Problem 31 - Algebra and Trigonometry

Question

The terminal point \(P(x, y)\) determined by a real number \(t\) is given. Find \(\sin t, \cos t,\) and \(\tan t\) $$ \left(-\frac{6}{7}, \frac{\sqrt{13}}{7}\right) $$

Step-by-Step Solution

Answer

\(\sin t = \frac{\sqrt{13}}{7}, \cos t = -\frac{6}{7}, \tan t = -\frac{\sqrt{13}}{6}\).

Step 1: Identify Definitions

The terminal point \(P(x, y)\) is associated with a real number \(t\) on the unit circle, such that \(x = \cos t\) and \(y = \sin t\). We'll use these coordinates to find \(\sin t\) and \(\cos t\).

Step 2: Find \(\sin t\)

Since the y-coordinate of the terminal point is given as \(y = \frac{\sqrt{13}}{7}\), we have \(\sin t = \frac{\sqrt{13}}{7}\).

Step 3: Find \(\cos t\)

Since the x-coordinate of the terminal point is given as \(x = -\frac{6}{7}\), we have \(\cos t = -\frac{6}{7}\).

Step 4: Find \(\tan t\)

To find \(\tan t\), use the formula \(\tan t = \frac{\sin t}{\cos t}\). So, \[\tan t = \frac{\frac{\sqrt{13}}{7}}{-\frac{6}{7}} = -\frac{\sqrt{13}}{6}\].

Key Concepts

Unit Circle

The unit circle is a powerful tool in trigonometry. It is a circle centered at the origin \(0,0\) on a coordinate plane with a radius of 1. This circle allows us to define the trigonometric functions sine, cosine, and tangent in terms of the coordinates of points on the circle.

When we talk about the unit circle, we often deal with angles measured in radians. A full circle is \(2\pi\) radians, which corresponds to 360 degrees. The circle makes it easy to visualize the angles and the corresponding points.

For any angle \(t\), the terminal point \(P\) on the unit circle is determined, giving us coordinates \(x\) and \(y\). These coordinates are directly related to the trigonometric functions: \(x = \cos t\) and \(y = \sin t\). These connections make solving trigonometric problems intuitive and visual.

Sine

The sine function relates the angle \(t\) to the y-coordinate of a point on the unit circle. In other words, \(\sin t\) is the vertical distance from the x-axis to the point \(P(x, y)\).

### Calculation of SineIn the problem, we are given a terminal point \(P\left(x, \frac{\sqrt{13}}{7}\right)\). Here, the y-coordinate is \(\frac{\sqrt{13}}{7}\), therefore \(\sin t = \frac{\sqrt{13}}{7}\).

The range of the sine function lies between -1 and 1, as the circle’s radius limits the movement of any point. This function is periodic with a period of \(2\pi\), meaning that it repeats every \(2\pi\) radians.

Cosine

The cosine function relates the angle \(t\) to the x-coordinate of a point on the unit circle. \(\cos t\) tells us the horizontal distance from the y-axis to the point \(P(x, y)\).

### Calculation of CosineFor the problem’s terminal point \(P\left(-\frac{6}{7}, y\right)\), the x-coordinate is \(-\frac{6}{7}\), implying \(\cos t = -\frac{6}{7}\).

Like sine, the range of the cosine function is between -1 and 1 due to the unit circle’s radius. Cosine also has a period of \(2\pi\), making it a periodic function. This periodicity allows the pattern to repeat every complete circle.

Tangent

The tangent function is defined differently. It is the ratio of the sine of an angle to its cosine. Thus, \(\tan t = \frac{\sin t}{\cos t}\). This function can be thought of as the slope of the line that passes through the origin and extends out to the point \(P(x, y)\) on the unit circle.

### Calculation of TangentTo find \(\tan t\) in the given problem, we use the previously calculated sine and cosine values:

• \(\sin t = \frac{\sqrt{13}}{7}\)
• \(\cos t = -\frac{6}{7}\)

Substitute these into the tangent formula: \[\tan t = \frac{\frac{\sqrt{13}}{7}}{-\frac{6}{7}} = -\frac{\sqrt{13}}{6}\]

The range of tangent differs and can extend to any real number. It is an important function, especially useful in understanding angles and slopes.