Problem 28 - Precalculus

Question

Find the indicated trigonometric function values. If \(\cos \theta=\frac{40}{41},\) and the terminal side of \(\theta\) lies in quadrant IV, find tan \(\theta\)

Step-by-Step Solution

Answer

\(\tan \theta = -\frac{9}{40}\)

Step 1: Understanding the trigonometric identity

We know from the Pythagorean identity that for any angle \(\theta\), \(\cos^2 \theta + \sin^2 \theta = 1\). In this problem, we are given \(\cos \theta = \frac{40}{41}\).

Step 2: Solve for \(\sin \theta\)

Substitute \(\cos \theta\) into the Pythagorean identity: \(\left(\frac{40}{41}\right)^2 + \sin^2 \theta = 1\). Calculate \(\left(\frac{40}{41}\right)^2 = \frac{1600}{1681}\). Thus, \(\sin^2 \theta = 1 - \frac{1600}{1681} = \frac{81}{1681}\). Take the square root: \(\sin \theta = -\frac{9}{41}\) since \(\theta\) is in quadrant IV where sine is negative.

Step 3: Calculate \(\tan \theta\)

Use the definition of tangent: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Substitute the values found: \(\tan \theta = \frac{-\frac{9}{41}}{\frac{40}{41}} = -\frac{9}{40}\).

Key Concepts

Understanding the Pythagorean Identity

The Pythagorean identity is a fundamental concept in trigonometry. It establishes the relationship between the square of the sine and cosine functions for any angle \( \theta \). The identity is given by:

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

This equation tells us that if you know one of the trigonometric functions, you can always find the other. In simpler terms, as long as you know either the sine or the cosine of an angle, you can use this identity to compute the missing value.
In the exercise, we are given the value of \( \cos \theta = \frac{40}{41} \), and we need to use this identity to find \( \sin \theta \). This involves substituting the given cosine value into the identity as follows:

• \( \left(\frac{40}{41}\right)^2 + \sin^2 \theta = 1 \)

By solving for \( \sin^2 \theta \), we find that \( \sin^2 \theta = \frac{81}{1681} \). We then take the square root to find \( \sin \theta = -\frac{9}{41} \), choosing the negative root because of the specific properties of the fourth quadrant.

Positioning in Quadrant IV

In trigonometry, the coordinate plane is divided into four quadrants. Each quadrant has different sign conventions for the sine, cosine, and tangent functions.Quadrant IV, where our angle \( \theta \) lies, is known for having the cosine value positive and the sine value negative. This affects what values we use when taking roots.So when we originally had \( \sin^2 \theta \) equal to \( \frac{81}{1681} \) and took the square root, why did we choose \(-\frac{9}{41}\) over \(\frac{9}{41}\)? It's because in quadrant IV:

• Cosine (\( \cos \)) is positive.
• Sine (\( \sin \)) is negative.

Therefore, even though the mathematical operation produces two potential results, the quadrant information clarifies which sign to use. Understanding quadrants helps you choose the correct trigonometric sign, especially for angles close to the axis such as \( \theta \) in this situation.

Calculating the Tangent Function

Tangent is another critical trigonometric function often required for angle calculations. It connects the sine and cosine functions via the equation:

• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).

This definition gives the ratio of the sine to the cosine for any angle \( \theta \).With our findings -- \( \sin \theta = -\frac{9}{41} \) and \( \cos \theta = \frac{40}{41} \), calculating \( \tan \theta \) becomes straightforward. Replace the sine and cosine in the tangent equation with the values we've derived:

• \( \tan \theta = \frac{-\frac{9}{41}}{\frac{40}{41}} = -\frac{9}{40} \).

This shows the tangent function's dependence on both sine and cosine and highlights its property of sign change when compared to sine and cosine individually across different quadrants. In quadrant IV, where sine is negative and cosine is positive, the tangent is negative.