Problem 12 - Precalculus

Question

The terminal side of an angle \(\theta\) in standard position passes through values of the six trigonometric functions for angle \(\theta\) $$(-\sqrt{6},-\sqrt{5})$$

Step-by-Step Solution

Answer

The six trigonometric functions are: \(\sin\theta = -\frac{\sqrt{5}}{\sqrt{11}}\), \(\cos\theta = -\frac{\sqrt{6}}{\sqrt{11}}\), \(\tan\theta = \frac{\sqrt{5}}{\sqrt{6}}\), \(\csc\theta = -\frac{\sqrt{11}}{\sqrt{5}}\), \(\sec\theta = -\frac{\sqrt{11}}{\sqrt{6}}\), \(\cot\theta = \frac{\sqrt{6}}{\sqrt{5}}\).

Step 1: Identify the Coordinates

The given point is \((-\sqrt{6}, -\sqrt{5})\). Here, \(x = -\sqrt{6}\) and \(y = -\sqrt{5}\). We will use these to find the trigonometric functions for angle \(\theta\).

Step 2: Calculate the Radius (r)

The radius \(r\) is calculated using the distance formula: \(r = \sqrt{x^2 + y^2}\). Plugging in the values, we get \(r = \sqrt{(-\sqrt{6})^2 + (-\sqrt{5})^2} = \sqrt{6 + 5} = \sqrt{11}\).

Step 3: Find Sine and Cosine

The sine of the angle is given by \(\sin\theta = \frac{y}{r} = \frac{-\sqrt{5}}{\sqrt{11}}\). The cosine of the angle is \(\cos\theta = \frac{x}{r} = \frac{-\sqrt{6}}{\sqrt{11}}\).

Step 4: Calculate Tangent

The tangent of the angle is calculated as \(\tan\theta = \frac{y}{x} = \frac{-\sqrt{5}}{-\sqrt{6}} = \frac{\sqrt{5}}{\sqrt{6}}\).

Step 5: Derive Cosecant

The cosecant is the reciprocal of sine: \(\csc\theta = \frac{r}{y} = \frac{\sqrt{11}}{-\sqrt{5}}\).

Step 6: Determine Secant

The secant is the reciprocal of cosine: \(\sec\theta = \frac{r}{x} = \frac{\sqrt{11}}{-\sqrt{6}}\).

Step 7: Identify Cotangent

The cotangent is the reciprocal of tangent: \(\cot\theta = \frac{x}{y} = \frac{-\sqrt{6}}{-\sqrt{5}} = \frac{\sqrt{6}}{\sqrt{5}}\).

Key Concepts

Standard Position

In trigonometry, angles are often discussed in terms of their standard position on the coordinate plane. When an angle is in its standard position, its vertex sits at the origin of the Cartesian coordinate system, and its initial side extends along the positive x-axis. This forms the basis for defining trigonometric functions using a coordinate grid.
The terminal side of the angle is the side which "moves" or "rotates" to form the angle from its initial side. For the given problem, the terminal side of angle \( \theta \) passes through the point \((-\sqrt{6}, -\sqrt{5})\).

• Quadrants: With the angle in standard position, the coordinate plane is divided into four quadrants. The sign of \(x\) and \(y\) determines the quadrant where the terminal side lies. In this case, both \(x\) and \(y\) are negative, placing the terminal side in the third quadrant.
• Angle Measurement: Angles in trigonometry, especially in standard position, can be measured in degrees or radians, although no specific value might be given without further context.

Coordinate Geometry

Coordinate geometry is a critical part of understanding trigonometric functions in the coordinate plane.

In this scenario, the problem involves the point \((-\sqrt{6}, -\sqrt{5})\) which lies on the terminal side of the angle \( \theta \). Calculating various trigonometric functions involves using geometric techniques within this coordinate system:

• Radius Calculation: The radius or hypotenuse \( r \) is the distance from the origin to the point on the terminal side. This can be computed using the distance formula \( r = \sqrt{x^2 + y^2} \). This formula is derived from the Pythagorean theorem.
• Trigonometric Ratios: By understanding the position of a point in the coordinate plane, we can derive other trigonometric functions like sine, cosine, and tangent using the relationships:
  • Sine (\( \sin\theta \)) = opposite/hypotenuse = \( \frac{y}{r} \)
  • Cosine (\( \cos\theta \)) = adjacent/hypotenuse = \( \frac{x}{r} \)
  • Tangent (\( \tan\theta \)) = opposite/adjacent = \( \frac{y}{x} \)

Reciprocal Functions

Reciprocal trigonometric functions are an essential part of trigonometry. They are defined as the inverse of the basic trigonometric functions and provide additional relationships based on the original functions.

In the given problem, you need to find:

• Cosecant (\( \csc\theta \)): This function is the reciprocal of sine, which means \( \csc\theta = \frac{1}{\sin\theta} = \frac{r}{y} \).
• Secant (\( \sec\theta \)): Secant is the reciprocal of cosine, so \( \sec\theta = \frac{1}{\cos\theta} = \frac{r}{x} \).
• Cotangent (\( \cot\theta \)): This is the reciprocal of tangent, thus \( \cot\theta = \frac{1}{\tan\theta} = \frac{x}{y} \).

Understanding these reciprocal functions rounds out your grasp of trigonometric functions and gives you the tools to solve a variety of problems involving angles and coordinates.