Problem 121 - A Graphical Approach to Precalculus with Limits

Question

Find all trigonometric function values for each angle \(\boldsymbol{\theta}\). $$\sin \theta=\frac{\sqrt{2}}{6}, \text { given that } \cos \theta<0$$

Step-by-Step Solution

Answer

In the second quadrant, \( \cos \theta = -\sqrt{\frac{17}{18}} \), \( \tan \theta = -\frac{\sqrt{2}}{2\sqrt{17}} \), \( \csc \theta = 3\sqrt{2} \), \( \sec \theta = -\sqrt{\frac{18}{17}} \), \( \cot \theta = -2\sqrt{17} \).

Step 1: Determine Quadrants

Since \( \sin \theta = \frac{\sqrt{2}}{6} \) and \( \cos \theta < 0 \), we analyze the possible quadrants. The sine function is positive in the first and second quadrants. However, \( \cos \theta < 0 \) only in the second quadrant. Therefore, \( \theta \) is in the second quadrant.

Step 2: Use the Pythagorean Identity

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), substitute \( \sin \theta = \frac{\sqrt{2}}{6} \). Calculate \( \cos \theta \): \[ \left(\frac{\sqrt{2}}{6}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{2}{36} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{1}{18} = \frac{17}{18} \] \[ \cos \theta = -\sqrt{\frac{17}{18}} \] We choose the negative square root because \( \cos \theta < 0 \) in the second quadrant.

Step 3: Find Other Trigonometric Functions

To find \( \tan \theta \), use \( \tan \theta = \frac{\sin \theta}{\cos \theta} \): \[ \tan \theta = \frac{\frac{\sqrt{2}}{6}}{-\sqrt{\frac{17}{18}}} = -\frac{\sqrt{2} \times \sqrt{18}}{6\sqrt{17}} = -\frac{3\sqrt{2}}{6\sqrt{17}} = -\frac{\sqrt{2}}{2\sqrt{17}} \]For \( \csc \theta \), this is the reciprocal of \( \sin \theta \): \[ \csc \theta = \frac{6}{\sqrt{2}} = 3\sqrt{2} \] For \( \sec \theta \), the reciprocal of \( \cos \theta \): \[ \sec \theta = -\sqrt{\frac{18}{17}} \]For \( \cot \theta \), the reciprocal of \( \tan \theta \): \[ \cot \theta = -2\sqrt{17}\]

Key Concepts

Pythagorean Identity

The Pythagorean Identity is one of the fundamental formulas in trigonometry. It states that for any angle \( \theta \), the sum of the square of the sine and cosine functions is always equal to 1. Mathematically, it is expressed as: \[ \sin^2 \theta + \cos^2 \theta = 1 \] This identity is derived from the Pythagorean theorem and is incredibly useful for determining unknown trigonometric values when one is given. For example, in this specific problem, we start by knowing \( \sin \theta = \frac{\sqrt{2}}{6} \). Using the Pythagorean Identity, we substitute this value in to find \( \cos \theta \). When applying this identity:

• Square the given sine (or cosine) value.
• Subtract this value from 1 to get the cosine (or sine) squared.
• Take the square root of the result to find the cosine (or sine) value.

This robust formula is applicable to all angles and is a staple in solving trigonometric problems.

Trigonometric Functions in Quadrants

Trigonometric functions manifest differently across the four quadrants of the coordinate plane. This means that the sign (positive or negative) of each function varies depending on the quadrant in which the angle lies. The sine, cosine, and tangent functions depend on the angle's position:- **First Quadrant:** All trigonometric functions are positive.- **Second Quadrant:** Sine is positive, but cosine and tangent are negative.- **Third Quadrant:** Tangent is positive while sine and cosine are negative.- **Fourth Quadrant:** Cosine is positive, whereas sine and tangent are negative.In the exercise, given \( \sin \theta = \frac{\sqrt{2}}{6} \) (positive) and \( \cos \theta < 0 \) (negative), \( \theta \) must be in the second quadrant since sine is positive there, and cosine is negative. Understanding the behavior of these functions in different quadrants is crucial for identifying where the angle is located.

Reciprocal Trigonometric Functions

Reciprocal trigonometric functions are simply the inverses of the basic sine, cosine, and tangent functions. They provide more flexibility and options when solving trigonometric equations. The main reciprocal functions are:- **Cosecant (\( \csc \theta \))**: The reciprocal of sine, calculated as \( \csc \theta = \frac{1}{\sin \theta} \).- **Secant (\( \sec \theta \))**: The reciprocal of cosine, calculated as \( \sec \theta = \frac{1}{\cos \theta} \).- **Cotangent (\( \cot \theta \))**: The reciprocal of tangent, calculated as \( \cot \theta = \frac{1}{\tan \theta} \).For this exercise, the values are derived as follows: - \( \csc \theta = 3\sqrt{2} \) because it is the reciprocal of \( \sin \theta = \frac{\sqrt{2}}{6} \).- \( \sec \theta = -\sqrt{\frac{18}{17}} \) since \( \cos \theta = -\sqrt{\frac{17}{18}} \) in the second quadrant.- \( \cot \theta = -2\sqrt{17} \) as the reciprocal of \( \tan \theta \).These reciprocal relationships are important for solving equations that involve the original trigonometric functions.

Quadrant Analysis

Quadrant Analysis involves determining which quadrant an angle belongs to based on the signs of its trigonometric functions. Each quadrant is defined by the sign of the trigonometric functions within it, impacting how these functions are interpreted. To decide the quadrant:

• Identify the signs of both \( \sin \theta \) and \( \cos \theta \).
• Match these signs to the quadrants as outlined previously.

For instance, in our exercise, knowing that \( \sin \theta \) is positive and \( \cos \theta \) is negative, points us clearly to the second quadrant. This quadrant specific understanding helps in predicting and confirming the behavior and solutions of trigonometric equations.It also aids in computing additional angles in relation to a reference angle, considering angles differ due to their specific quadrant placements. Thus, quadrant analysis is key in ensuring accuracy in trigonometric calculations and solving.