Problem 42 - A Graphical Approach to Precalculus with Limits

Question

An equation of the terminal side of an angle \(\theta\) in standard position is given with a restriction on \(x\). Sketch the least positive angle \(\theta\), and find the values of the six trigonometric functions of \(\theta\). $$6 x-5 y=0, x \geq 0$$

Step-by-Step Solution

Answer

The six trigonometric functions are: \(\sin \theta = \frac{6}{\sqrt{61}}\), \(\cos \theta = \frac{5}{\sqrt{61}}\), \(\tan \theta = \frac{6}{5}\), \(\csc \theta = \frac{\sqrt{61}}{6}\), \(\sec \theta = \frac{\sqrt{61}}{5}\), \(\cot \theta = \frac{5}{6}\).

Step 1: Rewrite the Equation in Slope-Intercept Form

To better understand the line represented by the equation, we convert it into slope-intercept form. Start by rewriting the equation \(6x - 5y = 0\) into the form \(y = mx + b\). Solving for \(y\), we have \[5y = 6x\] so \[y = \frac{6}{5}x\]. This indicates a line through the origin with a slope of \(\frac{6}{5}\).

Step 2: Determine the Coordinates of a Point on the Terminal Side

Since the line \(y = \frac{6}{5}x\) passes through the origin, we choose another point where \(x = 5\) to avoid fractions. Substituting \(x = 5\) into the equation gives \(y = 6\). Hence, another point on the line is \((5, 6)\).

Step 3: Understand the Position of the Angle

The terminal side starts at the origin \((0,0)\) and passes through \((5,6)\). The x-coordinate is non-negative \( (x \geq 0)\), which implies the angle \(\theta\) is in the first quadrant. The angle is measured counterclockwise from the positive x-axis to the line.

Step 4: Calculate the Radius (Hypotenuse)

The radius (or hypotenuse of the triangle formed) is the distance from the origin to point \((5,6)\). Using the distance formula, this is \[r = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61} \].

Step 5: Find the Sine, Cosine, and Tangent of \(\theta\)

The basic trigonometric functions can be found as follows: - \(\sin \theta = \frac{opposite}{hypotenuse} = \frac{6}{\sqrt{61}}\) - \(\cos \theta = \frac{adjacent}{hypotenuse} = \frac{5}{\sqrt{61}}\) - \(\tan \theta = \frac{opposite}{adjacent} = \frac{6}{5}\)

Step 6: Calculate the Reciprocal Trigonometric Functions

Using the reciprocals of sine, cosine, and tangent, we find the other three trigonometric functions: - \(\csc \theta = \frac{1}{\sin \theta} = \frac{\sqrt{61}}{6}\)- \(\sec \theta = \frac{1}{\cos \theta} = \frac{\sqrt{61}}{5}\)- \(\cot \theta = \frac{1}{\tan \theta} = \frac{5}{6}\)

Key Concepts

Slope-Intercept Form

To understand a line's equation completely, converting it to the slope-intercept form is very useful. This form allows us to easily identify the slope and the y-intercept of the line. It's given as \( y = mx + b \), where \( m \) represents the slope, and \( b \) represents the y-intercept. In the exercise, we started with the equation \( 6x - 5y = 0 \). By reorganizing it to get \( y = \frac{6}{5}x \), we see that the slope \( m \) is \( \frac{6}{5} \) and the y-intercept \( b \) is 0. This tells us the line passes through the origin, and for every 5 units we move horizontally, the line rises by 6 units vertically. Converting equations to this form makes the geometry of the line evident, especially helpful when sketching or analyzing the line's behavior.

Standard Position Angle

An angle is in standard position if its vertex is at the origin and its initial side lies along the positive x-axis. Measuring the angle involves rotating counterclockwise from this axis to the terminal side. In our problem, the terminal side is represented by the line \( y = \frac{6}{5}x \), meaning it extends through the origin. Because the condition \( x \geq 0 \) stipulates that we are only considering the right half of the coordinate plane, the angle \( \theta \) lies in the first quadrant. This is the quadrant where all sine, cosine, and tangent values are positive. Recognizing standard position angles helps to determine the direction and quadrant of the terminal side, which is crucial in analyzing trigonometric functions.

Reciprocal Trigonometric Functions

The reciprocal trigonometric functions complement their primary trigonometric counterparts, providing more insights into the angle's properties. These include cosecant \( \csc \), secant \( \sec \), and cotangent \( \cot \). They are defined as follows:

• \( \csc \theta = \frac{1}{\sin \theta} \), meaning it is the reciprocal of sine.
• \( \sec \theta = \frac{1}{\cos \theta} \), the reciprocal of cosine.
• \( \cot \theta = \frac{1}{\tan \theta} \), the reciprocal of tangent.

Using the calculated sine, cosine, and tangent values for the angle in our exercise, we can find that:

• \( \csc \theta = \frac{\sqrt{61}}{6} \)
• \( \sec \theta = \frac{\sqrt{61}}{5} \)
• \( \cot \theta = \frac{5}{6} \)

By understanding these reciprocal functions, additional relationships and properties of the angle can be explored.

Distance Formula

The distance formula is a valuable tool in determining the distance between two points in a coordinate plane. This is especially useful in calculating the hypotenuse when thinking in terms of right-triangle trigonometry. The formula is \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]. For our task, we took the points \((0,0)\) and \((5,6)\) to calculate the distance, which represents the hypotenuse \( r \) in the related right triangle. Using this formula, we found \( r = \sqrt{61} \). Knowing the hypotenuse is critical when calculating sine, cosine, and other trigonometric functions of an angle, as it forms the basis of defining these ratios in right triangles.