Problem 73 - Precalculus: Functions and Graphs

Question

Find the exact values of the six trigonometric functions of \(\boldsymbol{\theta}\) if \(\boldsymbol{\theta}\) is in standard position and \(P\) is on the terminal side. $$P(-2,-5)$$

Step-by-Step Solution

Answer

In the third quadrant, \(\sin(\theta) = \frac{-5\sqrt{29}}{29}\), \(\cos(\theta) = \frac{-2\sqrt{29}}{29}\), \(\tan(\theta) = \frac{5}{2}\), \(\csc(\theta) = -\frac{\sqrt{29}}{5}\), \(\sec(\theta) = -\frac{\sqrt{29}}{2}\), \(\cot(\theta) = \frac{2}{5}\).

Step 1: Identify the Quadrant

The point \(P(-2, -5)\) is located in the third quadrant because both the \(x\) and \(y\) coordinates are negative. In the third quadrant, sine and cosine are both negative, and tangent is positive.

Step 2: Calculate the Radius (r)

The radius (hypotenuse) \(r\) is the distance from the origin to the point \((-2, -5)\). Use the Pythagorean theorem:\[ r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \]

Step 3: Define the Six Trigonometric Functions

Using the point \(P(-2, -5)\) and radius \(r = \sqrt{29}\), calculate the trigonometric functions as follows:- \( \sin(\theta) = \frac{y}{r} = \frac{-5}{\sqrt{29}} \)- \( \cos(\theta) = \frac{x}{r} = \frac{-2}{\sqrt{29}} \)- \( \tan(\theta) = \frac{y}{x} = \frac{-5}{-2} = \frac{5}{2} \)- \( \csc(\theta) = \frac{r}{y} = \frac{\sqrt{29}}{-5} \)- \( \sec(\theta) = \frac{r}{x} = \frac{\sqrt{29}}{-2} \)- \( \cot(\theta) = \frac{x}{y} = \frac{-2}{-5} = \frac{2}{5} \)

Step 4: Rationalize the Denominators

In trigonometric values, it's standard to rationalize the denominators:- \( \sin(\theta) = \frac{-5\sqrt{29}}{29} \)- \( \cos(\theta) = \frac{-2\sqrt{29}}{29} \)- \( \csc(\theta) = -\frac{\sqrt{29}}{5} \)- \( \sec(\theta) = -\frac{\sqrt{29}}{2} \)

Key Concepts

Quadrants

Understanding quadrants is essential when dealing with trigonometric functions, as it helps determine the signs of these functions. The coordinate plane is divided into four quadrants:

• The first quadrant where both x and y are positive.
• The second quadrant where x is negative and y is positive.
• The third quadrant where both x and y are negative.
• The fourth quadrant where x is positive and y is negative.

For the point \(P(-2, -5)\), both x and y are negative, placing it in the third quadrant. In this quadrant:

• Sine and cosine values are negative.
• Tangent is positive because it's the ratio of y to x, and two negatives make a positive.

This understanding is crucial for correctly determining the signs of the trigonometric functions.

Pythagorean theorem

The Pythagorean theorem is a fundamental principle in geometry that helps us find the distance between two points on a plane. For a right triangle, the theorem states:\[a^2 + b^2 = c^2\]where \(a\) and \(b\) are the legs, and \(c\) is the hypotenuse.
In the context of trigonometry, the hypotenuse is the line joining the origin to the point \(P(x, y)\) on the coordinate plane. By assigning the x-coordinate to one leg and the y-coordinate to the other, the radius \(r\) or hypotenuse is computed as:\[r = \sqrt{x^2 + y^2}\]For our point \((-2, -5)\), this becomes:\[r = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}\]Using the Pythagorean theorem in this way allows us to find the radius needed for calculating trigonometric functions.

Rationalizing Denominators

Rationalizing the denominator is an essential process in simplifying fractions, especially those involving square roots. We aim to eliminate radicals (square roots) in the denominator to make expressions more standard and easier to interpret.
For trigonometric functions, if the denominator is \(\sqrt{29}\), we multiply both the numerator and the denominator by \(\sqrt{29}\) to get rid of the square root in the denominator:

• \(\sin(\theta) = \frac{-5}{\sqrt{29}} = \frac{-5 \cdot \sqrt{29}}{29}\)
• \(\cos(\theta) = \frac{-2}{\sqrt{29}} = \frac{-2 \cdot \sqrt{29}}{29}\)

This step is crucial for achieving a mathematically accepted form of the trigonometric function values, making them simpler for others to work with or compare. Thus, rationalizing helps in maintaining a clear and concise presentation of results.