Problem 57 - Algebra and Trigonometry

Question

Find the exact value of the following under the given conditions: a. \(\cos (\alpha+\beta)\) b. \(\sin (\alpha+\beta)\) c. \(\tan (\alpha+\beta)\) \(\sin \alpha=\frac{3}{5}, \alpha\) lies in quadrant I, and \(\sin \beta=\frac{5}{13}, \beta\) lies in quadrant II.

Step-by-Step Solution

Answer

a. \(\cos (\alpha+\beta) = -\frac{63}{65}\), b. \(\sin (\alpha+\beta) = -\frac{16}{65}\),c. \(\tan (\alpha+\beta) = \frac{16}{63}\)

Step 1: Applying Pythagorean Identity

First, use the Pythagorean identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to find the cosine of \(\alpha\) and \(\beta\). As \(\alpha\) is in quadrant I where cosine is positive, \(\cos \alpha=\sqrt{1-\sin^2 \alpha} = \sqrt{1-\frac{9}{25}} = \frac{4}{5}\). On the other hand, \(\beta\) is in quadrant II where cosine is negative, so \(\cos \beta=-\sqrt{1-\sin^2 \beta} = -\sqrt{1-\frac{25}{169}} = -\frac{12}{13}\).

Step 2: Applying Cosine Addition Formula

Use the cosine addition formula \(\cos(\alpha+\beta) = \cos \alpha\cos \beta - \sin \alpha\sin \beta\). Substitute the values we calculated and simplify to get \(\cos(\alpha+\beta) = \frac{4}{5}*(-\frac{12}{13}) - \frac{3}{5}*\frac{5}{13} = -\frac{48}{65} - \frac{15}{65} = -\frac{63}{65}\).

Step 3: Applying Sine Addition Formula

Use the sine addition formula \(\sin(\alpha+\beta) = \sin \alpha\cos \beta + \cos \alpha\sin \beta\). Substitute the values we calculated and simplify to get \(\sin(\alpha+\beta) = \frac{3}{5}*(-\frac{12}{13}) + \frac{4}{5}*\frac{5}{13} = -\frac{36}{65} + \frac{20}{65} = -\frac{16}{65}\).

Step 4: Computing Tangent

Finally, use the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) to find the tangent of \(\alpha+\beta\). So, \(\tan(\alpha+\beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \frac{-\frac{16}{65}}{-\frac{63}{65}} = \frac{16}{63}\).

Key Concepts

Cosine Addition Formula

The cosine addition formula is a crucial identity in trigonometry used to determine the cosine of the sum of two angles. It is expressed as:

• \(\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\)

This formula lets us break down the cosine of a combined angle using the trigonometric values of the individual angles. It uses both the sines and cosines of the angles, making it a powerful tool for solving problems in various mathematical fields.
In our exercise, we calculated \(\cos \alpha = \frac{4}{5}\) and \(\cos \beta = -\frac{12}{13}\) using the Pythagorean identity. By plugging these values into the cosine addition formula, along with \(\sin \alpha = \frac{3}{5}\) and \(\sin \beta = \frac{5}{13}\), we found that \(\cos(\alpha + \beta) = -\frac{63}{65}\).
This result exemplifies how the formula works accurately even with angles located in different quadrants.

Sine Addition Formula

The sine addition formula is similarly essential to understanding angle sums. It is given by:

• \(\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta\)

This formula allows us to compute the sine of the sum of two angles by using the sines and cosines of each angle individually. It is particularly useful when angles belong to different quadrants, as in our case.
For our task, we used the values we derived: \(\sin \alpha = \frac{3}{5}\), \(\cos \beta = -\frac{12}{13}\), \(\cos \alpha = \frac{4}{5}\), and \(\sin \beta = \frac{5}{13}\). Substituting these values into the formula, we calculated \(\sin(\alpha + \beta) = -\frac{16}{65}\).
This computation demonstrates how the formula precisely accounts for the quadrants of \(\alpha\) and \(\beta\), influencing the sign and magnitude of the result.

Pythagorean Identity

The Pythagorean identity is one of the fundamental identities in trigonometry. It states that for any angle \(\theta\), the square of the sine plus the square of the cosine equals one:

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

This identity is foundational because it connects the sine and cosine functions through their intrinsic geometric relationship in the unit circle. It allows us to find one value when the other is known.
In our exercise, we found \(\cos \alpha\) and \(\cos \beta\) using the Pythagorean identity from the given values of \(\sin \alpha = \frac{3}{5}\) and \(\sin \beta = \frac{5}{13}\). Knowing \(\sin \alpha\), we computed \(\cos \alpha = \sqrt{1-\left(\frac{3}{5}\right)^2} = \frac{4}{5}\). Similarly, having \(\sin \beta\), we derived \(\cos \beta = -\sqrt{1-\left(\frac{5}{13}\right)^2} = -\frac{12}{13}\), noting the negative sign due to \(\beta\) being in quadrant II where cosine is negative.
This process highlights the versatility of the Pythagorean identity in solving trigonometric problems.

Tangent Identity

The tangent identity relates the sine and cosine of an angle to its tangent. It is expressed as:

• \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)

This identity provides a straightforward way to find the tangent of an angle once you have its sine and cosine. In trigonometric contexts, tangents are crucial for finding slopes and angles in right triangles.
In the given exercise, after determining \(\sin(\alpha + \beta) = -\frac{16}{65}\) and \(\cos(\alpha + \beta) = -\frac{63}{65}\), we used the tangent identity to find \(\tan(\alpha + \beta) = \frac{-\frac{16}{65}}{-\frac{63}{65}}= \frac{16}{63}\).
This demonstrates how the tangent identity directly uses the results from applying the sine and cosine addition formulas, thus simplifying the calculation of the angle's tangent efficiently.