Problem 83 - A Graphical Approach to Precalculus with Limits

Question

Use the identity \(\cos ^{2} s+\sin ^{2} s=1\) to find the value of \(x\) or \(y,\) as appropriate. Then, assuming that \(s\) corresponds to the given point on the unit circle, find the six circular function values for \(s\). $$\left(-\frac{1}{3}, y\right), y<0$$

Step-by-Step Solution

Answer

\( y = -\frac{2\sqrt{2}}{3} \). Circular functions: \( \cos s = -\frac{1}{3} \), \( \sin s = -\frac{2\sqrt{2}}{3} \), \( \tan s = 2\sqrt{2} \), \( \csc s = -\frac{3\sqrt{2}}{4} \), \( \sec s = -3 \), \( \cot s = \frac{\sqrt{2}}{4} \).

Step 1: Understand the identity

We are given the identity \( \cos^2 s + \sin^2 s = 1 \). This identity states that for any angle \( s \) on the unit circle, the square of the cosine plus the square of the sine is always equal to 1.

Step 2: Plug in the Given Cosine Value

We know \( \cos s = -\frac{1}{3} \). Substituting this into the identity gives us: \[\left(-\frac{1}{3}\right)^2 + \sin^2 s = 1\] which simplifies to \[ \frac{1}{9} + \sin^2 s = 1 \].

Step 3: Solve for \( \sin^2 s \)

Rearrange the equation from Step 2:\[ \sin^2 s = 1 - \frac{1}{9} \]\[ \sin^2 s = \frac{9}{9} - \frac{1}{9} \]\[ \sin^2 s = \frac{8}{9} \].

Step 4: Find \( \sin s \)

Since \( y < 0 \), this means \( \sin s < 0 \). Therefore, we take the negative square root:\[ \sin s = -\sqrt{\frac{8}{9}} \]\[ \sin s = -\frac{\sqrt{8}}{3} \]Since \( \sqrt{8} = 2\sqrt{2} \), we have \[ \sin s = -\frac{2\sqrt{2}}{3} \].

Step 5: Find All Six Trigonometric Functions

Now we find the six circular functions:1. \( \cos s = -\frac{1}{3} \)2. \( \sin s = -\frac{2\sqrt{2}}{3} \)3. \( \tan s = \frac{\sin s}{\cos s} = \frac{-\frac{2\sqrt{2}}{3}}{-\frac{1}{3}} = 2\sqrt{2} \)4. \( \csc s = \frac{1}{\sin s} = -\frac{3}{2\sqrt{2}} = -\frac{3\sqrt{2}}{4} \) (rationalizing denominator)5. \( \sec s = \frac{1}{\cos s} = -3 \)6. \( \cot s = \frac{1}{\tan s} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \) (rationalizing denominator)

Key Concepts

Unit Circle

The unit circle is a fundamental concept in trigonometry, primarily used to understand the relationships between angles and lengths. It is a circle with a radius of 1, centered at the origin of a coordinate plane. This simplicity makes it incredibly useful for defining trigonometric functions.

• On the unit circle, each point \( (x, y) \) corresponds to an angle \( s \) from the x-axis, measured in radians.
• The x-coordinate of a point on the unit circle gives us \( \cos s \) and the y-coordinate gives us \( \sin s \).
• For any angle \( s \), the coordinates satisfy the identity \( \cos^2 s + \sin^2 s = 1 \) because they are derived from the Pythagorean Theorem.

A point on the unit circle helps visualize how the cosine and sine values change as the angle rotates through different quadrants. Since our given point is \( (-\frac{1}{3}, y) \) and \( y < 0 \), this places our angle in the third or fourth quadrant, where sine values are negative.

Sine and Cosine

Sine and cosine are critical circular functions associated with an angle. They help us understand the position of a point on the unit circle, which directly links to the angle's trigonometric properties.

• In this exercise, we are told \( \cos s = -\frac{1}{3} \). This means the angle \( s \) lies in a part of the unit circle where the x-value (cosine) is negative.
• To find the sine value of the angle, we use the identity \( \cos^2 s + \sin^2 s = 1 \). By substituting the cosine value, we find the sine to be \( \sin s = -\frac{2\sqrt{2}}{3} \).
• This corresponding sine value tells us the y-coordinate is negative, further confirming the angle is in the third or fourth quadrant.

The sine and cosine values give primary insights into the angle's behavior and help determine other trigonometric functions like tangent, secant, cosecant, and cotangent.

Circular Functions

Circular functions are the trigonometric functions derived from the unit circle and include sine, cosine, and tangent, among others. They are essential for various calculations involving angles.

• The cosine function, derived from the x-coordinate on the unit circle, deals with horizontal displacement, \( \cos s = -\frac{1}{3} \) in our example.
• The sine function, representing vertical displacement, is \( \sin s = -\frac{2\sqrt{2}}{3} \).
• Tangent, secant, cosecant, and cotangent are other circular functions that provide deeper values:
  • \( \tan s = \frac{\sin s}{\cos s} = 2\sqrt{2} \)
  • \( \sec s = \frac{1}{\cos s} = -3 \)
  • \( \csc s = \frac{1}{\sin s} = -\frac{3\sqrt{2}}{4} \)
  • \( \cot s = \frac{1}{\tan s} = \frac{\sqrt{2}}{4} \)

These functions help in various fields such as physics, engineering, and computer science for analyzing periodic phenomena and solving angular problems.