Problem 68 - Algebra and Trigonometry

Question

\(P=\left(-\frac{\sqrt{7}}{3}, \frac{\sqrt{2}}{3}\right)\) is the point on the unit circle that corresponds to a real number \(t\). Find the exact values of the six trigonometric functions of \(t\).

Step-by-Step Solution

Answer

\(\cos t = -\frac{\sqrt{7}}{3}\), \(\sin t = \frac{\sqrt{2}}{3}\), \(\tan t = -\frac{\sqrt{14}}{7}\), \(\cot t = -\frac{\sqrt{14}}{2}\), \(\sec t = -\frac{3\sqrt{7}}{7}\), \(\csc t = \frac{3\sqrt{2}}{2}\)

Step 1: Identify the Coordinates

The given point is \(P = \left( -\frac{\sqrt{7}}{3}, \frac{\sqrt{2}}{3} \right)\). This point corresponds to \( (\cos t, \sin t) \) on the unit circle.

Step 2: Determine \(\cos t\)

Since the x-coordinate corresponds to \(\cos t\), we have \(\cos t = -\frac{\sqrt{7}}{3}\).

Step 3: Determine \(\sin t\)

Since the y-coordinate corresponds to \(\sin t\), we have \(\sin t = \frac{\sqrt{2}}{3}\).

Step 4: Calculate \(\tan t\)

\(\tan t\) is given by the ratio \(\frac{\sin t}{\cos t}\): \[\tan t = \frac{\sin t}{\cos t} = \frac{\frac{\sqrt{2}}{3}}{-\frac{\sqrt{7}}{3}} = - \frac{\sqrt{2}}{\sqrt{7}} = -\frac{\sqrt{14}}{7}\]

Step 5: Calculate \(\cot t\)

\(\cot t\) is the reciprocal of \(\tan t\): \[\cot t = \frac{1}{\tan t} = -\frac{\sqrt{7}}{\sqrt{2}} = -\frac{\sqrt{14}}{2}\]

Step 6: Calculate \(\sec t\)

\(\sec t\) is the reciprocal of \(\cos t\): \[\sec t = \frac{1}{\cos t} = \frac{1}{-\frac{\sqrt{7}}{3}} = -\frac{3}{\sqrt{7}} = -\frac{3\sqrt{7}}{7}\]

Step 7: Calculate \(\csc t\)

\(\csc t\) is the reciprocal of \(\sin t\): \[\csc t = \frac{1}{\sin t} = \frac{1}{\frac{\sqrt{2}}{3}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}\]

Key Concepts

Unit Circle

The unit circle is a crucial concept in trigonometry. It is a circle with a radius of 1, centered at the origin of a coordinate plane. The standard equation for the unit circle is: \ x^2 + y^2 = 1. The unit circle makes it easy to define the trigonometric functions. Every point on the unit circle represents a unique angle, t, measured from the positive x-axis. For instance, a point \(\cos t, \sin t\) on the unit circle corresponds to the cosine and sine of the angle t. In our exercise, the point \( -\frac{\sqrt{7}}{3}, \frac{\sqrt{2}}{3} \) lies on the unit circle. This means we can directly use the coordinates to find trigonometric relationships.

Sine and Cosine

Sine and cosine functions are fundamental in trigonometry. For any angle t on the unit circle:

• \( \cos t \) provides the x-coordinate.
• \( \sin t \) gives the y-coordinate.

In our problem, the given coordinates are \( -\frac{\sqrt{7}}{3}, \frac{\sqrt{2}}{3} \). So:

• \( \cos t = -\frac{\sqrt{7}}{3} \)
• \( \sin t = \frac{\sqrt{2}}{3} \)

These values help us determine other trigonometric functions. Remember that the sine and cosine are often used to describe oscillatory motion, such as waves.

Tangent and Cotangent

Tangent and cotangent are the ratios of sine and cosine functions. They describe the slope of the line created by the angle t on the unit circle. The tangent function is defined as: \( \tan t = \frac{\sin t}{\cos t} \). Using our coordinates: \[ \tan t = \frac{\frac{\sqrt{2}}{3}}{-\frac{\sqrt{7}}{3}} = -\frac{\sqrt{2}}{\sqrt{7}} = -\frac{\sqrt{14}}{7} \]Cotangent is the reciprocal of tangent: \[ \cot t = \frac{1}{\tan t} = -\frac{\sqrt{7}}{\sqrt{2}} = -\frac{\sqrt{14}}{2} \]Understanding these functions helps with more complex waveforms and circular motion problems.

Secant and Cosecant

Secant and cosecant are the reciprocals of cosine and sine, respectively. They are used less frequently but are essential for solving certain trigonometric problems. The secant function is defined as: \( \sec t = \frac{1}{\cos t} \). Using our given coordinates: \[ \sec t = \frac{1}{-\frac{\sqrt{7}}{3}} = -\frac{3}{\sqrt{7}} = -\frac{3\sqrt{7}}{7} \]Similarly, the cosecant function is: \[ \csc t = \frac{1}{\sin t} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \]Secant and cosecant are particularly useful in problems involving vertical and horizontal distances. They also appear in hyperbolic functions and advanced calculus.