Problem 15 - Calculus Single Variable

Question

\(\theta\) is a number between 0 and \(\pi / 2\). Calculate the unevaluated trigonometric function from the given information. \(\cos (\theta) ; \sin (\theta)=1 / 3\)

Step-by-Step Solution

Answer

\(\cos(\theta) = \frac{2\sqrt{2}}{3}\)

Step 1: Identify Pythagorean Identity

To find \(\cos(\theta)\), we use the Pythagorean identity for sine and cosine: \(\sin^2(\theta) + \cos^2(\theta) = 1\). This identity holds for any angle and is particularly useful here.

Step 2: Substitute Known Values

Substitute the known value \(\sin(\theta) = \frac{1}{3}\) into the identity. We have: \(\left(\frac{1}{3}\right)^2 + \cos^2(\theta) = 1\).

Step 3: Calculate \(\sin^2(\theta)\)

Calculate \(\sin^2(\theta)\) by squaring \(\frac{1}{3}\), which gives \(\left(\frac{1}{3}\right)^2 = \frac{1}{9}\).

Step 4: Solve for \(\cos^2(\theta)\)

Substitute \(\frac{1}{9}\) into the Pythagorean identity: \(\frac{1}{9} + \cos^2(\theta) = 1\). Rearrange to find \(\cos^2(\theta) = 1 - \frac{1}{9}\).

Step 5: Simplify \(\cos^2(\theta)\)

Simplify \(1 - \frac{1}{9}\) to get \(\cos^2(\theta) = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}\).

Step 6: Calculate \(\cos(\theta)\)

Since \(\theta\) is between 0 and \(\pi/2\), \(\cos(\theta)\) must be positive. Take the positive square root of \(\frac{8}{9}\) to find \(\cos(\theta) = \frac{\sqrt{8}}{3}\).

Step 7: Simplify Square Root

Simplify \(\sqrt{8}\) to \(2\sqrt{2}\), so \(\cos(\theta) = \frac{2\sqrt{2}}{3}\).

Key Concepts

Pythagorean Identity

The Pythagorean identity is a fundamental concept in trigonometry, capturing the intrinsic relationship between the sine and cosine functions. It is expressed as:\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]This identity is incredibly useful because it allows us to determine one trigonometric function if the other is known. Bringing this powerful tool into practice, if you know either \( \sin(\theta) \) or \( \cos(\theta) \), you can quickly find the other using algebra.
Imagine you have been given \( \sin(\theta) = \frac{1}{3} \) and you need to find \( \cos(\theta) \). You would simply substitute into the Pythagorean identity:- Square the sine value: \( \left( \frac{1}{3} \right)^2 = \frac{1}{9} \).- Rearrange the identity: \( \cos^2(\theta) = 1 - \frac{1}{9} \).- Simplify: \( \cos^2(\theta) = \frac{8}{9} \).This identity confirms that the sum of squares of sine and cosine for any angle \( \theta \) will always equal 1. It's handy for solving many trigonometric problems.

Sine Function

The sine function, denoted as \( \sin(\theta) \), is a basic trigonometric function that represents the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. Here are some key points:- The range of \( \sin(\theta) \) is from -1 to 1.- In the interval \( [0, \pi/2] \), \( \sin(\theta) \) is always positive.
While solving the problem, we know that \( \sin(\theta) = \frac{1}{3} \). This value tells us that \( \theta \) is an acute angle where the opposite side is much smaller compared to the hypotenuse.
Computing the square: \( \left(\frac{1}{3}\right)^2 = \frac{1}{9} \), is a necessary step in using the Pythagorean identity. This value represents the squared proportion of the opposite side to the hypotenuse. Understanding how \( \sin(\theta) \) relates to a right triangle helps build the foundation for more complex trigonometric concepts.

Cosine Function

The cosine function, represented as \( \cos(\theta) \), shows the ratio of the adjacent side's length to the hypotenuse in a right-angled triangle. Let's delve into its properties and application from our problem:- The range of \( \cos(\theta) \) is also from -1 to 1.- In the interval \( [0, \pi/2] \), \( \cos(\theta) \) is positive.
In our task, after finding that \( \cos^2(\theta) = \frac{8}{9} \), we take its square root to determine \( \cos(\theta) = \frac{\sqrt{8}}{3} \). Since we know \( \sqrt{8} \) simplifies further:- \( \sqrt{8} = 2\sqrt{2} \), thus:- \( \cos(\theta) = \frac{2\sqrt{2}}{3} \)
This result tells us the adjacent side's proportion to the hypotenuse in our right triangle. Understanding \( \cos(\theta) \) aids in numerous applications, especially when using other trigonometric identities and functions in more advanced mathematics.