Problem 31 - Algebra 2

Question

Find the exact values of \(\sin 2 \theta, \cos 2 \theta, \sin \frac{\theta}{2},\) and \(\cos \frac{\theta}{2}\) for each of the following. $$ \cos \theta=-\frac{12}{13} ; 180^{\circ}<\theta<270^{\circ} $$

Step-by-Step Solution

Answer

\(\sin 2\theta = \frac{120}{169}\), \(\cos 2\theta = \frac{119}{169}\), \(\sin \frac{\theta}{2} = \sqrt{\frac{25}{26}}\), \(\cos \frac{\theta}{2} = -\sqrt{\frac{1}{26}}\).

Step 1: Identify the Quadrants

Given that \(180^{\circ} < \theta < 270^{\circ}\), \(\theta\) is in the third quadrant. In this quadrant, \(\sin \theta\) is negative and \(\cos \theta\) is also negative.

Step 2: Use Pythagorean Identity to Find \(\sin \theta\)

We know that \(\cos \theta = -\frac{12}{13}\) and \(\sin^2 \theta + \cos^2 \theta = 1\). Substitute \(\cos \theta\) to find \(\sin \theta\):\[ \sin^2 \theta + \left(-\frac{12}{13}\right)^2 = 1 \]\[ \sin^2 \theta + \frac{144}{169} = 1 \]\[ \sin^2 \theta = \frac{25}{169} \]\[ \sin \theta = -\frac{5}{13} \]Choose the negative root because \(\theta\) is in the third quadrant, where \(\sin\) is negative.

Step 3: Find \(\sin 2\theta\) and \(\cos 2\theta\)

Use the double-angle identities:\(\sin 2\theta = 2 \sin \theta \cos \theta\)and\(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\).Calculate:\[ \sin 2\theta = 2 \times -\frac{5}{13} \times -\frac{12}{13} = \frac{120}{169} \]\[ \cos 2\theta = \left(-\frac{12}{13}\right)^2 - \left(-\frac{5}{13}\right)^2 = \frac{144}{169} - \frac{25}{169} = \frac{119}{169} \]

Step 4: Use Half-Angle Formulas

Now find \(\sin \frac{\theta}{2}\) and \(\cos \frac{\theta}{2}\) using the half-angle formulas:\(\sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}}\)and\(\cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}}\).Since \(90^{\circ} < \frac{\theta}{2} < 135^{\circ}\) (second quadrant), \(\sin \frac{\theta}{2}\) is positive and \(\cos \frac{\theta}{2}\) is negative:\[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \left(-\frac{12}{13}\right)}{2}} = \sqrt{\frac{25}{26}} \]\[ \cos \frac{\theta}{2} = -\sqrt{\frac{1 + \left(-\frac{12}{13}\right)}{2}} = -\sqrt{\frac{1}{26}} \].

Step 5: Conclusion

The exact values are \(\sin 2\theta = \frac{120}{169}\), \(\cos 2\theta = \frac{119}{169}\), \(\sin \frac{\theta}{2} = \sqrt{\frac{25}{26}}\), and \(\cos \frac{\theta}{2} = -\sqrt{\frac{1}{26}}\).

Key Concepts

Double-Angle Identities

Double-angle identities are handy tools in trigonometry, helping us to find the sine and cosine of double angles using simpler trigonometric functions. They are highly beneficial for simplifying complex expressions or solving trigonometric equations. The double-angle identity for sine is

• \( \sin 2\theta = 2 \sin \theta \cos \theta \)

This identity shows that the sine of double an angle can be calculated using the values of sine and cosine for a single angle. Also, the double-angle identity for cosine is given by

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

This formula presents alternative forms using the Pythagorean identity and can also be written as:

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

Each form offers unique advantages depending on the problem requirements. In the provided exercise, these identities helped determine the values of \( \sin 2\theta \) and \( \cos 2\theta \).

Half-Angle Formulas

Half-angle formulas are extremely useful when dealing with angles that are bisected. These identities are derived from double-angle identities and allow us to express trigonometric functions of half-angles using whole angles. The half-angle formula for sine is given by:

• \( \sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}} \)

This identity allows us to calculate the sine of half an angle if the cosine value of the whole angle is known. Similarly, the half-angle formula for cosine is:

• \( \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} \).

It is important to recognize that these square roots can have positive or negative values depending on which quadrant the angle lies in. In the exercise, the quadrants help determine the correct sign for each value.

Pythagorean Identity

The Pythagorean identity is a fundamental principle in trigonometry that connects sine and cosine functions. It states that for any angle \( \theta \):

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

This identity underpins many calculations and can be algebraically manipulated to determine one function if the other is known. In the provided exercise, the Pythagorean identity was crucial for finding the sine value given the cosine value:

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

by substituting \( \cos \theta = -\frac{12}{13} \) and solving for \( \sin \theta \) as \( -\frac{5}{13} \). Using its variants, it can quickly adjust to other trigonometric expressions, making it a versatile part of any trigonometric toolkit.

Quadrantal Angles

Quadrantal angles are specific angles corresponding to positions on the unit circle where sine and cosine take on values of 0, 1, or -1. Do not be confused by the term; these angles occur exactly at the four cardinal directions: 0°, 90°, 180°, 270°, and so on. In exercises, understanding the quadrant where an angle resides is key because it dictates the sign of sine, cosine, or tangent for those angles. Each quadrant influences the sign of trigonometric functions:

• **First Quadrant:** All functions are positive.
• **Second Quadrant:** Sine is positive; cosine and tangent are negative.
• **Third Quadrant:** Tangent is positive; sine and cosine are negative.
• **Fourth Quadrant:** Cosine is positive; sine and tangent are negative.

In the given problem, knowing \(180^{\circ} < \theta < 270^{\circ}\) placed the angle in the third quadrant. This dictated that \( \sin \theta \) and \( \cos \theta \) were negative, assisting in developing proper trigonometric conclusions for the problem.