Problem 3 - Precalculus

Question

The terminal side of an angle \(\theta\) in standard position passes through values of the six trigonometric functions for angle \(\theta\) $$\left(\frac{1}{2}, \frac{2}{5}\right)$$

Step-by-Step Solution

Answer

The six trig functions are: \( \sin(\theta) = \frac{4\sqrt{41}}{41} \), \( \cos(\theta) = \frac{5\sqrt{41}}{41} \), \( \tan(\theta) = \frac{4}{5} \), \( \cot(\theta) = \frac{5}{4} \), \( \sec(\theta) = \frac{\sqrt{41}}{5} \), \( \csc(\theta) = \frac{\sqrt{41}}{4} \).

Step 1: Determine the Distance (r)

To find the trigonometric functions, we first need to calculate the distance from the origin to the point \(\left(\frac{1}{2}, \frac{2}{5}\right)\). This distance is represented as \(r\) and calculated using the formula: \[ r = \sqrt{x^2 + y^2} \] where \(x = \frac{1}{2}\) and \(y = \frac{2}{5}\). Substitute the values: \[ r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{2}{5}\right)^2} = \sqrt{\frac{1}{4} + \frac{4}{25}} = \sqrt{\frac{25}{100} + \frac{16}{100}} = \sqrt{\frac{41}{100}} = \frac{\sqrt{41}}{10} \]

Step 2: Calculate Sin(θ), Cos(θ), and Tan(θ)

With \(r\) known, use the definitions of sine, cosine, and tangent:\[ \sin(\theta) = \frac{y}{r} \]\[ \cos(\theta) = \frac{x}{r} \]\[ \tan(\theta) = \frac{y}{x} \]Substitute the values:\[ \sin(\theta) = \frac{\frac{2}{5}}{\frac{\sqrt{41}}{10}} = \frac{2}{5} \cdot \frac{10}{\sqrt{41}} = \frac{4\sqrt{41}}{41} \]\[ \cos(\theta) = \frac{\frac{1}{2}}{\frac{\sqrt{41}}{10}} = \frac{1}{2} \cdot \frac{10}{\sqrt{41}} = \frac{5\sqrt{41}}{41} \]\[ \tan(\theta) = \frac{\frac{2}{5}}{\frac{1}{2}} = \frac{4}{5} \]

Step 3: Calculate Cot(θ), Sec(θ), and Csc(θ)

Use the reciprocal identities for cotangent, secant, and cosecant:\[ \cot(\theta) = \frac{x}{y} \]\[ \sec(\theta) = \frac{1}{\cos(\theta)} \]\[ \csc(\theta) = \frac{1}{\sin(\theta)} \]Substitute the values:\[ \cot(\theta) = \frac{1/2}{2/5} = \frac{5}{4} \]\[ \sec(\theta) = \frac{1}{\frac{5\sqrt{41}}{41}} = \frac{\sqrt{41}}{5} \]\[ \csc(\theta) = \frac{1}{\frac{4\sqrt{41}}{41}} = \frac{\sqrt{41}}{4} \]

Key Concepts

Standard Position

When we talk about angles in trigonometry, particularly in the Cartesian coordinate system, understanding the concept of "standard position" is fundamental. An angle is said to be in standard position if its vertex is at the origin of the coordinate plane, and its initial side lies on the positive x-axis. The measurement function then swings the terminal side counterclockwise for positive angles and clockwise for negative angles.

• Vertex at the Origin: This is the fixed point where the two line segments forming the angle meet.
• Initial Side on Positive x-axis: This is where the angle starts. Any rotation from this point is the angle measurement.

Understanding angles in standard position aids in visualizing and computing trigonometric functions as it sets a consistent reference point. As was applied in the original exercise, the point \((\frac{1}{2}, \frac{2}{5})\) lies on the plane, denoting where the terminal side comes to rest after measuring the angle.

Reciprocal Identities

Trigonometric functions are interconnected in many ways, and one very useful set of relationships between them are known as reciprocal identities. These identities express basic trig functions in terms of each other. This is especially straightforward given by:

• Cosecant (\(\csc(\theta)\)): The reciprocal of sine, defined as \(\csc(\theta) = \frac{1}{\sin(\theta)}\).
• Secant (\(\sec(\theta)\)): The reciprocal of cosine, defined as \(\sec(\theta) = \frac{1}{\cos(\theta)}\).
• Cotangent (\(\cot(\theta)\)): The reciprocal of tangent, defined as \(\cot(\theta) = \frac{1}{\tan(\theta)}\).

In the example from the original problem, these identities helped find the remaining three trigonometric functions once sine, cosine, and tangent were established. For instance, knowing \(\sin(\theta)\) immediately allowed for finding \(\csc(\theta)\) as demonstrated.

Right Triangle

Right triangles form the backbone of trigonometry, providing visual and practical understanding for calculating the primary trigonometric functions. A right triangle features one 90-degree angle, forming a neat geometric shape that allows for defining sine, cosine, and tangent.When you break down trigonometric functions using a right triangle, these terms help:

• Hypotenuse: The longest side opposite the right angle. It is used as the divisor in sine and cosine.
• Adjacent Side: Side next to the angle of interest. It helps form cosine along with the hypotenuse.
• Opposite Side: Side opposite the angle of interest. It works with the hypotenuse to calculate sine.

While the original task relied on coordinate geometry, the distance calculated (\(r\)) acts similar to the hypotenuse. Each trigonometric function sees its origin here, rooted in the simple yet powerful right triangle.