Problem 62 - Precalculus

Question

For the following exercises, find the requested information. Given that \(\sin a=\frac{2}{3}\) and \(\cos b=-\frac{1}{4},\) with \(a\) and \(b\) both in the interval \(\left[\frac{\pi}{2}, \pi\right),\) find \(\sin (a+b)\) and \(\cos (a-b)\)

Step-by-Step Solution

Answer

\(\sin(a+b) = -\frac{2 + 5\sqrt{3}}{12}\) and \(\cos(a-b) = \frac{\sqrt{5} + 2\sqrt{15}}{12}\).

Step 1: Use Sine Addition Formula

We need to find \(\sin(a+b)\) using the formula \(\sin(a+b) = \sin a \cos b + \cos a \sin b\). Given \(\sin a = \frac{2}{3}\) and \(\cos b = -\frac{1}{4}\), we must first find \(\cos a\) and \(\sin b\) given the interval \(\left[\frac{\pi}{2}, \pi\right)\).

Step 2: Find \(\cos a\) from \(\sin a\)

Since \(\sin^2 a + \cos^2 a = 1\), we solve for \(\cos a\) as follows:\[\cos a = -\sqrt{1 - \sin^2 a} = -\sqrt{1 - \left(\frac{2}{3}\right)^2} = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}\]We choose the negative because \(a\) is in the second quadrant, where cosine is negative.

Step 3: Find \(\sin b\) from \(\cos b\)

Since \(\sin^2 b + \cos^2 b = 1\), we solve for \(\sin b\) as follows:\[\sin b = \sqrt{1 - \cos^2 b} = \sqrt{1 - \left(-\frac{1}{4}\right)^2} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}\]We choose the positive because \(b\) is in the second quadrant, where sine is positive.

Step 4: Calculate \(\sin(a+b)\)

Substitute \(\cos a\) and \(\sin b\) into the sine addition formula:\[\sin(a+b) = \frac{2}{3}\left(-\frac{1}{4}\right) + \left(-\frac{\sqrt{5}}{3}\right)\left(\frac{\sqrt{15}}{4}\right) = -\frac{2}{12} - \frac{\sqrt{75}}{12} = -\frac{2 + 5\sqrt{3}}{12}\]

Step 5: Use Cosine Subtraction Formula

We need to find \(\cos(a-b)\) using the formula \(\cos(a-b) = \cos a \cos b + \sin a \sin b\). We already have \(\cos a = -\frac{\sqrt{5}}{3}\), \(\cos b = -\frac{1}{4}\), \(\sin a = \frac{2}{3}\), and \(\sin b = \frac{\sqrt{15}}{4}\).

Step 6: Calculate \(\cos(a-b)\)

Substitute the values into the cosine subtraction formula:\[\cos(a-b) = \left(-\frac{\sqrt{5}}{3}\right)\left(-\frac{1}{4}\right) + \left(\frac{2}{3}\right)\left(\frac{\sqrt{15}}{4}\right) = \frac{\sqrt{5}}{12} + \frac{2\sqrt{15}}{12} = \frac{\sqrt{5} + 2\sqrt{15}}{12}\]

Key Concepts

sine addition formula

The sine addition formula is a critical tool in trigonometry. It's used to find the sine of the sum of two angles. The formula is given by:

• \[\sin(a + b) = \sin a \cdot \cos b + \cos a \cdot \sin b\]

This formula combines the sine and cosine of individual angles to determine the sine of their sum. For example, if we know \(\sin a\) and \(\cos b\), we can use this formula to find \(\sin(a+b)\) as outlined in our exercise.

In our exercise, we were given \(\sin a=\frac{2}{3}\) and \(\cos b=-\frac{1}{4}\), and needed to find \(\sin(a+b)\). To do this, we first found \(\cos a\) and \(\sin b\) using the Pythagorean identity, bearing in mind the quadrant constraints. Then, substituting these values into the sine addition formula, allowed us to calculate \(\sin(a+b)\). Understanding this step-by-step application reinforces the importance of the sine addition formula in solving trigonometric problems.

cosine subtraction formula

The cosine subtraction formula is vital when dealing with trigonometric functions and their applications. This formula allows us to determine the cosine of the difference between two angles. It is expressed as:

• \[\cos(a - b) = \cos a \cdot \cos b + \sin a \cdot \sin b\]

This formula relies on the separate calculations of the cosine and sine for individual angles to find the final result.

In our problem, we already calculated \(\cos a\), \(\cos b\), \(\sin a\), and \(\sin b\). Using these values, we applied the cosine subtraction formula to find \(\cos(a-b)\). This demonstrates how the formula directly leverages previously computed trigonometric values, underscoring its utility in complex calculations involving differences of angles.

trigonometric functions

Trigonometric functions—sine, cosine, and tangent—are fundamental to understanding the relationships between the angles and sides of triangles.

• **Sine**: Measures the ratio of the length of the opposite side to the hypotenuse in a right triangle.
• **Cosine**: Measures the ratio of the adjacent side to the hypotenuse.
• **Tangent**: The ratio of the sine to the cosine, which is the opposite to the adjacent side.

In the context of our exercise, we primarily dealt with sine and cosine. These two functions were critical in applying the sine addition and cosine subtraction formulas.

The interval \(\left[\frac{\pi}{2}, \pi\right)\) indicates that angles \(a\) and \(b\) lie in the second quadrant. Here, sine values are positive, and cosine values are negative. This quadrant-based understanding of trigonometric functions is crucial for determining the correct signs of calculated values, as seen in our solutions. Overall, mastering these functions is key to solving more complex trigonometric problems efficiently.