Problem 20 - Algebra 2

Question

Find the exact values of the six trigonometric functions of \(\theta\) if the terminal side of \(\theta\) in standard position contains the given point. \((\sqrt{2},-\sqrt{2})\)

Step-by-Step Solution

Answer

The six trigonometric functions are: \(\sin(\theta) = -\frac{\sqrt{2}}{2}\), \(\cos(\theta) = \frac{\sqrt{2}}{2}\), \(\tan(\theta) = -1\), \(\csc(\theta) = -\sqrt{2}\), \(\sec(\theta) = \sqrt{2}\), \(\cot(\theta) = -1\).

Step 1: Understand the Point and the Coordinate System

The point given is \((\sqrt{2}, -\sqrt{2})\). This point lies in the fourth quadrant of the coordinate plane, as the x-coordinate is positive and the y-coordinate is negative.

Step 2: Calculate the Radius (Hypotenuse)

To find the hypotenuse \(r\), use the formula \(r = \sqrt{x^2 + y^2}\). Substituting our values, we get:\[r = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2\]

Step 3: Find the Sine of \(\theta\)

The sine of \(\theta\) is given by \( \sin(\theta) = \frac{y}{r} \). Substitute the values:\[\sin(\theta) = \frac{-\sqrt{2}}{2}\]

Step 4: Find the Cosine of \(\theta\)

The cosine of \(\theta\) is given by \( \cos(\theta) = \frac{x}{r} \). Substitute the values:\[\cos(\theta) = \frac{\sqrt{2}}{2}\]

Step 5: Find the Tangent of \(\theta\)

The tangent of \(\theta\) is given by \( \tan(\theta) = \frac{y}{x} \). Substitute the values:\[\tan(\theta) = \frac{-\sqrt{2}}{\sqrt{2}} = -1\]

Step 6: Find the Cosecant of \(\theta\)

The cosecant is the reciprocal of sine, \( \csc(\theta) = \frac{r}{y} \). Substitute the values:\[\csc(\theta) = \frac{2}{-\sqrt{2}} = -\sqrt{2}\]

Step 7: Find the Secant of \(\theta\)

The secant is the reciprocal of cosine, \( \sec(\theta) = \frac{r}{x} \). Substitute the values:\[\sec(\theta) = \frac{2}{\sqrt{2}} = \sqrt{2}\]

Step 8: Find the Cotangent of \(\theta\)

The cotangent is the reciprocal of tangent, \( \cot(\theta) = \frac{x}{y} \). Substitute the values:\[\cot(\theta) = \frac{\sqrt{2}}{-\sqrt{2}} = -1\]

Key Concepts

Coordinate System

When working with trigonometric functions, understanding the coordinate system is crucial. It helps us to visualize and solve problems involving angles and points. In this exercise, we're dealing with the Cartesian coordinate system, which is a two-dimensional plane divided into four quadrants. The quadrants are numbered counterclockwise starting from the positive x-axis.

Each quadrant has distinct signs for x and y coordinates:

• Quadrant I: Both x and y coordinates are positive.
• Quadrant II: x is negative, y is positive.
• Quadrant III: Both x and y are negative.
• Quadrant IV: x is positive, y is negative.

For the given point \((\sqrt{2}, -\sqrt{2})\), it lies in Quadrant IV because the x-coordinate (\(\sqrt{2}\)) is positive and the y-coordinate (-\(\sqrt{2}\)) is negative. This quadrant placement significantly influences the sign and value of trigonometric functions.

Radius Calculation

Finding the radius, or hypotenuse, of the triangle formed in the coordinate system is a fundamental step in determining trigonometric functions. The radius can be found using the distance formula, which in trigonometric terms is expressed as:

\[ r = \sqrt{x^2 + y^2} \]

This formula is derived from the Pythagorean theorem. Here, \(x\) and \(y\) are the coordinates of a point on the plane. In the exercise, the given point is \((\sqrt{2}, -\sqrt{2})\). Plugging these into the formula, we have:

\[ r = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \]

The radius \(r\) is important as it serves as the denominator in sine and cosine calculations, helping us accurately describe the angle's position in its respective quadrant.

Reciprocal Identities

Reciprocal identities are relationships between trigonometric functions and their reciprocals, offering alternative ways to express these functions. These identities are especially useful in calculations and manipulations involving trigonometric functions.

Here are the basic reciprocal identities:

• Cosecant is the reciprocal of sine: \( \csc(\theta) = \frac{1}{\sin(\theta)} \)
• Secant is the reciprocal of cosine: \( \sec(\theta) = \frac{1}{\cos(\theta)} \)
• Cotangent is the reciprocal of tangent: \( \cot(\theta) = \frac{1}{\tan(\theta)} \)

Applying these to the exercise, suppose we have \( \sin(\theta) = \frac{-\sqrt{2}}{2} \), \( \cos(\theta) = \frac{\sqrt{2}}{2} \), and \( \tan(\theta) = -1 \).
Then from the reciprocal relationships, the formulas become:

\[ \csc(\theta) = \frac{2}{-\sqrt{2}} = -\sqrt{2} \]
\[ \sec(\theta) = \frac{2}{\sqrt{2}} = \sqrt{2} \]
\[ \cot(\theta) = \frac{\sqrt{2}}{-\sqrt{2}} = -1 \]

These reciprocal functions complete the set of six standard trigonometric functions based on a single angle \( \theta \).