Problem 47 - Precalculus with Limits

Question

In Exercises \(43-50\) , find the exact value of the trigonometric function given that \(\sin u=\frac{5}{13}\) and \(\cos v=-\frac{3}{5} .\) Both \(u\) and \(v\) are in Quadrant II.) \(\tan (u+v)\)

Step-by-Step Solution

Answer

The exact value of \( \tan(u+v) \) is \( \frac{33}{56} \).

Step 1: Identify the sine and cosine of the given angles

The sine of u, that is \( \sin u \), is given as \( \frac{5}{13} \). Using the Pythagorean identity \( \sin^2u + \cos^2u = 1 \), we can solve for \( \cos u \) which gives us \( \cos u = \sqrt{1 - \sin^2 u} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = - \frac{12}{13} \). The negative value is chosen because cosine is negative in quadrant II. Similarly, given the cosine of v, that is \( \cos v = - \frac{3}{5} \), we find \( \sin v \) using the Pythagorean identity to get \( \sin v = - \sqrt{1 - \cos^2 v} = - \sqrt{1 - \left(- \frac{3}{5}\right)^2} = - \frac{4}{5} \). The negative value is chosen because sine is negative in quadrant II.

Step 2: Apply the formula for the tangent of the sum of two angles

The formula for the tangent of the sum of two angles states that \( \tan(u+v) = \frac{\sin u \cos v + \cos u \sin v}{\cos u \cos v - \sin u \sin v} \). In this case we substitute the given values of \( \sin u, \sin v, \cos u, \cos v \) into the formula which gives us: \( \tan(u+v) = \frac{\frac{5}{13} \times - \frac{3}{5} + - \frac{12}{13} \times -\frac{4}{5}}{- \frac{12}{13} \times - \frac{3}{5} - \frac{5}{13} \times -\frac{4}{5}} \).

Step 3: Simplify your equation

Solve the equation to get the final answer. After simplifying, \( \tan(u+v) = \frac{33}{56} \).