angles in trigonometric solutions.

General Solutions

Trigonometric equations often have multiple solutions because of the periodic nature of trigonometric functions. General solutions help by expressing these infinitely many solutions in compact forms:

• For the tangent function with \(\tan(\theta) = a\), the general solution is \(\theta = \tan^{-1}(a) + n\pi\).
• For the sine function with \(\sin(\theta) = b\), solutions include \(\theta = \sin^{-1}(b) + 2k\pi\) or \(\theta = \pi - \sin^{-1}(b) + 2k\pi\).

This is because of the periodicity of the sine and tangent functions, resulting in infinite solutions spaced regularly by their periods. Understanding general solutions is essential for solving trigonometric equations comprehensively.