Problem 43 - Algebra 2

Question

ACT/SAT Find the exact value of \(\cos 2 \theta\) if \(\sin \theta=\frac{-\sqrt{5}}{3}\) and \(180^{\circ}<\theta<270^{\circ}\) $$ \begin{array}{l}{\text { A } \frac{-\sqrt{6}}{6}} \\ {\text { B } \frac{-\sqrt{30}}{6}} \\ {\text { C } \frac{-4 \sqrt{5}}{9}} \\ {\text { D } \frac{-1}{9}}\end{array} $$

Step-by-Step Solution

Answer

\(\cos 2\theta = \frac{-1}{9}\) (Option D).

Step 1: Determine Quadrant

Since \(180^{\circ}<\theta<270^{\circ}\), \(\theta\) is in the third quadrant where both sine and cosine are negative.

Step 2: Find \(\cos \theta\) Using Pythagorean Identity

Use the identity \(\sin^2 \theta + \cos^2 \theta = 1\), and substitute \(\sin \theta = \frac{-\sqrt{5}}{3}\). This gives \(\left(\frac{-\sqrt{5}}{3}\right)^2 + \cos^2 \theta = 1\).

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

Calculate \(\left(\frac{-\sqrt{5}}{3}\right)^2 = \frac{5}{9}\). Substitute into the equation \(\frac{5}{9} + \cos^2 \theta = 1\) to find \(\cos^2 \theta = 1 - \frac{5}{9} = \frac{4}{9}\). Thus, \(\cos \theta = -\frac{2}{3}\) (negative because in the third quadrant).

Step 4: Use Double Angle Formula for Cosine

Apply the double angle formula: \(\cos 2\theta = 2\cos^2 \theta - 1\). Substitute \(\cos \theta = -\frac{2}{3}\) to find \(\cos 2\theta = 2\left(-\frac{2}{3}\right)^2 - 1\).

Step 5: Calculate \(\cos 2\theta\)

Calculate \(\left(-\frac{2}{3}\right)^2 = \frac{4}{9}\), then substitute to get \(2\left(\frac{4}{9}\right) - 1 = \frac{8}{9} - 1 = \frac{8}{9} - \frac{9}{9} = \frac{-1}{9}\).

Key Concepts

Double Angle Formula

The double angle formula is a useful trigonometric identity that helps in simplifying expressions involving angles that are twice another angle. For cosine, the double angle identity is given by:

• \( \cos 2\theta = 2\cos^2 \theta - 1 \)

This formula is derived from the angle addition formulas. It is handy when you know the value of \( \cos \theta \) and need to find \( \cos 2\theta \), which is precisely what we did in the solution provided.

Using the given \( \cos \theta = -\frac{2}{3} \), the double angle formula simplifies our calculation process. By squaring \( \cos \theta \) and substituting into the formula, the result is \( \frac{-1}{9} \), which is the exact value of \( \cos 2\theta \).

This approach is especially useful in trigonometry exams like the ACT and SAT as it allows a quick computation once you correctly identify \( \cos \theta \).

Pythagorean Identity

The Pythagorean identity is one of the fundamental identities in trigonometry. It relates the squares of sine and cosine functions:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

This identity holds true for any angle \( \theta \). In our exercise, knowing \( \sin \theta = \frac{-\sqrt{5}}{3} \), we can find \( \cos \theta \) by rearranging the identity to:

• \( \cos^2 \theta = 1 - \sin^2 \theta \)

Substitute \( \sin \theta \) in, and solve for \( \cos^2 \theta \).

It’s important to consider the quadrant of \( \theta \). Here, \( \theta \) is in the third quadrant where cosine is negative. Hence, we take the negative root of \( \cos^2 \theta = \frac{4}{9} \), resulting in \( \cos \theta = -\frac{2}{3} \).

This consistent relationship helps solve many trigonometric problems with confidence!

Trigonometric Functions

Trigonometric functions form the backbone of trigonometry. They include sine, cosine, tangent, and their reciprocals - cosecant, secant, and cotangent. Each function is associated with an angle within the context of a right triangle or the unit circle.

In this problem, we focus on the sine and cosine functions:

• \( \sin \theta \) - gives the ratio of the opposite side to the hypotenuse in a right triangle.
• \( \cos \theta \) - gives the ratio of the adjacent side to the hypotenuse.

These functions not only describe ratios in triangles but are essential components in modeling periodic phenomena like waves.

On the unit circle, which is a circle with radius 1 centered at the origin in the coordinate plane, these functions can be defined for any angle \( \theta \). The sine and cosine of an angle correspond to the coordinates of a point on the circle.

For the given problem, understanding these basic trigonometric functions and their relationships allows for the application of trigonometric identities, such as the double angle formula and Pythagorean identity, to derive exact values of angles and functions, making problem-solving more streamlined and intuitive.