Problem 69 - Precalculus : Building Concepts and Connections

Question

Fill in the given table with the missing information. Approximate all nonexact anstoers to four decimal places. $$\begin{array}{|c|c|c|c|c|} \hline \text { Quadrant } & \sin t & \cos t & \tan t \\ \hline II & \ &-\frac{5}{13}\\\ \hline \end{array}$$

Step-by-Step Solution

Answer

The missing values in the table are: \( \sin t = \frac{12}{13} \) and \( \tan t = -\frac{12}{5} \).

Step 1: Find the Value of \( \sin t \)

We know that \( \sin^2 t + \cos^2 t = 1 \). So, to find \( \sin t \), we rearrange the formula to get \( \sin t = \sqrt{1 - \cos^2 t} \). Since \( \cos t = -\frac{5}{13} \), we get \( \sin t = \sqrt{1 -\left(-\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13} \). Since \( t \) is in the second quadrant, \( \sin t \) is positive, so we don't need to consider the negative square root.

Step 2: Find the Value of \( \tan t \)

We use the formula \( \tan t = \frac{\sin t}{\cos t} \) to find the value of \( \tan t \). Substituting the values of \( \sin t \) which is \( \frac{12}{13} \) and \( \cos t \) which is \( -\frac{5}{13} \) into the formula, we get \( \tan t = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5} \).

Key Concepts

Understanding Sin t Calculation

When we're dealing with trigonometric functions in various quadrants, it's crucial to remember the identity that holds for all angles, which is \( \sin^2 t + \cos^2 t = 1 \). This identity allows us to find one trigonometric function if we know the other, by rearranging the formula. In this exercise, we know \( \cos t \) and aim to find \( \sin t \).

Given that \( \cos t = -\frac{5}{13} \) and using the Pythagorean identity, we can rearrange it to \( \sin t = \sqrt{1 - \cos^2 t} \). With values plugged in, we find \( \sin t = \sqrt{1 - \left(-\frac{5}{13}\right)^2} = \sqrt{\frac{144}{169}} = \frac{12}{13} \).

It's important to identify in which quadrant the angle is, because this determines the sign of the trigonometric function. Since \( t \) is in the second quadrant, where sine is positive, we use the positive value of \( \sin t \). This detail is vital when solving problems that involve calculating trigonometric functions.

Exploring Cos t Calculation

In trigonometry, determining the values of functions based on their placement in the quadrants is essential to solve problems accurately. The cosine function, \( \cos t \), gives us the horizontal distance from the origin to the point on the unit circle, and its value varies based on the quadrant the angle is located in.

For angles residing in quadrant II, the cosine is always negative. This is because the x-coordinate of any point in the second quadrant on the unit circle is negative, verifying that \( \cos t = -\frac{5}{13} \) is consistent with the quadrant rule for cosine.

Understanding the signs of trigonometric functions depending on their quadrant is crucial. For quadrant II, remember:

• \( \sin t \): positive
• \( \cos t \): negative
• \( \tan t \): negative

Knowing these sign rules simplifies the process of calculating functions efficiently and correctly.

Calculating Tan t Effectively

To find the value of \( \tan t \), we use the relationship \( \tan t = \frac{\sin t}{\cos t} \). With our values, \( \sin t = \frac{12}{13} \) and \( \cos t = -\frac{5}{13} \), we substitute them into the formula to compute \( \tan t = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5} \).

It's crucial to recognize that the tangent function is the result of dividing sine by cosine, which means the signs and values of sine and cosine directly affect the value of tangent. In quadrant II, since sine is positive and cosine is negative, the tangent becomes negative, which aligns with what we know about the behavior of angles in the second quadrant.

Overall, understanding how to calculate \( \tan t \) using simple division helps demystify its computation. By knowing the sign rules and correctly applying the quotient of sine and cosine, one can easily calculate the tangent without error.