Problem 62 - Precalculus : Building Concepts and Connections

Question

Fill in the given table with the missing information. Approximate all nonexact anstoers to four decimal places. $$\begin{array}{|c|c|c|c|c|} \hline \text { Quadrant } & \sin t & \cos t & \tan t \\ \hline IV & & \frac{1}{2} & \\ \hline \end{array}$$

Step-by-Step Solution

Answer

\(\sin t \approx -0.8660\) , \(\tan t \approx -1.7320\)

Step 1: Determine the value of sin t

The Pythagorean identity in trigonometry states that \( \sin^{2}t + \cos^{2}t = 1 \). This implies that given the value of cos t, we can find sin t. From the given table, \(\cos t = \frac{1}{2}\). So, \(\sin t = \sqrt{1 - \cos^{2}t} = \sqrt{1 - (\frac{1}{2})^2} \). Since we are in quadrant IV and sin is negative, \(\sin t = -\sqrt{\frac{3}{4}} = -\frac{\sqrt{3}}{2}\)

Step 2: Determine the value of tan t

The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side. We can also determine the tangent from sin and cos as \( \tan t = \frac{\sin t}{\cos t} \). From the table, we know \(\cos t = \frac{1}{2}\) and from Step 1, we found \(\sin t = -\frac{\sqrt{3}}{2}\). Substituting these values in the formula, we get \(\tan t = -\frac{\sqrt{3}}{2} \div \frac{1}{2} = -\sqrt{3}\).

Step 3: Approximate the answers to four decimal places

Although the obtained answers for sin t and tan t are exact, the problem requires us to approximate the answers to four decimal places. Calculating, we find that \(-\frac{\sqrt{3}}{2} \approx -0.8660\) and \(-\sqrt{3} \approx -1.7320\).

Key Concepts

Pythagorean identity

The Pythagorean identity is a crucial concept in trigonometry, playing an important role in understanding the relationship between the different trigonometric functions of an angle. This identity is represented by the equation \( \sin^2 t + \cos^2 t = 1 \). This implies that if you know the cosine of an angle, you can calculate its sine, and vice versa. The Pythagorean identity stems from the Pythagorean theorem, which describes relationships within a right triangle.

In the given exercise, we use this identity to find the value of \( \sin t \). Given \( \cos t = \frac{1}{2} \), we can rearrange the identity as follows: \( \sin t = \pm \sqrt{1 - \cos^2 t} \). By inserting the value of \( \cos t \), we have \( \sin t = \pm \sqrt{1 - \left(\frac{1}{2}\right)^2} \). The solution proceeds to select the correct sign based on the quadrant information.

Understanding how to apply the Pythagorean identity is essential, as it forms the backbone of many trigonometric equations, allowing you to calculate one trigonometric function if you know another.

Trigonometric functions

Trigonometric functions are a fundamental part of math that relate angles of a triangle to the lengths of its sides. The three basic functions are sine (\( \sin\)), cosine (\( \cos\)), and tangent (\( \tan\)). Each function represents a different aspect of an angle's properties and has unique applications in various fields, including physics, architecture, and even music.

• Sine (\( \sin \)): In the context of a right triangle, sine of an angle is the ratio of the length of the side opposite to the angle to the hypotenuse.
• Cosine (\( \cos \)): Cosine is the ratio of the length of the adjacent side to the hypotenuse of the triangle.
• Tangent (\( \tan \)): Tangent relates to sine and cosine and is the ratio of \( \sin t \) to \( \cos t \), hence \( \tan t = \frac{\sin t}{\cos t} \).

In the exercise, we specifically deal with these trigonometric functions to fill in the table for quadrant IV. Each function has specific positive and negative signs depending on which quadrant the angle is in.

Approximation

Approximation is the process of finding a value that is close enough to the correct answer, which provides a practical solution when the exact answer is either too complicated or impossible to obtain. In trigonometry, approximation allows for simplification in calculations where precision beyond a certain decimal point doesn't affect the larger context of a problem.

For this exercise, the problem requires us to approximate certain trigonometric values to four decimal places. While we initially derive values like \( \sin t = -\frac{\sqrt{3}}{2} \) and \( \tan t = -\sqrt{3} \), which are exact, approximating them as \( -0.8660 \) and \( -1.7320 \) respectively allows for easier communication and application of these values. Approximations are especially handy in settings like engineering and computer calculations where only a finite decimal representation is feasible.