Problem 33 - Precalculus Mathematics for Calculus

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

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

Step 1: Identify the coordinates of the terminal point

The coordinates of the terminal point are given as \(P\left(-\frac{6}{7}, \frac{\sqrt{13}}{7}\right)\), where \(x = -\frac{6}{7}\) and \(y = \frac{\sqrt{13}}{7}\).

Step 2: Recognize the formula for \\( \sin t \\)

For a point \((x, y)\) on a unit circle, \(\sin t\) corresponds to the \(y\)-coordinate of the point \(P\). Therefore, \(\sin t = \frac{\sqrt{13}}{7}\).

Step 3: Recognize the formula for \\ ( \cos t \\)

For a point \((x, y)\) on a unit circle, \(\cos t\) corresponds to the \(x\)-coordinate of the point \(P\). Therefore, \(\cos t = -\frac{6}{7}\).

Step 4: Determine \\ ( \tan t \\) using \\ ( \sin t \\) and \\ ( \cos t \\)

\(\tan t = \frac{\sin t}{\cos t}\). Substitute the values of \(\sin t\) and \(\cos t\):\[\tan t = \frac{\frac{\sqrt{13}}{7}}{-\frac{6}{7}} = -\frac{\sqrt{13}}{6}.\]

Key Concepts

Unit Circle

The unit circle is a fundamental concept in trigonometry. It is a circle with a radius of 1, centered at the origin of a coordinate plane. This simple structure allows for easy calculation of trigonometric functions based on angles. A point \( P(x, y) \) on the unit circle is used to determine the sine, cosine, and tangent values for a given angle \( t \). Understanding the unit circle is crucial because it connects trigonometric functions with geometry and allows visualizing angles and their respective trigonometric values.

The unit circle conveniently features angles measured in radians around it. The coordinates \((x, y)\) for any point on this circle fulfill the equation \( x^2 + y^2 = 1 \). From these coordinates, we can directly extract the sine and cosine values:

• \( x \) represents \( \cos t \)
• \( y \) represents \( \sin t \)

The unit circle helps predict the behavior of trigonometric functions through their symmetry and periodicity.

Sine Function

The sine function, represented as \( \sin t \), is one of the primary trigonometric functions. It gives the vertical coordinate \( y \) of a point on the unit circle for a given angle \( t \). This means \( \sin t \) indicates how far up or down the point is from the x-axis, reflecting the angle's height in the circle.

For the terminal point \( P \left( -\frac{6}{7}, \frac{\sqrt{13}}{7} \right) \), the sine value is the \( y \)-coordinate:

• \( \sin t = \frac{\sqrt{13}}{7} \)

The sine function is periodic, with a period of \( 2\pi \), meaning it repeats its values every \( 2\pi \) radians. Observing these patterns on the unit circle helps grasp how the sine function behaves over different quadrants.

Cosine Function

The cosine function, noted as \( \cos t \), is another central trigonometric function. It provides the horizontal coordinate \( x \) of a point on the unit circle for a specific angle \( t \). This represents how far left or right the point is from the y-axis, reflecting the base or width concerning the circle.

In the exercise's terminal point \( P \left( -\frac{6}{7}, \frac{\sqrt{13}}{7} \right) \), the cosine value is obtained from the \( x \)-coordinate:

• \( \cos t = -\frac{6}{7} \)

Like sine, the cosine function is periodic with a period of \( 2\pi \). It has a special symmetry, being an even function: \( \cos(-t) = \cos(t) \). This characteristic is illustrated clearly by the behavior of the points on the unit circle.

Tangent Function

The tangent function, denoted as \( \tan t \), is determined by dividing the sine by the cosine values \( \left( \tan t = \frac{\sin t}{\cos t} \right) \). It represents the slope of the line that connects the origin to the point \( P(x, y) \) on the unit circle. This means the tangent gives insight into the angle's steepness or incline.

For our specific terminal point \( P \left( -\frac{6}{7}, \frac{\sqrt{13}}{7} \right) \), the tangent is calculated as:

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

The tangent function's periodicity differs from sine and cosine, repeating every \( \pi \) radians instead of \( 2\pi \). The understanding of the tangent function is crucial for analyzing angles that result in vertical asymptotes, evident at points like \( \frac{\pi}{2} \) within the trigonometric cycle.