Problem 117 - A Graphical Approach to Precalculus with Limits

Question

Find all trigonometric function values for each angle \(\boldsymbol{\theta}\). \(\sin \theta=\frac{\sqrt{5}}{7},\) given that \(\theta\) is in quadrant \(I\)

Step-by-Step Solution

Answer

\( \sin \theta = \frac{\sqrt{5}}{7} \), \( \cos \theta = \frac{2\sqrt{11}}{7} \), \( \tan \theta = \frac{\sqrt{55}}{22} \), \( \csc \theta = \frac{7\sqrt{5}}{5} \), \( \sec \theta = \frac{7\sqrt{11}}{22} \), \( \cot \theta = \frac{2\sqrt{55}}{5} \).

Step 1: Understand the Given Information

We are given that \( \sin \theta = \frac{\sqrt{5}}{7} \) and \( \theta \) is in the first quadrant. In the first quadrant, all trigonometric functions are positive.

Step 2: Find the Cosine Function Value

To find \( \cos \theta \), use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \]Substitute \( \sin \theta = \frac{\sqrt{5}}{7} \) into the identity:\[ \left(\frac{\sqrt{5}}{7}\right)^2 + \cos^2 \theta = 1 \]\[ \frac{5}{49} + \cos^2 \theta = 1 \]\[ \cos^2 \theta = 1 - \frac{5}{49} = \frac{44}{49} \]Thus, \( \cos \theta = \sqrt{\frac{44}{49}} = \frac{\sqrt{44}}{7} = \frac{2\sqrt{11}}{7} \). In the first quadrant, \( \cos \theta \) is positive.

Step 3: Find the Remaining Trigonometric Function Values

Calculate \( \tan \theta \) using the ratio of sine and cosine:\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{\sqrt{5}}{7}}{\frac{2\sqrt{11}}{7}} = \frac{\sqrt{5}}{2\sqrt{11}} \]Rationalize the denominator:\[ \tan \theta = \frac{\sqrt{5}}{2\sqrt{11}} \cdot \frac{\sqrt{11}}{\sqrt{11}} = \frac{\sqrt{55}}{22} \]Calculate \( \csc \theta \) as the reciprocal of sine:\[ \csc \theta = \frac{1}{\sin \theta} = \frac{7}{\sqrt{5}} \]Rationalize the denominator:\[ \csc \theta = \frac{7\sqrt{5}}{5} \]Calculate \( \sec \theta \) as the reciprocal of cosine:\[ \sec \theta = \frac{1}{\cos \theta} = \frac{7}{2\sqrt{11}} \]Rationalize the denominator:\[ \sec \theta = \frac{7\sqrt{11}}{22} \]Calculate \( \cot \theta \) as the reciprocal of tangent:\[ \cot \theta = \frac{1}{\tan \theta} = \frac{22}{\sqrt{55}} \]Rationalize the denominator:\[ \cot \theta = \frac{22\sqrt{55}}{55} = \frac{2\sqrt{55}}{5} \]

Step 4: Final Result Summary

The trigonometric function values are:\( \sin \theta = \frac{\sqrt{5}}{7} \) \( \cos \theta = \frac{2\sqrt{11}}{7} \) \( \tan \theta = \frac{\sqrt{55}}{22} \)\( \csc \theta = \frac{7\sqrt{5}}{5} \) \( \sec \theta = \frac{7\sqrt{11}}{22} \)\( \cot \theta = \frac{2\sqrt{55}}{5} \)

Key Concepts

Pythagorean Identity

When dealing with trigonometric functions, one of the most essential concepts is the Pythagorean identity. This identity links the sine and cosine functions in a neat equation:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

This equation holds true for any angle \( \theta \), making it an incredibly useful tool when you know one of the sine or cosine values and need to find the other. In our given exercise, we knew that \( \sin \theta = \frac{\sqrt{5}}{7} \). Leveraging the Pythagorean identity, we can easily find \( \cos \theta \) by rearranging and substituting values into this equation:

• \( \cos^2 \theta = 1 - \sin^2 \theta \)

By plugging in our sine value:

• \( \cos^2 \theta = 1 - \left(\frac{\sqrt{5}}{7}\right)^2 = \frac{44}{49} \)

The above calculation results in \( \cos \theta = \frac{2\sqrt{11}}{7} \), ensuring all trigonometric functions remain positive in the first quadrant.
Remembering the Pythagorean identity can simplify solving many trigonometric problems.

Reciprocal Identities

The reciprocal identities give us another critical insight into trigonometric functions. Understanding these identities can help you find values for the lesser-known functions like cosecant (csc), secant (sec), and cotangent (cot).Let's quickly review them:

• \( \csc \theta = \frac{1}{\sin \theta} \)
• \( \sec \theta = \frac{1}{\cos \theta} \)
• \( \cot \theta = \frac{1}{\tan \theta} \)

In our exercise, we were able to find \( \csc \theta \), \( \sec \theta \), and \( \cot \theta \) by using these reciprocal identities. As an example, let's look at \( \csc \theta \):

• \( \csc \theta = \frac{1}{\sin \theta} = \frac{7}{\sqrt{5}} \)

Each function relates directly to another, by simply taking the reciprocal. These identities are straightforward but powerful tools, especially when tackling complex problems that require precise answers.

Rationalizing the Denominator

Rationalizing the denominator is a common mathematical process, especially helpful when dealing with trigonometric functions and fractions. The goal is to eliminate any square roots (or irrational numbers) from the denominator, simplifying the expression.This process includes multiplying both the numerator and the denominator by a suitable value that removes the root from the denominator. Let's revisit it with the tangent from the exercise:

• \( \tan \theta = \frac{\sqrt{5}}{2\sqrt{11}} \)
• To rationalize: multiply both top and bottom by \( \sqrt{11} \)
• \( \frac{\sqrt{5} \cdot \sqrt{11}}{2\sqrt{11} \cdot \sqrt{11}} = \frac{\sqrt{55}}{22} \)

Rationalizing doesn't change the value of the expression, but it makes it cleaner and generally more acceptable in standard math practices. This step is particularly essential in exams or situations where simplicity and preciseness are paramount.
Always ensure that your final answers in trigonometry are presented in their simplest forms.