Problem 15 - Algebra 2 and Trigonometry

Question

In \(11-18, P\) is a point on the terminal side of an angle in standard position with measure \(\theta\) and on a circle with center at the origin and radius \(r .\) For each point \(P,\) find: a. \(r\) b. \(\csc \theta\) c. \(\sec \theta\) d. \(\cot \theta\) \((-6,6)\)

Step-by-Step Solution

Answer

a. \(r = 6\sqrt{2}\), b. \(\csc \theta = \sqrt{2}\), c. \(\sec \theta = -\sqrt{2}\), d. \(\cot \theta = -1\).

Step 1: Calculate the radius (r)

To find the radius \(r\) of the circle, we use the formula for the distance from the origin \((0, 0)\) to the point \((-6, 6)\). The formula is \(r = \sqrt{x^2 + y^2}\). Substitute \(x = -6\) and \(y = 6\) to get: \[r = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}.\] So, the radius \(r\) is \(6\sqrt{2}\).

Step 2: Determine \(\sin \theta\) and \(\cos \theta\)

Using the coordinates \((-6, 6)\), we find \(\sin \theta\) and \(\cos \theta\). The formula for \(\sin \theta\) is \(\frac{y}{r}\) and for \(\cos \theta\) is \(\frac{x}{r}\). Thus, \(\sin \theta = \frac{6}{6\sqrt{2}} = \frac{\sqrt{2}}{2}\) and \(\cos \theta = \frac{-6}{6\sqrt{2}} = -\frac{\sqrt{2}}{2}\).

Step 3: Find \(\csc \theta\)

The cosecant function is the reciprocal of the sine function. Therefore, \(\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}.\) So, \(\csc \theta = \sqrt{2}.\)

Step 4: Find \(\sec \theta\)

The secant function is the reciprocal of the cosine function. Therefore, \(\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{\sqrt{2}}{2}} = -\sqrt{2}.\) So, \(\sec \theta = -\sqrt{2}.\)

Step 5: Find \(\cot \theta\)

The cotangent function is the ratio of \(\cos \theta\) and \(\sin \theta\). Therefore, \(\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1.\) So, \(\cot \theta = -1.\)

Key Concepts

Cosecant

The cosecant function is one of the lesser-known trigonometric functions, often abbreviated as \(\csc\). It's essentially the reciprocal of the sine function. If you're familiar with sine, calculating cosecant becomes very straightforward. When given a point \((x, y)\) on the terminal side of an angle in standard position, you can find the sine as \( \sin \theta = \frac{y}{r} \), where \( r \) is the radius of the circle, which, in this case, is \( 6\sqrt{2} \). Therefore, the cosecant is:

• \( \csc \theta = \frac{1}{\sin \theta} \)
• This can be expressed as \( \frac{r}{y} \)

\( \csc \theta \) provides the length of the hypotenuse over the length of the opposite side from the angle \( \theta \). In our example, the cosecant works out to \( \sqrt{2} \). It becomes especially useful in solving triangles when you have right-angled scenarios.

Secant

The secant function, often written as \( \sec \), is the reciprocal of the cosine function. It isn't used as frequently as sine or cosine, but it's essential, especially in certain mathematical and engineering applications. Determining secant follows the pattern of using coordinates on the unit circle:

• Given a point \((-6, 6)\), find \( \cos \theta = \frac{x}{r} \)
• Thus, \( \sec \theta = \frac{1}{\cos \theta} = \frac{r}{x} \)

This reciprocal relationship tells us that secant equals the length of the hypotenuse divided by the adjacent side to the angle \(\theta\). In the scenario from our step-by-step solution, \(\sec \theta\) evaluates to \(-\sqrt{2}\). It's critical, especially in applications involving waves and oscillations.

Cotangent

The cotangent function, shown as \( \cot \), is the reciprocal of the tangent function and is fundamental in trigonometry. It describes the relationship between the adjacent side and the opposite side of a right triangle:

• Calculate \( \cot \theta \) as \( \frac{\cos \theta}{\sin \theta} \)
• This can be represented as \( \frac{x}{y} \)

For our example point \((-6, 6)\), the cotangent becomes \(-1\). This result arises because the cosine and sine are equal in magnitude and opposite in sign. The cotangent function is incredibly useful in finding slope and solving problems involving right triangles.

Standard Position Angle

A standard position angle helps us understand trigonometric functions visually. An angle is in standard position when its vertex is at the origin of the coordinate plane, and its initial side is along the positive x-axis. This setup forms the basis for the unit circle, where:

• The coordinates \((x, y)\) represent points on the circle's circumference
• The angle \(\theta\) is formed by the rotation from the x-axis

Such angles facilitate the exploration of trigonometric ratios, allowing students to connect angle measures with circle properties effortlessly. This foundational concept is crucial to understand how angles and side lengths interact in trigonometric functions, like sine, cosine, and their reciprocals.