Problem 32 - 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

The angle \(\theta = \frac{2\pi}{3}\) with: \(\sin = \frac{\sqrt{3}}{2}\), \(\cos = -\frac{1}{2}\), \(\tan = -\sqrt{3}\); \(\csc = \frac{2\sqrt{3}}{3}\), \(\sec = -2\), \(\cot = -\frac{\sqrt{3}}{3}\).

Step 1: Determine the Quadrant

In standard position, the angle \(\theta\) starts from the positive \(x\)-axis. The given point is \((-1, \sqrt{3})\), which is located in the second quadrant where \(x < 0\) and \(y > 0\).

Step 2: Calculate the Reference Angle

The reference angle is calculated using the arctangent of the ratio \(\frac{|y|}{|x|}\). For the point \((-1, \sqrt{3})\), the reference angle is \(\tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}\;\text{(or } 60^\circ\text{)}\).

Step 3: Calculate the Angle in Standard Position

In Quadrant II, the angle \(\theta\) can be found using \(180^\circ - \text{reference angle}\). Hence, \(\theta = 180^\circ - 60^\circ = 120^\circ\). In radians, \(\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\).

Step 4: Find the Hypotenuse

The length of the hypotenuse \(r\) is found using the distance formula: \(r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2\).

Step 5: Calculate Sine, Cosine, and Tangent

Use the formula \(\sin(\theta) = \frac{y}{r}\), \(\cos(\theta) = \frac{x}{r}\), \(\tan(\theta) = \frac{y}{x}\):\(\sin(\theta) = \frac{\sqrt{3}}{2}\),\(\cos(\theta) = \frac{-1}{2}\),\(\tan(\theta) = \frac{\sqrt{3}}{-1} = -\sqrt{3}\).

Step 6: Calculate Cosecant, Secant, and Cotangent

Cosecant, secant, and cotangent are the reciprocals of sine, cosine, and tangent respectively:\(\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \;\text{(Rationalized)}\),\(\sec(\theta) = \frac{1}{\cos(\theta)} = -2\),\(\cot(\theta) = \frac{1}{\tan(\theta)} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \;\text{(Rationalized)}\).

Key Concepts

Standard Position Angle

A standard position angle is fundamental in trigonometry. When discussing angles in this position, it's essential to understand that the angle starts from the positive x-axis.

Key points include:

• The vertex of the angle is at the origin of the coordinate system (0,0).


• The initial side is the positive x-axis.


• The angle is measured counterclockwise for positive angles.

In our example, the point (-1, \(\sqrt{3}\)) lies in the second quadrant.

This tells us that \(\theta\) not only begins at the positive x-axis but also stretches towards the y-axis and lands in the second quadrant where x is negative, and y is positive. This positioning gives us the necessary context to calculate the reference angle and the actual angle respecting the quadrant's constraints.

Reference Angle

The reference angle is a crucial component in trigonometry, making the evaluation of trigonometric functions simpler, especially across different quadrants.

Calculating the reference angle involves using the arctangent function sometimes, like in our problem.

Here are some handy details about reference angles:

• It is always positive and is defined as the acute angle between the terminal side of the angle and the x-axis.


• In our example, the reference angle for the point (-1, \(\sqrt{3}\)) is \(\tan^{-1}\left( \frac{\sqrt{3}}{1} \right)\), which equates to \(\frac{\pi}{3}\) or \(60^\circ\).


• This step allows us to find the correct angle \(\theta\) for any standard position angle across different quadrants.

Having the reference angle means we can determine \(\theta\) using the directionality of angles in each quadrant. In Quadrant II, where our point lies, \(\theta\) can be determined as \(180^\circ - 60^\circ = 120^\circ\) or \(\pi - \frac{\pi}{3} = \frac{2\pi}{3}\).

Reciprocal Trigonometric Functions

Trigonometric functions have reciprocals that frequently appear in various problem-solving scenarios. Understanding these reciprocal functions is crucial for complete trigonometric problem-solving.

Here they are explained:

• Cosecant (\(\csc\)) is the reciprocal of sine. This means \(\csc\,\theta = \frac{1}{\sin\,\theta}\).


• Secant (\(\sec\)) is the reciprocal of cosine. Therefore, \(\sec\,\theta = \frac{1}{\cos\,\theta}\).


• Cotangent (\(\cot\)) is the reciprocal of tangent, giving \(\cot\,\theta = \frac{1}{\tan\,\theta}\).

In our scenario, the values of \(\sin(\theta) = \frac{\sqrt{3}}{2}\), \(\cos(\theta) = \frac{-1}{2}\), and \(\tan(\theta) = -\sqrt{3}\) lead us to:

• \(\csc(\theta) = \frac{2\sqrt{3}}{3}\) after rationalizing, because we prefer rational denominators.


• \(\sec(\theta) = -2\), a straightforward reciprocal when the denominator is already rationalized.


• \(\cot(\theta) = -\frac{\sqrt{3}}{3}\), once the irrational denominator is rationalized.

These computations highlight not only the relationships between the trigonometric functions and their reciprocals but also emphasize the importance of rationalizing where necessary.