Problem 33 - 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 the values of x and y from point P

The terminal point given is \( P(x, y) = \left(-\frac{6}{7}, \frac{\sqrt{13}}{7}\right) \). Thus, \( x = -\frac{6}{7} \) and \( y = \frac{\sqrt{13}}{7} \).

Step 2: Determine the radius r

In the unit circle, the radius \( r \) is always 1, but let's verify the length of the radius given by the point with the formula \( r = \sqrt{x^2 + y^2} \). Substituting \( x \) and \( y \) into this formula:\[ r = \sqrt{\left(-\frac{6}{7}\right)^2 + \left(\frac{\sqrt{13}}{7}\right)^2} = \sqrt{\frac{36}{49} + \frac{13}{49}} = \sqrt{\frac{49}{49}} = 1\]This confirms that the radius is 1.

Step 3: Calculate \( \sin t \)

\( \sin t \) is equal to the \( y \)-coordinate of the point on the unit circle, which is \( y = \frac{\sqrt{13}}{7} \). Thus, \( \sin t = \frac{\sqrt{13}}{7} \).

Step 4: Calculate \( \cos t \)

\( \cos t \) is equal to the \( x \)-coordinate of the point on the unit circle, which is \( x = -\frac{6}{7} \). Thus, \( \cos t = -\frac{6}{7} \).

Step 5: Calculate \( \tan t \)

\( \tan t \) is the ratio of \( y \) to \( x \), given by the formula \( \tan t = \frac{y}{x} \). Substituting the values for \( y \) and \( x \), we have:\[ \tan t = \frac{\frac{\sqrt{13}}{7}}{-\frac{6}{7}} = \frac{\sqrt{13}}{-6} = -\frac{\sqrt{13}}{6}.\]

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 the coordinate plane (0,0). This circle helps us understand how the trigonometric functions, sine, cosine, and tangent, relate to angles.

In the unit circle, any point on the circle is represented by the coordinates \(x, y\), where \(x = \cos t\) and \(y = \sin t\) for a given angle \(t\). These coordinates originate from the unit circle definition: the hypotenuse (or radius) of any triangle formed is 1.

This simplifies calculations significantly because when computing trigonometric values like sine and cosine, you only need to look at the y and x coordinates, respectively. The radius is always 1, making it possible to directly associate these coordinates with standard trigonometric functions.

Sine

The sine function measures the vertical coordinate of a point on the unit circle.

When you have a terminal point \(P(x, y)\) on the unit circle, \( ext{sin} \, t\) is simply the y-coordinate of this point. For an angle \(t\), this point is determined, and the value of sine is its vertical distance from the origin, which lies between -1 and 1.

If given the point \(P = \left(-\frac{6}{7}, \frac{\sqrt{13}}{7}\right)\), then \( ext{sin} \, t = \frac{\sqrt{13}}{7}\). This means at the angle \(t\), the height directly above or below the origin is \frac{\sqrt{13}}{7}\.

Cosine

Cosine is associated with the horizontal coordinate of a point on the unit circle. The function, much like sine, is calculated using the coordinates of the terminal point on the unit circle.

For any terminal point \(P(x, y)\) on the unit circle, \(\cos t\) is the x-coordinate. This makes \(\cos t\) equivalent to the horizontal distance from the origin.

In the example where \(P = \left(-\frac{6}{7}, \frac{\sqrt{13}}{7}\right)\), the \(\cos t = -\frac{6}{7}\). This indicates that along a horizontal axis, the point lies \(\frac{6}{7}\) units to the left of the origin at the angle \(t\). Cosine values, like sine, also range between -1 and 1.

Tangent

The tangent function is derived from the sine and cosine functions. It represents the slope of the line formed by the segment connecting the point to the origin.

The formula for tangent is \(\tan t = \frac{\sin t}{\cos t} = \frac{y}{x}\). This provides the ratio of sine to cosine for that particular angle \(t\).

If we continue with the same point \(P = \left(-\frac{6}{7}, \frac{\sqrt{13}}{7}\right)\), we calculate tangent by: \(\tan t = \frac{\frac{\sqrt{13}}{7}}{-\frac{6}{7}} = -\frac{\sqrt{13}}{6}\). The value of tangent can express the steepness or inclination of the angle \(t\). It is not restricted to -1 and 1 and can take on any real number value.