Problem 31 - A Graphical Approach to Precalculus with Limits

Question

Sketch an angle \(\theta\) in standard position such that \(\theta\) has the least possible positive measure, and the given point is on the terminal side of \(\theta .\) Find the values of the six trigonometric functions for each angle. Rationalize denominators when applicable. Do not use a calculator. $$(1, \sqrt{3})$$

Step-by-Step Solution

Answer

\(\sin \theta = \frac{\sqrt{3}}{2}\), \(\cos \theta = \frac{1}{2}\), \(\tan \theta = \sqrt{3}\); reciprocals: \(\csc \theta = \frac{2\sqrt{3}}{3}\), \(\sec \theta = 2\), \(\cot \theta = \frac{\sqrt{3}}{3}\).

Step 1: Understand the Problem

We need to sketch an angle \( \theta \) in standard position (with its initial side on the positive x-axis) such that the terminal side of this angle passes through the point \((1, \sqrt{3})\). We are also required to find the six trigonometric functions related to this angle.

Step 2: Sketch the Angle

Plot the point \((1, \sqrt{3})\) on the coordinate plane. Since \(1\) and \(\sqrt{3}\) are both positive, the point is in the first quadrant. Draw a line from the origin to this point. The angle \(\theta\) is formed between the positive x-axis and this line.

Step 3: Determine the Reference Triangle

Notice that we can form a right triangle by dropping a perpendicular from the point \((1, \sqrt{3})\) to the x-axis. The sides of this right triangle are \(1\) (along the x-axis), \(\sqrt{3}\) (along the y-axis), and the hypotenuse from the origin to the point, which is calculated in the next step.

Step 4: Calculate the Hypotenuse

Use the Pythagorean theorem to find the hypotenuse: \[ r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2. \] So, the hypotenuse is \(2\).

Step 5: Find the Trigonometric Functions

Now, use the sides \(1\), \(\sqrt{3}\), and \(2\) to find the trigonometric functions: \[\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2}, \quad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{2}, \quad \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{3}}{1} = \sqrt{3}.\]For the reciprocal functions:\[\csc \theta = \frac{1}{\sin \theta} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}, \quad \sec \theta = \frac{1}{\cos \theta} = 2, \quad \cot \theta = \frac{1}{\tan \theta} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.\]

Step 6: Conclusion

The minimum positive angle \(\theta\) in standard position that ends at \((1, \sqrt{3})\) is in the first quadrant. The trigonometric functions are properly calculated based on the side lengths of the triangle formed.

Key Concepts

Standard Position

In trigonometry, understanding the concept of an angle in standard position is fundamental. An angle is said to be in standard position when its vertex is at the origin of the coordinate plane, and its initial side lies along the positive x-axis. The angle's terminal side is determined by rotating in either the counterclockwise or clockwise direction. In the current exercise, the point \((1, \sqrt{3})\) serves as a point on the terminal side of the angle. This means the angle is positioned in the first quadrant since both coordinates are positive. In summary, always remember that an angle in standard position starts on the positive x-axis and extends to wherever the terminal side points.

Trigonometric Functions

Trigonometric functions are key to understanding angles and triangles. They relate the ratios of the sides of right triangles to their angles. The primary trigonometric functions are:

• Sine (\(\sin\))
• Cosine (\(\cos\))
• Tangent (\(\tan\))

Each of these functions corresponds to a ratio of sides in a right triangle.

• The sine of an angle is the ratio of the length of the opposite side to the hypotenuse.
• The cosine is the ratio of the adjacent side to the hypotenuse.
• The tangent is the ratio of the opposite side to the adjacent side.

In our exercise example, we used the specific known values of the sides of the reference triangle to determine the values of these functions: \(\sin \theta = \frac{\sqrt{3}}{2}\), \(\cos \theta = \frac{1}{2}\), and \(\tan \theta = \sqrt{3}\). Understanding these basics provides the groundwork for exploring more complex trigonometric relationships.

Reference Triangle

When working with angles in a coordinate plane, the reference triangle is a very useful tool. It is essentially a right triangle associated with an angle in standard position. This triangle helps in computing trigonometric functions by using its sides. The reference triangle is formed using the x-axis and a perpendicular dropped from the terminal side to the x-axis.In this exercise, the reference triangle was formed using the point \((1, \sqrt{3})\) and the x-axis. The base (adjacent side) is 1, the height (opposite side) is \(\sqrt{3}\), and the hypotenuse, calculated using the Pythagorean Theorem, is 2. This setup enables us to appropriately assign ratios to each trigonometric function. Reference triangles always simplify finding trigonometric values and are a staple in trigonometric problem-solving.

Reciprocal Functions

Reciprocal functions are derived from the basic trigonometric functions and are just as important in fully understanding trigonometry. They offer more ways to explore the relationships between side lengths and angles.

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

In the given exercise, these functions yielded: \(\csc \theta = \frac{2\sqrt{3}}{3}\), \(\sec \theta = 2\), and \(\cot \theta = \frac{\sqrt{3}}{3}\). These relationships help in solving trigonometric equations and in understanding the symmetry and periodicity of the standard trig functions.