Problem 45 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the exact value of the trigonometric expression when \(\sin u=\frac{5}{13}\) and \(\cos v=-\frac{3}{5} \cdot(\text { Both } u\) and \(v\) are in Quadrant II.) $$\cos (u-v)$$

Step-by-Step Solution

Answer

The exact value of the trigonometric expression \(\cos (u-v)\) is \(\frac{16}{65}\).

Step 1: Assert the Given Values

We are given that \(\sin u = \frac{5}{13}\) and \(\cos v = -\frac{3}{5}\). Both \(u\) and \(v\) are in Quadrant II, which indicates that \(\cos u\) and \(\sin v\) are negative as cosine is negative in Quadrant II, while sine is positive.

Step 2: Use Pythagorean Identity

Use the Pythagorean identity \(\sin^2 θ + \cos^2 θ = 1\) to find the values of \(\cos u\) and \(\sin v\). By substituting the given values into the identity, we have:\( \cos u = -\sqrt{1-(\frac{5}{13})^2} = -\frac{12}{13}\) and \( \sin v = \sqrt{1-(-\frac{3}{5})^2} = -\frac{4}{5}\). The negative signs are because cosine and sine are negative in the 2nd quadrant.

Step 3: Use Cosine Difference Identity

Use the cosine difference identity \(\cos(u-v) = \cos u \cos v + \sin u \sin v\), and insert the identified values for \(\sin u\), \(\sin v\), \(\cos u\), \(\cos v\). By substituting, we get: \(\cos(u-v) = \cos u \cos v + \sin u \sin v = -\frac{12}{13} \cdot -\frac{3}{5} + \frac{5}{13} \cdot -\frac{4}{5} = \frac{36}{65} - \frac{20}{65} = \frac{16}{65}\).

Key Concepts

Sine and Cosine Functions

Understanding sine and cosine functions is crucial for solving trigonometric expressions. Sine and cosine are functions of an angle often used to model periodic phenomena.
The sine function, \( \sin \theta \), represents the ratio of the opposite side to the hypotenuse in a right triangle. In the unit circle, it's the y-coordinate of a point corresponding to the angle \( \theta \).
On the other hand, cosine, \( \cos \theta \), is the ratio of the adjacent side to the hypotenuse. In the unit circle, it's the x-coordinate.

• In Quadrant II, \( \sin \theta \) is positive and \( \cos \theta \) is negative.

This understanding helps interpret given values. If \( \sin u = \frac{5}{13} \), we know the angle's opposite side is 5 and hypotenuse is 13. Hence, \( \cos u \) must be negative and resolved using the Pythagorean identity, leading to \( \cos u = -\frac{12}{13} \).
Similarly, if \( \cos v = -\frac{3}{5} \), its sine value calculated will be \( \sin v = -\frac{4}{5} \) by applying the Pythagorean identity.

Pythagorean Identity

The Pythagorean Identity is a fundamental theorem in trigonometry and helps to determine unknown trigonometric function values. It states:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
This equation represents the relationship between the sine and cosine of the same angle. When one value is known, the other can be found by rearranging and solving the identity.

• For \( \sin u = \frac{5}{13} \), applying the identity gives \( \cos u = -\sqrt{1 - \left( \frac{5}{13} \right)^2} = -\frac{12}{13} \).
• Similarly, for \( \cos v = -\frac{3}{5} \), it leads to \( \sin v = \sqrt{1 - \left( -\frac{3}{5} \right)^2} = -\frac{4}{5} \).

The negative sign considerations are due to the angles being in the 2nd quadrant, where cosine is negative. These calculations are essential to solve trigonometric identities accurately.

Cosine Difference Identity

The Cosine Difference Identity is a powerful tool for calculating the cosine of the difference between two angles. It is given by:
\[ \cos(u-v) = \cos u \cos v + \sin u \sin v \]
This formula allows combining individual trigonometric functions of two different angles to find the trigonometric function of their difference.
For the given problem, using the values:

• \( \cos u = -\frac{12}{13} \) and \( \cos v = -\frac{3}{5} \)
• \( \sin u = \frac{5}{13} \) and \( \sin v = -\frac{4}{5} \)

Substituting into the identity results in:
\[ \cos(u-v) = \left(-\frac{12}{13}\right) \left(-\frac{3}{5}\right) + \left(\frac{5}{13}\right) \left(-\frac{4}{5}\right) = \frac{16}{65} \]
This calculation demonstrates how compound angles can be evaluated using basic trigonometric values and identities.