Problem 11 - Algebra 2

Question

Find the value of each expression. \(\sin \theta,\) if \(\cos \theta=\frac{2}{3} ; 0^{\circ} \leq \theta<90^{\circ}\)

Step-by-Step Solution

Answer

\( \sin \theta = \frac{\sqrt{5}}{3} \)

Step 1: Understand the Problem

We are given that \( \cos \theta = \frac{2}{3} \) and we need to find \( \sin \theta \) where \( 0^{\circ} \leq \theta < 90^{\circ} \). Since \( \theta \) is in the first quadrant, \( \sin \theta \) will be positive.

Step 2: Use the Pythagorean Identity

The trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \) can be used. Given \( \cos \theta = \frac{2}{3} \), we substitute it into the identity: \[ \sin^2 \theta + \left( \frac{2}{3} \right)^2 = 1 \] Simplify it to:\[ \sin^2 \theta + \frac{4}{9} = 1 \]

Step 3: Simplify and Solve for \( \sin^2 \theta \)

Subtract \( \frac{4}{9} \) from both sides of the equation:\[ \sin^2 \theta = 1 - \frac{4}{9} \] Convert 1 into a fraction with a denominator of 9:\[ \sin^2 \theta = \frac{9}{9} - \frac{4}{9} \] This simplifies to:\[ \sin^2 \theta = \frac{5}{9} \]

Step 4: Find \( \sin \theta \)

To find \( \sin \theta \), take the square root of both sides of the equation:\[ \sin \theta = \sqrt{\frac{5}{9}} \] Since \( \theta \) is in the first quadrant, \( \sin \theta \) is positive. Therefore:\[ \sin \theta = \frac{\sqrt{5}}{3} \]

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a cornerstone of trigonometry and relates the sine and cosine of an angle in a right triangle. This identity is expressed as:\[\sin^2 \theta + \cos^2 \theta = 1\]This equation is essential because it applies to any angle \(\theta\), not just those in right triangles. It's particularly useful in converting between sine and cosine values. By knowing one of these values, you can always find the other.For example, if you have \( \cos \theta = \frac{2}{3} \) and you need to calculate \( \sin \theta \), you simply substitute the cosine value into the Pythagorean Identity:

• Calculate \( \cos^2 \theta\): \( \left(\frac{2}{3}\right)^2 = \frac{4}{9} \)
• Substitute into the identity: \( \sin^2 \theta + \frac{4}{9} = 1 \)
• Simplify to find \( \sin^2 \theta = \frac{5}{9} \)

This makes the Pythagorean Identity an invaluable tool for solving trigonometric problems, helping you find missing ratios with ease.

Sine Function

The Sine Function is one of the primary functions used in trigonometry to describe the ratio of the opposite side to the hypotenuse of a right triangle. It is denoted as \( \sin \theta \), where \( \theta \) is the angle in question.When using the Pythagorean Identity to find \( \sin \theta \), once you have calculated \( \sin^2 \theta \), the next step is to solve for \( \sin \theta \) itself. This involves taking the square root:- If \( \sin^2 \theta = \frac{5}{9} \), then \( \sin \theta = \sqrt{\frac{5}{9}} \)- Simplifying gives \( \sin \theta = \frac{\sqrt{5}}{3} \)Understanding that the sine function varies between -1 and 1, and that it is always positive in the first quadrant \((0^{\circ} \leq \theta < 90^{\circ})\) is crucial. This ensures that we take the positive root when evaluating \( \sin \theta \).

Cosine Function

The Cosine Function is represented as \( \cos \theta \) and is defined as the ratio of the adjacent side to the hypotenuse in a right triangle. Like sine, cosine values range between -1 and 1.In trigonometric identities, cosine plays a vital role. By knowing \( \cos \theta \), as given \( \cos \theta = \frac{2}{3} \), you can determine the sine using the Pythagorean Identity. Cosine's relation to sine in the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) is fundamental to solving equations where one trigonometric ratio is unknown.Since in our example \( \theta \) is in the first quadrant:- \( \cos \theta = \frac{2}{3} \) signifies that both sine and cosine are positive- This property is used to choose the positive root when calculating \( \sin \theta \)The cosine function's clear role helps students find unknown values in exercises by applying these simple, elegant relationships between trigonometric functions.