Find Exact Value Of Tan(α-β): Trig Problem Solved!

Hey guys! Today, we're diving into a fun trigonometric problem where we need to find the exact value of $\operatorname{tan}(\alpha - \beta)$ given some specific conditions for $\alpha$ and $\beta$. This is a classic problem that combines our understanding of trigonometric identities, quadrant rules, and the Pythagorean theorem. Let's break it down step by step to make sure we grasp every concept. So, grab your calculators (though we won't need them much for exact values!) and let's get started!

Problem Statement

We are given the following information:

• $\sin \alpha = -\frac{7}{25}$ and $\alpha$ is in Quadrant IV.
• $\cos \beta = -\frac{4}{5}$ and $\beta$ is in Quadrant II.

Our mission is to find the exact value of $\operatorname{tan}(\alpha - \beta)$.

Understanding the Problem

Before we jump into calculations, let's make sure we understand what the problem is asking. We need to use the given information about $\sin \alpha$ and $\cos \beta$, along with their respective quadrants, to find the value of $\operatorname{tan}(\alpha - \beta)$. This involves several key concepts:

1. Trigonometric Identities: We'll need the tangent subtraction formula, which is:

   $$\operatorname{tan}(\alpha - \beta) = \frac{\operatorname{tan} \alpha - \operatorname{tan} \beta}{1 + \operatorname{tan} \alpha \operatorname{tan} \beta}$$

2. Quadrant Rules: The quadrant in which an angle lies tells us the signs of the trigonometric functions. Remember the handy acronym ASTC (All Students Take Calculus):
   • Quadrant I: All trigonometric functions are positive.
   • Quadrant II: Sine (and its reciprocal, cosecant) is positive.
   • Quadrant III: Tangent (and its reciprocal, cotangent) is positive.
   • Quadrant IV: Cosine (and its reciprocal, secant) is positive.
3. Pythagorean Theorem: We'll use this to find the missing sides of our right triangles, which will help us determine the values of the other trigonometric functions.

Step 1: Finding $\cos \alpha$ and $\operatorname{tan} \alpha$

We know $\sin \alpha = -\frac{7}{25}$ and $\alpha$ is in Quadrant IV. In Quadrant IV, cosine is positive, and sine is negative (which aligns with our given information). To find $\cos \alpha$, we can use the Pythagorean identity:

$$\sin^2 \alpha + \cos^2 \alpha = 1$$

Plugging in the value of $\sin \alpha$:

$$\left(-\frac{7}{25}\right)^2 + \cos^2 \alpha = 1$$

$$\frac{49}{625} + \cos^2 \alpha = 1$$

$$\cos^2 \alpha = 1 - \frac{49}{625} = \frac{576}{625}$$

Taking the square root (and remembering that cosine is positive in Quadrant IV):

$$\cos \alpha = \sqrt{\frac{576}{625}} = \frac{24}{25}$$

Now that we have $\sin \alpha$ and $\cos \alpha$, we can find $\operatorname{tan} \alpha$ using the identity:

$$\operatorname{tan} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{7}{25}}{\frac{24}{25}} = -\frac{7}{24}$$

Step 2: Finding $\sin \beta$ and $\operatorname{tan} \beta$

We know $\cos \beta = -\frac{4}{5}$ and $\beta$ is in Quadrant II. In Quadrant II, sine is positive, and cosine is negative (which aligns with our given information). To find $\sin \beta$, we use the Pythagorean identity again:

$$\sin^2 \beta + \cos^2 \beta = 1$$

Plugging in the value of $\cos \beta$:

$$\sin^2 \beta + \left(-\frac{4}{5}\right)^2 = 1$$

$$\sin^2 \beta + \frac{16}{25} = 1$$

$$\sin^2 \beta = 1 - \frac{16}{25} = \frac{9}{25}$$

Taking the square root (and remembering that sine is positive in Quadrant II):

$$\sin \beta = \sqrt{\frac{9}{25}} = \frac{3}{5}$$

Now we can find $\operatorname{tan} \beta$:

$$\operatorname{tan} \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}$$

Step 3: Applying the Tangent Subtraction Formula

Now we have all the pieces we need! We know:

• $\operatorname{tan} \alpha = -\frac{7}{24}$
• $\operatorname{tan} \beta = -\frac{3}{4}$

Let's plug these values into the tangent subtraction formula:

$$\operatorname{tan}(\alpha - \beta) = \frac{\operatorname{tan} \alpha - \operatorname{tan} \beta}{1 + \operatorname{tan} \alpha \operatorname{tan} \beta}$$

$$\operatorname{tan}(\alpha - \beta) = \frac{-\frac{7}{24} - \left(-\frac{3}{4}\right)}{1 + \left(-\frac{7}{24}\right)\left(-\frac{3}{4}\right)}$$

First, let's simplify the numerator:

$$-\frac{7}{24} + \frac{3}{4} = -\frac{7}{24} + \frac{18}{24} = \frac{11}{24}$$

Now, let's simplify the denominator:

$$1 + \left(-\frac{7}{24}\right)\left(-\frac{3}{4}\right) = 1 + \frac{21}{96} = 1 + \frac{7}{32} = \frac{32}{32} + \frac{7}{32} = \frac{39}{32}$$

Finally, let's divide the numerator by the denominator:

$$\operatorname{tan}(\alpha - \beta) = \frac{\frac{11}{24}}{\frac{39}{32}} = \frac{11}{24} \cdot \frac{32}{39} = \frac{11 \cdot 32}{24 \cdot 39} = \frac{11 \cdot 4}{3 \cdot 39} = \frac{44}{117}$$

Final Answer

Therefore, the exact value of $\operatorname{tan}(\alpha - \beta)$ is $\frac{44}{117}$.

Key Takeaways

• Trigonometric Identities are Your Friends: The tangent subtraction formula was crucial in solving this problem. Make sure you know your trigonometric identities! Trigonometric identities are very important and useful in solving any kind of mathematical problem.
• Quadrant Rules Matter: Knowing the quadrant of an angle helps determine the signs of the trigonometric functions. Without this information, we couldn't have correctly found $\cos \alpha$ and $\sin \beta$. It is very important to determine the sings of trigonometric function and put it in the formula.
• Pythagorean Identity is Essential: The Pythagorean identity ($\sin^2 \theta + \cos^2 \theta = 1$) is a fundamental tool for finding missing trigonometric values. The Pythagorean theorem can help in finding the missing trigonometric ratios if one ratio is available and also using ASTC rule we can determine other ratios.
• Simplify, Simplify, Simplify: Don't be afraid to simplify fractions and expressions. It makes the final calculation much easier. Simplification is also very important as it reduces the errors in the solution.

Practice Makes Perfect

To really master these concepts, try working through similar problems. Change the given trigonometric values or the quadrants, and see if you can still find the exact value of $\operatorname{tan}(\alpha - \beta)$. The more you practice, the more confident you'll become! Practice is very important as it gives a clear understanding of the concept and formula used.

I hope this breakdown was helpful, guys! Keep practicing, and you'll be trig masters in no time!

$$

Hey there, math enthusiasts! Today, we're going to tackle a classic trigonometry problem: finding the exact value of $\operatorname{tan}(\alpha - \beta)$ when given some specific conditions for angles $\alpha$ and $\beta$. This type of problem is a fantastic way to solidify your understanding of trigonometric identities, quadrant rules, and the Pythagorean theorem. We'll break it down into easy-to-follow steps, so you can confidently solve similar problems in the future. Let's dive in and unlock the secrets of trigonometry!

Problem Setup: What We Know

Let's start by restating the problem. We're given the following information:

• $\sin \alpha = -\frac{7}{25}$, and $\alpha$ lies in Quadrant IV.
• $\cos \beta = -\frac{4}{5}$, and $\beta$ lies in Quadrant II.

Our goal is to determine the exact value of $\operatorname{tan}(\alpha - \beta)$.

The Trigonometric Toolkit: Essential Concepts

Before we start crunching numbers, let's review the key concepts we'll be using. This will help us approach the problem strategically:

1. The Tangent Subtraction Formula: This is the cornerstone of our solution. It states that:

   $$\operatorname{tan}(\alpha - \beta) = \frac{\operatorname{tan} \alpha - \operatorname{tan} \beta}{1 + \operatorname{tan} \alpha \operatorname{tan} \beta}$$

   This formula allows us to express the tangent of the difference of two angles in terms of the tangents of the individual angles.

2. Quadrant Rules (ASTC): Remember the acronym ASTC (All Students Take Calculus)? It's a handy way to recall which trigonometric functions are positive in each quadrant:

   • Quadrant I: All functions (sine, cosine, tangent) are positive.
   • Quadrant II: Sine (and its reciprocal, cosecant) is positive.
   • Quadrant III: Tangent (and its reciprocal, cotangent) is positive.
   • Quadrant IV: Cosine (and its reciprocal, secant) is positive. These rules are crucial for determining the signs of our trigonometric values.

3. The Pythagorean Identity: This identity is a fundamental relationship between sine and cosine:

   $$\sin^2 \theta + \cos^2 \theta = 1$$

   We'll use this to find missing trigonometric values, given one value and the quadrant.

Step-by-Step Solution: Breaking Down the Problem

Now that we have our toolkit ready, let's tackle the problem step-by-step:

Step 1: Finding $\cos \alpha$ and $\operatorname{tan} \alpha$

We're given $\sin \alpha = -\frac{7}{25}$ and that $\alpha$ is in Quadrant IV. In Quadrant IV, cosine is positive, and sine is negative (which matches our given information). To find $\cos \alpha$, we'll use the Pythagorean identity:

$$\sin^2 \alpha + \cos^2 \alpha = 1$$

Substitute the value of $\sin \alpha$:

$$\left(-\frac{7}{25}\right)^2 + \cos^2 \alpha = 1$$

$$\frac{49}{625} + \cos^2 \alpha = 1$$

Solve for $\cos^2 \alpha$:

$$\cos^2 \alpha = 1 - \frac{49}{625} = \frac{576}{625}$$

Take the square root (and remember that $\cos \alpha$ is positive in Quadrant IV):

$$\cos \alpha = \sqrt{\frac{576}{625}} = \frac{24}{25}$$

Now we can find $\operatorname{tan} \alpha$ using the identity:

$$\operatorname{tan} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{7}{25}}{\frac{24}{25}} = -\frac{7}{24}$$

Step 2: Finding $\sin \beta$ and $\operatorname{tan} \beta$

We know $\cos \beta = -\frac{4}{5}$ and that $\beta$ is in Quadrant II. In Quadrant II, sine is positive, and cosine is negative (again, matching our given information). Let's use the Pythagorean identity to find $\sin \beta$:

$$\sin^2 \beta + \cos^2 \beta = 1$$

Substitute the value of $\cos \beta$:

$$\sin^2 \beta + \left(-\frac{4}{5}\right)^2 = 1$$

$$\sin^2 \beta + \frac{16}{25} = 1$$

Solve for $\sin^2 \beta$:

$$\sin^2 \beta = 1 - \frac{16}{25} = \frac{9}{25}$$

Take the square root (and remember that $\sin \beta$ is positive in Quadrant II):

$$\sin \beta = \sqrt{\frac{9}{25}} = \frac{3}{5}$$

Now we can find $\operatorname{tan} \beta$:

$$\operatorname{tan} \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}$$

Step 3: Applying the Tangent Subtraction Formula

We've gathered all the ingredients! We now have:

• $\operatorname{tan} \alpha = -\frac{7}{24}$
• $\operatorname{tan} \beta = -\frac{3}{4}$

Let's plug these values into the tangent subtraction formula:

$$\operatorname{tan}(\alpha - \beta) = \frac{\operatorname{tan} \alpha - \operatorname{tan} \beta}{1 + \operatorname{tan} \alpha \operatorname{tan} \beta}$$

$$\operatorname{tan}(\alpha - \beta) = \frac{-\frac{7}{24} - \left(-\frac{3}{4}\right)}{1 + \left(-\frac{7}{24}\right)\left(-\frac{3}{4}\right)}$$

Simplify the numerator:

$$-\frac{7}{24} + \frac{3}{4} = -\frac{7}{24} + \frac{18}{24} = \frac{11}{24}$$

Simplify the denominator:

$$1 + \left(-\frac{7}{24}\right)\left(-\frac{3}{4}\right) = 1 + \frac{21}{96} = 1 + \frac{7}{32} = \frac{32}{32} + \frac{7}{32} = \frac{39}{32}$$

Divide the numerator by the denominator:

$$\operatorname{tan}(\alpha - \beta) = \frac{\frac{11}{24}}{\frac{39}{32}} = \frac{11}{24} \cdot \frac{32}{39} = \frac{11 \cdot 32}{24 \cdot 39} = \frac{11 \cdot 4}{3 \cdot 39} = \frac{44}{117}$$

The Grand Finale: Our Answer

Therefore, the exact value of $\operatorname{tan}(\alpha - \beta)$ is $\frac{44}{117}$.

Key Strategies for Trigonometry Success

• Master the Formulas: Knowing your trigonometric identities is crucial. Practice using them in different scenarios.
• Visualize the Quadrants: Understanding the ASTC rule and how it affects the signs of trigonometric functions is essential.
• Pythagorean Identity is Your Best Friend: Use it to find missing trigonometric values when given one value and the quadrant.
• Break It Down: Complex problems become manageable when you break them into smaller, logical steps. This is a very important tip to follow to make a complex problem easy.
• Simplify, Simplify, Simplify: Don't be afraid to simplify fractions and expressions. It reduces the chance of errors and makes the calculations easier. You need to make the expression simplified to get an accurate answer.

Level Up Your Trigonometry Skills

To truly master these concepts, try solving similar problems. Change the given trigonometric values or the quadrants, and see if you can still find the exact value of $\operatorname{tan}(\alpha - \beta)$. The more you practice, the more confident and proficient you'll become in trigonometry. This problem is very important and has a high chance of appearing in your exams.

I hope this comprehensive guide has been helpful! Keep practicing, keep exploring, and you'll become a trigonometry whiz in no time! These problem may seem very complex but after practice they will become very easy.

$$

Hey everyone! Let's tackle a classic trigonometry challenge today: finding the exact value of $\operatorname{tan}(\alpha - \beta)$. This problem is a fantastic workout for your trigonometric muscles, combining identities, quadrant knowledge, and a bit of Pythagorean magic. We'll break it down step by step, making sure you understand the 'why' behind each step. So, grab your metaphorical (or literal) pencil, and let's get started on this trigonometric adventure!

The Challenge: Problem Statement

Here's the scenario we're facing:

• We know that $\sin \alpha = -\frac{7}{25}$, and angle $\alpha$ resides in Quadrant IV.
• We also know that $\cos \beta = -\frac{4}{5}$, and angle $\beta$ is located in Quadrant II.

Our mission, should we choose to accept it (and we do!), is to find the exact value of $\operatorname{tan}(\alpha - \beta)$.

The Trigonometric Toolbox: Key Concepts

Before we dive into calculations, let's make sure our toolbox is well-stocked. We'll need these essential concepts:

1. Tangent Subtraction Formula: This is our primary weapon! It tells us:

   $$\operatorname{tan}(\alpha - \beta) = \frac{\operatorname{tan} \alpha - \operatorname{tan} \beta}{1 + \operatorname{tan} \alpha \operatorname{tan} \beta}$$

   This formula is a must-know for solving problems involving the tangent of a difference of angles.

2. Quadrant Rules (ASTC): Think ASTC (All Students Take Calculus) to remember which functions are positive in each quadrant:

   • Quadrant I: All (sine, cosine, tangent) are positive.
   • Quadrant II: Sine (and cosecant) is positive.
   • Quadrant III: Tangent (and cotangent) is positive.
   • Quadrant IV: Cosine (and secant) is positive. These rules are critical for determining the signs of trigonometric values.

3. Pythagorean Identity: This is the workhorse of trigonometry, relating sine and cosine:

   $$\sin^2 \theta + \cos^2 \theta = 1$$

   We'll use it to find missing trigonometric values.

Solving the Puzzle: A Step-by-Step Approach

Let's break down the problem into manageable steps:

Step 1: Unearthing $\cos \alpha$ and $\operatorname{tan} \alpha$

We're given $\sin \alpha = -\frac{7}{25}$ and that $\alpha$ is in Quadrant IV. Since cosine is positive in Quadrant IV, we can use the Pythagorean identity to find $\cos \alpha$:

$$\sin^2 \alpha + \cos^2 \alpha = 1$$

Substitute $\sin \alpha$:

$$\left(-\frac{7}{25}\right)^2 + \cos^2 \alpha = 1$$

$$\frac{49}{625} + \cos^2 \alpha = 1$$

Solve for $\cos^2 \alpha$:

$$\cos^2 \alpha = 1 - \frac{49}{625} = \frac{576}{625}$$

Take the square root (remembering $\cos \alpha$ is positive in Quadrant IV):

$$\cos \alpha = \sqrt{\frac{576}{625}} = \frac{24}{25}$$

Now, we can calculate $\operatorname{tan} \alpha$:

$$\operatorname{tan} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{7}{25}}{\frac{24}{25}} = -\frac{7}{24}$$

Step 2: Discovering $\sin \beta$ and $\operatorname{tan} \beta$

We're given $\cos \beta = -\frac{4}{5}$ and that $\beta$ is in Quadrant II. Sine is positive in Quadrant II, so let's use the Pythagorean identity again to find $\sin \beta$:

$$\sin^2 \beta + \cos^2 \beta = 1$$

Substitute $\cos \beta$:

$$\sin^2 \beta + \left(-\frac{4}{5}\right)^2 = 1$$

$$\sin^2 \beta + \frac{16}{25} = 1$$

Solve for $\sin^2 \beta$:

$$\sin^2 \beta = 1 - \frac{16}{25} = \frac{9}{25}$$

Take the square root (remembering $\sin \beta$ is positive in Quadrant II):

$$\sin \beta = \sqrt{\frac{9}{25}} = \frac{3}{5}$$

Now, we can calculate $\operatorname{tan} \beta$:

$$\operatorname{tan} \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}$$

Step 3: Unleashing the Tangent Subtraction Formula

We've gathered our forces! We have:

• $\operatorname{tan} \alpha = -\frac{7}{24}$
• $\operatorname{tan} \beta = -\frac{3}{4}$

Let's plug these into the tangent subtraction formula:

$$\operatorname{tan}(\alpha - \beta) = \frac{\operatorname{tan} \alpha - \operatorname{tan} \beta}{1 + \operatorname{tan} \alpha \operatorname{tan} \beta}$$

$$\operatorname{tan}(\alpha - \beta) = \frac{-\frac{7}{24} - \left(-\frac{3}{4}\right)}{1 + \left(-\frac{7}{24}\right)\left(-\frac{3}{4}\right)}$$

Simplify the numerator:

$$-\frac{7}{24} + \frac{3}{4} = -\frac{7}{24} + \frac{18}{24} = \frac{11}{24}$$

Simplify the denominator:

$$1 + \left(-\frac{7}{24}\right)\left(-\frac{3}{4}\right) = 1 + \frac{21}{96} = 1 + \frac{7}{32} = \frac{32}{32} + \frac{7}{32} = \frac{39}{32}$$

Divide the numerator by the denominator:

$$\operatorname{tan}(\alpha - \beta) = \frac{\frac{11}{24}}{\frac{39}{32}} = \frac{11}{24} \cdot \frac{32}{39} = \frac{11 \cdot 32}{24 \cdot 39} = \frac{11 \cdot 4}{3 \cdot 39} = \frac{44}{117}$$

The Triumphant Conclusion: Our Answer

The exact value of $\operatorname{tan}(\alpha - \beta)$ is $\frac{44}{117}$.

Pro Tips for Trigonometric Prowess

• Know Your Identities: The tangent subtraction formula (and other trigonometric identities) are your best friends. Memorize them and practice using them.
• Quadrant Awareness: Pay close attention to the quadrants! They dictate the signs of your trigonometric functions.
• Pythagorean Power: The Pythagorean identity is a powerful tool for finding missing trigonometric values.
• Step-by-Step Strategy: Break down complex problems into smaller, manageable steps. This makes the process less daunting.
• Simplify Ruthlessly: Simplify fractions and expressions whenever possible. This reduces the risk of errors and makes the calculations easier to handle.

Sharpen Your Skills: Practice Makes Perfect

To truly master these concepts, work through similar problems. Experiment with different trigonometric values and quadrants. The more you practice, the more confident you'll become in your trigonometry abilities. These are some very important tips to follow to enhance your trigonometry skill.

I hope this comprehensive walkthrough has been helpful and insightful! Keep exploring the world of trigonometry, and you'll unlock its beauty and power.