Problem 52 - Algebra and Trigonometry

Question

Evaluate each expression under the given conditions. \(\sin (\theta-\phi) ; \tan \theta=\frac{4}{3}, \theta\) in Quadrant III, \(\sin \phi=-\sqrt{10} / 10\) \(\phi\) in Quadrant IV

Step-by-Step Solution

Answer

\(-\frac{9\sqrt{10}}{25}\)

Step 1: Identify Trigonometric Identities

To evaluate \( \sin(\theta - \phi) \), we use the identity: \( \sin(\theta - \phi) = \sin\theta \cos\phi - \cos\theta \sin\phi \). Our goal is to find each of these trigonometric functions, given \( \tan\theta = \frac{4}{3} \) (in Quadrant III) and \( \sin \phi = -\frac{\sqrt{10}}{10} \) (in Quadrant IV).

Step 2: Determine \(\sin \theta\) and \(\cos \theta\) from \(\tan \theta\)

In Quadrant III, both sine and cosine are negative. For \( \tan\theta = \frac{4}{3} \), create a right triangle where \( \text{opposite} = 4 \) and \( \text{adjacent} = 3 \). Compute the hypotenuse using the Pythagorean theorem: \( \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5 \). Thus, \( \sin \theta = -\frac{4}{5} \) and \( \cos \theta = -\frac{3}{5} \).

Step 3: Determine \(\cos \phi\) from \(\sin \phi\)

Given \( \sin \phi = -\frac{\sqrt{10}}{10} \) in Quadrant IV, where cosine is positive. Use \( \sin^2 \phi + \cos^2 \phi = 1 \) to find \( \cos\phi \). \( \cos^2 \phi = 1 - \sin^2 \phi = 1 - \left(-\frac{\sqrt{10}}{10}\right)^2 = \frac{90}{100} = \frac{9}{10} \). Thus, \( \cos\phi = \frac{3}{\sqrt{10}} \).

Step 4: Apply the Identity and Compute the Expression

Substitute the values into the identity: \( \sin(\theta - \phi) = \sin\theta \cos\phi - \cos\theta \sin\phi \). This yields: \( \left(-\frac{4}{5}\right) \left( \frac{3}{\sqrt{10}}\right) - \left(-\frac{3}{5}\right)\left( -\frac{\sqrt{10}}{10} \right) \). Simplifying this gives: \(-\frac{12}{5\sqrt{10}} + \frac{3\sqrt{10}}{50} \). Combine and simplify these to get the final result.

Step 5: Simplify the Expression

Combine the fractions by finding a common denominator if needed. Simplify the expression \(-\frac{12}{5\sqrt{10}} + \frac{3\sqrt{10}}{50} \) by rationalizing the denominators and combining the fractions. This should give you the simplified answer. The result is \(-\frac{9\sqrt{10}}{25} \).

Key Concepts

Sine and Cosine

The trigonometric functions sine and cosine are fundamental when dealing with angles and triangles. They are defined based on the ratios of sides in a right-angled triangle. For an angle \( \theta \), \( \sin\theta \) is the ratio of the length of the opposite side to the hypotenuse, while \( \cos\theta \) is the ratio of the adjacent side to the hypotenuse. These values change depending on the angle's position in the coordinate plane.

In this exercise, we were given \( \tan\theta = \frac{4}{3} \). In Quadrant III, both sine and cosine are negative, resulting in \( \sin\theta = -\frac{4}{5} \) and \( \cos\theta = -\frac{3}{5} \). Similarly, we determined \( \cos\phi \) from \( \sin \phi \) using the Pythagorean identity for \( \phi \) in Quadrant IV, where \( \cos\phi = \frac{3}{\sqrt{10}} \).

Knowing how to calculate sine and cosine is crucial, especially when applying identities like \( \sin(\theta - \phi) = \sin\theta \cos\phi - \cos\theta \sin\phi \).

Quadrant Analysis

Quadrant analysis is vital when working with trigonometric functions, as the sign of the functions can change depending on the quadrant where the angle lies. The Cartesian plane is divided into four quadrants:

• Quadrant I: Both sine and cosine are positive.
• Quadrant II: Sine is positive, but cosine is negative.
• Quadrant III: Both sine and cosine are negative.
• Quadrant IV: Cosine is positive, but sine is negative.

For \( \theta \) in Quadrant III, both sine and cosine are negative, which affects the computation of \( \sin\theta \) and \( \cos\theta \). For \( \phi \) in Quadrant IV, cosine remains positive while sine is negative. This knowledge is used to determine the correct signs of the trigonometric functions, ensuring we compute correct values in trigonometric identities.

Tangent Function

The tangent function, represented as \( \tan\theta \), is the ratio of the sine to the cosine of an angle \( \theta \). Therefore, \( \tan\theta = \frac{\sin\theta}{\cos\theta} \).

In our example, \( \tan\theta \) was given as \( \frac{4}{3} \), which helped us find \( \sin\theta \) and \( \cos\theta \) as \( -\frac{4}{5} \) and \( -\frac{3}{5} \) respectively since \( \theta \) is in Quadrant III where both sine and cosine are negative.

Understanding the relationship between sine, cosine, and tangent is crucial when working with trigonometric identities. The tangential value provides a means to find the individual sine and cosine values, vital in further calculations and transformations.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle used to relate the sides of a right triangle. It states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. For our problem, knowing \( \tan\theta = \frac{4}{3} \), we construct a right triangle with opposite side 4 and adjacent side 3. Using the theorem:

\[ \text{hypotenuse} = \sqrt{4^2 + 3^2} = 5 \].

By applying this result, we find \( \sin\theta = -\frac{4}{5} \) and \( \cos\theta = -\frac{3}{5} \).

This principle also helps find \( \cos\phi \) from \( \sin\phi \) through the identity \( \sin^2\phi + \cos^2\phi = 1 \), leading to \( \cos\phi = \frac{3}{\sqrt{10}} \).

This method consolidates the connection between geometry and trigonometry, aiding in evaluating expressions involving trigonometric identities.