Problem 14 - 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. $$ \cos \theta=3 \sin 2 \theta $$

Step-by-Step Solution

Answer

\(\theta \approx 0.17 \) rad and \(\theta \approx 2.97 \) rad.

Step 1: Recall the Double Angle Identity for Sine

The double angle identity for sine states that \( \sin 2\theta = 2\sin\theta\cos\theta \). We will use this identity to rewrite the given equation \( \cos\theta = 3 \sin 2\theta \) as \( \cos\theta = 3 \times 2\sin\theta\cos\theta \).

Step 2: Simplify the Equation

Substituting the double angle identity into the equation, we get \( \cos\theta = 6\sin\theta\cos\theta \). Then, by dividing both sides of the equation by \( \cos\theta \) (assuming \( \cos\theta eq 0 \)), we get \( 1 = 6\sin\theta \).

Step 3: Solve for \(\sin\theta\)

From the equation \( 1 = 6\sin\theta \), divide both sides by 6 to find \( \sin\theta = \frac{1}{6} \).

Step 4: Find \(\theta\) Values

Since \( \sin\theta = \frac{1}{6} \), \(\theta\) can be found using the inverse sine function. Therefore, \( \theta = \arcsin\left( \frac{1}{6} \right) \) and by symmetry, \( \theta = \pi - \arcsin\left( \frac{1}{6} \right) \).

Step 5: Calculate and Express in Radians

To find the actual radians, compute \( \arcsin\left( \frac{1}{6} \right) \approx 0.1674 \) radians. Thus, \( \theta = 0.1674 \) radians and \( \theta = \pi - 0.1674 \approx 2.9742 \) radians.

Key Concepts

Radian Measure

Radian measure is a way to describe angles using the radius of a circle. Unlike degrees, radians measure how far around a circle's circumference the angle sweeps. One full rotation around a circle is equal to \(2\pi\) radians, which is equivalent to 360 degrees.

• A quarter rotation (or right angle) is measured as \(\frac{\pi}{2}\) radians.
• Half a rotation, moving from one side of a circle to the other, is measured as \(\pi\) radians.

Understanding radian measure is crucial when dealing with trigonometric equations like the one presented. Here, we are interested in finding all values of \(\theta\) that satisfy a given trigonometric equation, within the interval from 0 to \(2\pi\). This essentially means finding solutions that complete one full cycle or rotation around a circle.

So, when you see the instruction to "express answers in terms of \(\pi\)," it's asking you to leave constants of \(\pi\) in your answer rather than converting them to decimal places immediately. This keeps your answers precise and easier to interpret, especially in mathematics and physics.

Double Angle Identity

The double angle identity is a powerful tool in trigonometry, used to simplify expressions involving angles that are multiples of another. For sine, this identity is expressed as \(\sin 2\theta = 2\sin\theta\cos\theta\).

• This equation allows you to rewrite expressions involving \(\sin 2\theta\) in terms of \(\sin\theta\) and \(\cos\theta\).
• It is particularly useful when solving equations, as it turns a less familiar problem into one that is linear and easier to manipulate.

In the original exercise, the problem \(\cos\theta = 3 \sin 2\theta\) was simplified using this identity. By substituting \(\sin 2\theta\) with \(2\sin\theta\cos\theta\), the equation was transformed into \(\cos\theta = 6\sin\theta\cos\theta\). This new form allowed further simplification by dividing both sides by \(\cos\theta\), assuming \(\cos\theta eq 0\), leading to a straightforward linear equation \(1 = 6\sin\theta\).

Understanding and using identities like the double angle identity can make seemingly complex trigonometric equations much easier to solve.

Inverse Sine Function

The inverse sine function, usually written as \(\arcsin\), is used to find the angle whose sine is a given number.

• This function is essential when solving equations that require you to "undo" the sine operation to isolate the angle, as in our original exercise.
• The range of \(\arcsin\) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), giving the principal values of the angle within this interval.

In our solution, once we simplified the equation to \(\sin\theta = \frac{1}{6}\), we used the inverse sine function to determine \(\theta\). This gave us \(\theta = \arcsin\left( \frac{1}{6} \right)\). To ensure all solutions within the interval \(0 \leq \theta \leq 2\pi\) were identified, it was necessary to also consider \(\theta = \pi - \arcsin\left( \frac{1}{6} \right)\) due to the trigonometric symmetry.

The arc functions, like \(\arcsin\), are invaluable for solving for angles in trigonometry when given a specific ratio, helping to transform algebraic solutions into geometric understanding.