Problem 15 - Algebra 2 and Trigonometry

Question

In \(11-18,\) find all radian measures of \(\theta\) in the interval \(0 \leq \theta \leq 2 \pi\) that make each equation true. Express your answers in terms of \(\pi\) when possible; otherwise, to the nearest hundredth. $$ 5 \sin ^{2} \theta-4 \sin \theta+\cos 2 \theta=0 $$

Step-by-Step Solution

Answer

The solutions are \( \theta = \frac{\pi}{2}, 0.34, \) and \( 2.80 \) radians.

Step 1: Use Trigonometric Identity

We start by using the trigonometric identity for cosine of double angles. Recall that \( \cos 2\theta = 1 - 2 \sin^2 \theta \). Substitute this into the equation to get: \[ 5 \sin^2 \theta - 4 \sin \theta + (1 - 2 \sin^2 \theta) = 0. \] Simplify the expression.

Step 2: Simplify the Equation

After substituting the identity, simplify the equation: \[ 5\sin^2 \theta - 4\sin \theta + 1 - 2\sin^2 \theta = 0. \] Combine like terms to get a quadratic in terms of \( \sin \theta \): \[ 3\sin^2 \theta - 4\sin \theta + 1 = 0. \]

Step 3: Use the Quadratic Formula

The simplified equation is a quadratic equation of the form \( a\sin^2\theta + b\sin \theta + c = 0 \), where \( a=3 \), \( b=-4 \), and \( c=1 \). Use the quadratic formula \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Calculate: \[ \sin \theta = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 3 \times 1}}{2 \times 3}. \]

Step 4: Calculate the Discriminant and Solutions

Calculate the discriminant: \[ 16 - 12 = 4. \] Use it in the quadratic formula: \[ \sin \theta = \frac{4 \pm \sqrt{4}}{6} = \frac{4 \pm 2}{6}. \] This gives two solutions: \( \sin \theta = 1 \) and \( \sin \theta = \frac{1}{3} \).

Step 5: Find Angles for \( \sin \theta = 1 \)

For \( \sin \theta = 1 \), the angle \( \theta \) is \( \frac{\pi}{2} \), as \( \sin \theta \) equals 1 at \( \frac{\pi}{2} \) within the given interval \( 0 \leq \theta \leq 2\pi \).

Step 6: Find Angles for \( \sin \theta = \frac{1}{3} \)

Use inverse sine to find \( \theta = \arcsin\left(\frac{1}{3}\right) \). Also, consider the sine symmetry: \( \theta = \pi - \arcsin\left(\frac{1}{3}\right) \). Both angles are valid within \( 0 \leq \theta \leq 2\pi \). Use a calculator to compute the approximate values: \( \theta \approx 0.34 \) and \( \theta \approx 2.80 \) radians.

Key Concepts

Radian Measures

Radians are a fundamental way to measure angles using the radius of a circle. In this exercise, we are focusing on the interval from 0 to \( 2\pi \) radians. This is equivalent to a full circle, as \( 2\pi \) radians correspond to 360 degrees.
Radians offer a natural way of measuring angles and are often used in mathematical formulations because they seamlessly connect linear and angular distances.
For instance, an angle of \( \pi \) radians means the arc length is equal to the circle's radius, providing a straightforward relationship.
Understanding radian measures is essential for solving trigonometric equations, as they are frequently expressed in terms of \( \pi \). This allows for precise mathematical expressions without the need to convert between degrees and radians.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all angles. They are critical tools for simplifying and solving equations, as seen in this exercise.
One important identity used here is the double angle identity:
\[ \cos 2\theta = 1 - 2\sin^2 \theta. \]
This identity allows us to transform expressions involving \( \cos 2\theta \) into terms of \( \sin \theta \), simplifying the problem.
Another widely used set of identities are the Pythagorean identities, like \( \sin^2 \theta + \cos^2 \theta = 1 \), which relate the sine and cosine of an angle. Recognizing and applying these identities helps in reducing complex equations into manageable parts, making it easier to find solutions.

Inverse Trigonometric Functions

Inverse trigonometric functions enable us to find angle measures when the value of a trigonometric function is known. In this context, they are used to calculate angles based on given sine values.
For example, the inverse sine function, \( \arcsin \), returns an angle whose sine is a given number. In the exercise, to solve \( \sin \theta = \frac{1}{3} \), we use:
\[ \theta = \arcsin \left( \frac{1}{3} \right). \]
This calculates the principal angle of \( \theta \). However, because the sine function is positive in two quadrants, it is necessary to consider both the principal and supplementary angles, giving two potential solutions. Knowing how to leverage inverse functions enables us to translate between trigonometric function values and the angles themselves.

Sine and Cosine Functions

Sine and cosine functions are the backbone of trigonometry, describing oscillations like waves and circular movements. These functions have specific properties that make them crucial to solving trigonometric equations.
The sine function measures the vertical component of an angle on the unit circle. It is periodic with a period of \( 2\pi \), meaning it repeats its values every full circle rotation. Similarly, the cosine function measures the horizontal component and shares this periodicity.
In our problem, finding where \( \sin \theta = 1 \) corresponds to the angle \( \frac{\pi}{2} \), showcasing the sine function peak. Understanding where sine and cosine hit specific values helps solve equations by pinpointing exact angle solutions.

• Sine reaches its maximum at \( \ \frac{\pi}{2} \) radians.
• Cosine reaches its maximum at 0 and \( 2\pi \) radians.

Recognizing these key points assists in tackling various trigonometric problems effectively.