Problem 16 - 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. \(\cot \theta\) \((-4,8)\)

Step-by-Step Solution

Answer

a. \(r = 4\sqrt{5}\), b. \(\csc \theta = \frac{\sqrt{5}}{2}\), c. \(\sec \theta = -\sqrt{5}\), d. \(\cot \theta = -\frac{1}{2}\)

Step 1: Find the Radius

The point \(P(x,y)\) is \((-4, 8)\). The radius \(r\) of the circle can be found using the distance formula, which, since the center is at the origin \((0,0)\), becomes \(r = \sqrt{x^2 + y^2}\). Substitute \(x = -4\) and \(y = 8\): \[ r = \sqrt{(-4)^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5} \]

Step 2: Calculate Sine and Cosine

Using the coordinates \((-4, 8)\) and the radius \(r = 4\sqrt{5}\), find \(\sin \theta\) and \(\cos \theta\):\[ \sin \theta = \frac{y}{r} = \frac{8}{4\sqrt{5}} = \frac{2}{\sqrt{5}} \] \[ \cos \theta = \frac{x}{r} = \frac{-4}{4\sqrt{5}} = \frac{-1}{\sqrt{5}} \]

Step 3: Calculate CSC Theta

The cosecant of \(\theta\) is the reciprocal of sine. Thus, calculate \(\csc \theta\):\[ \csc \theta = \frac{1}{\sin \theta} = \frac{\sqrt{5}}{2} \]

Step 4: Calculate SEC Theta

The secant of \(\theta\) is the reciprocal of cosine. Calculate \(\sec \theta\):\[ \sec \theta = \frac{1}{\cos \theta} = -\sqrt{5} \]

Step 5: Calculate COT Theta

The cotangent of \(\theta\) is the cosine over sine. Find \(\cot \theta\):\[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{-1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \frac{-1}{2} \]

Key Concepts

Radius Calculation

To find the radius of a circle when given a point on its circumference, we use the distance formula. This formula finds the distance between the center of the circle, which is at the origin (0, 0), and the point P(x, y). For the point (-4, 8), the formula is \(r = \sqrt{(-4)^2 + 8^2}\). Here’s how it breaks down:

• The x-coordinate is -4, and when squared, (-4)^2 equals 16.
• The y-coordinate is 8, and when squared, 8^2 equals 64.
• Add these results: 16 + 64 = 80.
• Take the square root of 80, which simplifies to \(4\sqrt{5}\).

By following these steps, you can easily calculate any radius when you know the x and y coordinates of a point on the circle's edge.

Cosecant

Cosecant is a trigonometric function that is the reciprocal of sine. It is noted as \(\csc \theta = \frac{1}{\sin \theta}\). To understand its use, let's consider the previous calculation of \(\sin \theta\) with the point \((-4, 8)\). The sine was \(\sin \theta = \frac{2}{\sqrt{5}}\). Thus, \(\csc \theta\) is:

• \(\csc \theta = \frac{1}{\left(\frac{2}{\sqrt{5}}\right)}\)
• Which simplifies to \(\frac{\sqrt{5}}{2}\).

Understanding cosecant is important, especially in contexts where ratios of a triangle or circle are needed. It helps to remind that cosecant and sine are inverses.

Secant

The secant function, denoted as \(\sec \theta\), is the inverse of cosine. It shows up often in various problems involving circles and triangles. Here, the formula is \(\sec \theta = \frac{1}{\cos \theta}\). From our previous example, where \(\cos \theta = \frac{-1}{\sqrt{5}}\), finding the secant becomes:

• \(\sec \theta = \frac{1}{\left(\frac{-1}{\sqrt{5}}\right)}\)
• This simplifies to \(-\sqrt{5}\).

Secant provides critical insights into angles, especially those over a full rotation or radian measure.

Cotangent

Cotangent, represented as \(\cot \theta\), describes the ratio of the adjacent side over the opposite side in a right triangle or alternatively as cosine over sine. It’s best viewed as:\(\cot \theta = \frac{\cos \theta}{\sin \theta}\). Using our calculated values, with \(\cos \theta = \frac{-1}{\sqrt{5}}\) and \(\sin \theta = \frac{2}{\sqrt{5}}\), we find:

• \(\cot \theta = \frac{\left(\frac{-1}{\sqrt{5}}\right)}{\left(\frac{2}{\sqrt{5}}\right)}\)
• This simplifies to \(-\frac{1}{2}\).

The cotangent function is integral for understanding angles in trigonometry, especially when swapping between function forms to solve problems.