Problem 71 - Precalculus with Calculus Previews

Question

If \(P\left(x_{1}\right)\) and \(P\left(x_{2}\right)\) are points in quadrant II on the terminal side of the angles \(x_{1}\) and \(x_{2}\), respectively, with \(\cos x_{1}=-\frac{1}{3}\) and \(\sin x_{2}=\frac{2}{3}\), find (a) \(\sin \left(x_{1}+x_{2}\right),(b) \cos \left(x_{1}+x_{2}\right),(\mathbf{c}) \sin \left(x_{1}-x_{2}\right)\), and (d) \(\cos \left(x_{1}-x_{2}\right)\).

Step-by-Step Solution

Answer

\(\sin(x_1 + x_2) = -\frac{2(\sqrt{10} + 1)}{9}\), \(\cos(x_1 + x_2) = \frac{\sqrt{5} - 4\sqrt{2}}{9}\), \(\sin(x_1 - x_2) = \frac{-2\sqrt{10} + 2}{9}\), \(\cos(x_1 - x_2) = \frac{\sqrt{5} + 4\sqrt{2}}{9}\).

Step 1: Understanding the Problem

We need to find trigonometric values for sum and difference of angles given specific trigonometric values in quadrant II. For points in quadrant II, sine is positive and cosine is negative.

Step 2: Identify Known Values

We have \(\cos x_{1} = -\frac{1}{3}\) and \(\sin x_{2} = \frac{2}{3}\). From the Pythagorean identity, we can find the missing sine and cosine values. Since \(x_1\) and \(x_2\) are in quadrant II.

Step 3: Calculate \(\sin x_{1}\) from \(\cos x_{1}\)

Use the identity \(\sin^2 x_{1} + \cos^2 x_{1} = 1\). Substitute to find \(\sin x_{1}\):\[\sin^2 x_{1} = 1 - \left(-\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9}\]\(\sin x_{1} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\).

Step 4: Calculate \(\cos x_{2}\) from \(\sin x_{2}\)

Use the identity \(\sin^2 x_{2} + \cos^2 x_{2} = 1\). Substitute to find \(\cos x_{2}\):\[\cos^2 x_{2} = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9}\]\(\cos x_{2} = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}\).

Step 5: Applying Sum Formulas

Use the identities: \(\sin(a+b) = \sin a \cos b + \cos a \sin b\) and \(\cos(a+b) = \cos a \cos b - \sin a \sin b\).

Step 6: Find \(\sin(x_1 + x_2)\)

Using the sum identity:\[\sin(x_1 + x_2) = \frac{2\sqrt{2}}{3} \left(-\frac{\sqrt{5}}{3}\right) + \left(-\frac{1}{3}\right)\frac{2}{3}\]\[\sin(x_1 + x_2) = -\frac{2\sqrt{10}}{9} - \frac{2}{9} = -\frac{2(\sqrt{10} + 1)}{9}\]

Step 7: Find \(\cos(x_1 + x_2)\)

Using the sum identity:\[\cos(x_1 + x_2) = \left(-\frac{1}{3}\right)\left(-\frac{\sqrt{5}}{3}\right) - \frac{2\sqrt{2}}{3} \frac{2}{3}\]\[\cos(x_1 + x_2) = \frac{\sqrt{5}}{9} - \frac{4\sqrt{2}}{9} = \frac{\sqrt{5} - 4\sqrt{2}}{9}\]

Step 8: Applying Difference Formulas

Use the identities: \(\sin(a-b) = \sin a \cos b - \cos a \sin b\) and \(\cos(a-b) = \cos a \cos b + \sin a \sin b\).

Step 9: Find \(\sin(x_1 - x_2)\)

Using the difference identity:\[\sin(x_1 - x_2) = \frac{2\sqrt{2}}{3} \left(-\frac{\sqrt{5}}{3}\right) - \left(-\frac{1}{3}\right)\frac{2}{3}\]\[\sin(x_1 - x_2) = -\frac{2\sqrt{10}}{9} + \frac{2}{9} = \frac{-2\sqrt{10} + 2}{9}\]

Step 10: Find \(\cos(x_1 - x_2)\)

Using the difference identity:\[\cos(x_1 - x_2) = \left(-\frac{1}{3}\right)\left(-\frac{\sqrt{5}}{3}\right) + \frac{2\sqrt{2}}{3} \frac{2}{3}\]\[\cos(x_1 - x_2) = \frac{\sqrt{5}}{9} + \frac{4\sqrt{2}}{9} = \frac{\sqrt{5} + 4\sqrt{2}}{9}\]

Key Concepts

Sum and Difference Formulas

To calculate the trigonometric values of angles like \(x_1 + x_2\) or \(x_1 - x_2\), we use sum and difference formulas. These formulas expand the sine and cosine of angle sums or differences into a combination of sine and cosine of the individual angles. This means that:

• The formula for \(\sin(a + b)\) is \(\sin a \cos b + \cos a \sin b\).
• The formula for \(\cos(a + b)\) is \(\cos a \cos b - \sin a \sin b\).
• The formula for \(\sin(a - b)\) is \(\sin a \cos b - \cos a \sin b\).
• The formula for \(\cos(a - b)\) is \(\cos a \cos b + \sin a \sin b\).

These identities are essential because they help transform angles that are not directly measurable into forms that can be easily computed using known values.

Pythagorean Identity

The Pythagorean identity is a key concept in trigonometry that connects the sine and cosine of an angle. It is rooted in the Pythagorean theorem and states:\[ \sin^2 x + \cos^2 x = 1 \]This identity allows us to find the missing value of either sine or cosine if we know the other value. For example, if you know \(\cos x\) is \(-\frac{1}{3}\), you can find \(\sin x\) by substituting in the Pythagorean identity:

• \(\sin^2 x = 1 - \cos^2 x = 1 - \left(-\frac{1}{3}\right)^2 = \frac{8}{9}\)
• \(\sin x = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\)

This formula is indispensable when determining sine or cosine values in various quadrants.

Quadrant II Trigonometric Values

In trigonometry, the unit circle is divided into four quadrants, with each having specific properties for sine, cosine, and other trigonometric functions. **Quadrant II**, covering angles between \(90^\circ\) to \(180^\circ\), has a distinct characteristic:- Cosine values are negative.- Sine values are positive.This is crucial because understanding these characteristics lets you determine the sign of your trigonometric values without needing to calculate from scratch every time. For instance, if we have \( \cos x_1 = -\frac{1}{3} \) and \( x_1 \) is in Quadrant II, \( \sin x_1\) must be positive, resulting in \( \frac{2\sqrt{2}}{3} \). This helps ensure calculations stay consistent and accurate.

Trigonometric Calculations

Calculating trigonometric values involves a few steps to ensure that you correctly apply identities and properties. Consider these steps to reliably find the values:1. **Identify Known Values**: Understand the trigonometric values given for each angle.2. **Use Pythagorean Identity**: Find the missing sine or cosine values if only one is given.3. **Apply Sum/Difference Formulas**: Use these formulas to find trigonometric values for angle sums or differences.

• For \( \sin(x_1 + x_2) \), use \( \sin x_1 \cos x_2 + \cos x_1 \sin x_2 \).
• For \( \cos(x_1 + x_2) \), use \( \cos x_1 \cos x_2 - \sin x_1 \sin x_2 \).
• For \( \sin(x_1 - x_2) \), use \( \sin x_1 \cos x_2 - \cos x_1 \sin x_2 \).
• For \( \cos(x_1 - x_2) \), use \( \cos x_1 \cos x_2 + \sin x_1 \sin x_2 \).

This systematic approach helps in seamlessly carrying out trigonometric calculations, eliminating errors, and resulting in precise outcomes.