Problem 138 - Calculus Volume 1

Question

\(P\) is a point on the unit circle. a. Find the (exact) missing coordinate value of each point and b. find the values of the six trigonometric functions for the angle \(\theta\) with a terminal side that passes through point P. Rationalize denominators. \(P\left(x, \frac{-\sqrt{15}}{4}\right), x>0\)

Step-by-Step Solution

Answer

The missing coordinate is \( x = \frac{1}{4} \). Trig functions: \( \sin \theta = \frac{-\sqrt{15}}{4} \), \( \cos \theta = \frac{1}{4} \), \( \tan \theta = -\sqrt{15} \), \( \sec \theta = 4 \), \( \csc \theta = -\frac{4\sqrt{15}}{15} \), \( \cot \theta = -\frac{\sqrt{15}}{15} \).

Step 1: Use the Pythagorean identity

On the unit circle, the coordinates of a point \( P(x, y) \) satisfy the equation \( x^2 + y^2 = 1 \). Given \( y = \frac{-\sqrt{15}}{4} \), substitute \( y \) into the equation: \[ x^2 + \left(\frac{-\sqrt{15}}{4}\right)^2 = 1 \].

Step 2: Simplify and solve for \( x^2 \)

Calculate \( \left(\frac{-\sqrt{15}}{4}\right)^2 = \frac{15}{16} \). The equation becomes \( x^2 + \frac{15}{16} = 1 \). Simplify to find \( x^2 \): \[ x^2 = 1 - \frac{15}{16} \].

Step 3: Calculate \( x^2 \)

Subtract \( \frac{15}{16} \) from 1. Use the equivalent fraction for 1: \( \frac{16}{16} \). So, \[ x^2 = \frac{16}{16} - \frac{15}{16} = \frac{1}{16} \].

Step 4: Solve for \( x \)

Since \( x^2 = \frac{1}{16} \), take the square root to find \( x \). Since \( x > 0 \), choose the positive root: \[ x = \frac{1}{4} \].

Step 5: Determine the six trigonometric functions

Using the coordinates \( x = \frac{1}{4} \) and \( y = \frac{-\sqrt{15}}{4} \):- \( \sin \theta = y = \frac{-\sqrt{15}}{4} \)- \( \cos \theta = x = \frac{1}{4} \)- \( \tan \theta = \frac{y}{x} = \frac{-\sqrt{15}/4}{1/4} = -\sqrt{15} \)- \( \cot \theta = \frac{1}{\tan \theta} = -\frac{1}{\sqrt{15}} = -\frac{\sqrt{15}}{15} \) (after rationalizing the denominator)- \( \sec \theta = \frac{1}{\cos \theta} = 4 \)- \( \csc \theta = \frac{1}{\sin \theta} = \frac{4}{-\sqrt{15}} = -\frac{4\sqrt{15}}{15} \) (after rationalizing the denominator).

Step 6: Final Calculation: Verify all trigonometric values

Ensure all calculations are consistent with the unit circle identity \( x^2 + y^2 = 1 \) and rationalize denominators where necessary.

Key Concepts

Unit Circle

The unit circle is a fundamental concept in trigonometry. It is a circle with a radius of one, centered at the origin of the coordinate plane. Each point on the unit circle can be described by its coordinates

• The x-coordinate represents the cosine of the angle \( \theta \)
• The y-coordinate represents the sine of the angle \( \theta \)

Because the radius of the unit circle is 1, every point on its circumference satisfies the equation \( x^2 + y^2 = 1 \). Thus, if you know one coordinate, such as \( y = \frac{-\sqrt{15}}{4} \), you can find the other by rearranging and solving this equation. The unit circle is a simple yet powerful tool for understanding trigonometric functions and identities.

Pythagorean Identity

The Pythagorean identity is derived from the Pythagorean theorem. It connects trigonometric functions with the geometry of the unit circle. The identity states:\[ \sin^2 \theta + \cos^2 \theta = 1 \]This identity is particularly useful because it applies to any angle \( \theta \).

• In the context of the unit circle, \( \sin(\theta) \) is the y-coordinate
• \( \cos(\theta) \) is the x-coordinate

To solve for a missing coordinate, plug the known value into this identity and rearrange the equation. For example, with \( y = \frac{-\sqrt{15}}{4} \), substitute into the equation to find \( x \), confirming that the identity holds true for these coordinates.

Rationalizing Denominators

Rationalizing denominators is an important process in simplifying expressions in mathematics. This involves converting a denominator containing a radical, such as \( \sqrt{15} \), into a rational number. The process improves clarity and can simplify further calculations.For example, if you have an expression like \( \frac{1}{\sqrt{15}} \), multiply the numerator and the denominator by \( \sqrt{15} \) to remove the radical:

• \( \frac{1}{\sqrt{15}} \times \frac{\sqrt{15}}{\sqrt{15}} = \frac{\sqrt{15}}{15} \)

This allows expressions like \( \cot \theta \) to become \( -\frac{\sqrt{15}}{15} \), which is easier to understand and more acceptable in mathematical writing. Rationalizing the denominator is crucial for clean, concise mathematical solutions.

Angle Theta

Understanding angle \( \theta \) within the context of trigonometric functions and the unit circle is critical to mastering these concepts. Angle \( \theta \) is typically measured in radians or degrees, but in unit circle applications, radians are often more common.

• The orientation of angle \( \theta \) determines the sign and values of the trigonometric functions
• For instance, if \( \theta \) is in the fourth quadrant, the sine function (y-coordinate) will be negative, and the cosine (x-coordinate) will be positive

This reflects directly on trigonometric calculations, such as \( \sin \theta = \frac{-\sqrt{15}}{4} \) and \( \cos \theta = \frac{1}{4} \), ensuring that you take the position of \( \theta \) into account when determining the signs of the trigonometric values. Understanding where \( \theta \) lies on the unit circle helps solve problems accurately and efficiently.