Finding The Exact Value Of Cos(α - Β) Trigonometry Problem Solution

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In this comprehensive guide, we will delve into the intricacies of trigonometry to find the exact value of $\cos (\alpha-\beta)$. We are given that $\sin (\alpha) = -\frac{4}{5}$ and $\cos (\beta) = -\frac{5}{8}$, with $\alpha$ residing in Quadrant III and $\beta$ in Quadrant II. To achieve our goal, we will leverage fundamental trigonometric identities, quadrant rules, and algebraic manipulations. This article aims to provide a step-by-step solution, ensuring clarity and understanding for readers of all backgrounds.

Understanding the Problem

Before we embark on the solution, let's dissect the problem statement. Our primary objective is to determine the precise value of $\cos (\alpha-\beta)$. This expression involves the cosine of the difference between two angles, $\alpha$ and $\beta$. We are furnished with the sine of $\alpha$ and the cosine of $\beta$, along with the crucial information that $\alpha$ lies in Quadrant III and $eta$ in Quadrant II. This quadrant information is paramount as it dictates the signs of trigonometric functions.

In Quadrant III, both sine and cosine are negative, while tangent is positive. Conversely, in Quadrant II, sine is positive, cosine is negative, and tangent is negative. These sign conventions will play a pivotal role in determining the correct signs of the trigonometric values we calculate.

To compute $\cos (\alpha-\beta)$, we will employ the cosine difference identity, which is a cornerstone of trigonometric manipulations. This identity elegantly expresses the cosine of the difference between two angles in terms of their individual sines and cosines. Understanding and applying this identity is crucial for solving the problem at hand.

Trigonometric Identities and Quadrant Rules

At the heart of solving trigonometric problems lies a deep understanding of trigonometric identities and quadrant rules. These are the fundamental tools that enable us to manipulate expressions and determine the signs of trigonometric functions in different quadrants. Let's delve into these essential concepts.

Trigonometric Identities

Trigonometric identities are equations that hold true for all values of the variables involved. They are the bedrock of trigonometric manipulations and allow us to rewrite expressions in different forms. The identity that is most pertinent to our problem is the cosine difference identity:

cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos (\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta

This identity expresses the cosine of the difference between two angles in terms of the cosines and sines of the individual angles. It is the key to unlocking the value of $\cos (\alpha-\beta)$ given the information we have.

Besides the cosine difference identity, other fundamental identities are worth mentioning. The Pythagorean identity, $\sin^2 \theta + \cos^2 \theta = 1$, is a cornerstone of trigonometry and will be instrumental in finding $\cos \alpha$ and $\sin \beta$ from the given information. Additionally, understanding reciprocal identities (e.g., $\csc \theta = \frac{1}{\sin \theta}$, $\sec \theta = \frac{1}{\cos \theta}$, $\cot \theta = \frac{1}{\tan \theta}$) and quotient identities (e.g., $\tan \theta = \frac{\sin \theta}{\cos \theta}$) can be beneficial in various trigonometric problems.

Quadrant Rules

Quadrant rules dictate the signs of trigonometric functions in each of the four quadrants of the Cartesian plane. These rules are a consequence of the definitions of trigonometric functions in terms of the coordinates of a point on the unit circle.

  • Quadrant I: All trigonometric functions (sine, cosine, tangent, cosecant, secant, cotangent) are positive.
  • Quadrant II: Sine and cosecant are positive; cosine, tangent, secant, and cotangent are negative.
  • Quadrant III: Tangent and cotangent are positive; sine, cosine, cosecant, and secant are negative.
  • Quadrant IV: Cosine and secant are positive; sine, tangent, cosecant, and cotangent are negative.

In our problem, we are given that $\alpha$ lies in Quadrant III, where both sine and cosine are negative. This means that when we calculate $\cos \alpha$, we must choose the negative square root. Similarly, $\beta$ lies in Quadrant II, where sine is positive and cosine is negative. This will influence our choice of sign when determining $\sin \beta$.

Step-by-Step Solution

Now that we have laid the groundwork by understanding the problem, trigonometric identities, and quadrant rules, let's embark on the step-by-step solution.

Step 1: Find cos α

We are given that $\sin \alpha = -\frac{4}{5}$. To find $\cos \alpha$, we can leverage the Pythagorean identity:

sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1

Substituting the given value of $\sin \alpha$, we get:

(−45)2+cos⁡2α=1\left(-\frac{4}{5}\right)^2 + \cos^2 \alpha = 1

1625+cos⁡2α=1\frac{16}{25} + \cos^2 \alpha = 1

Subtracting $\frac{16}{25}$ from both sides:

cos⁡2α=1−1625\cos^2 \alpha = 1 - \frac{16}{25}

cos⁡2α=925\cos^2 \alpha = \frac{9}{25}

Taking the square root of both sides:

cos⁡α=±35\cos \alpha = \pm \frac{3}{5}

Since $\alpha$ is in Quadrant III, where cosine is negative, we choose the negative value:

cos⁡α=−35\cos \alpha = -\frac{3}{5}

Step 2: Find sin β

We are given that $\cos \beta = -\frac{5}{8}$. To find $\sin \beta$, we again use the Pythagorean identity:

sin⁡2β+cos⁡2β=1\sin^2 \beta + \cos^2 \beta = 1

Substituting the given value of $\cos \beta$, we get:

sin⁡2β+(−58)2=1\sin^2 \beta + \left(-\frac{5}{8}\right)^2 = 1

sin⁡2β+2564=1\sin^2 \beta + \frac{25}{64} = 1

Subtracting $\frac{25}{64}$ from both sides:

sin⁡2β=1−2564\sin^2 \beta = 1 - \frac{25}{64}

sin⁡2β=3964\sin^2 \beta = \frac{39}{64}

Taking the square root of both sides:

sin⁡β=±398\sin \beta = \pm \frac{\sqrt{39}}{8}

Since $\beta$ is in Quadrant II, where sine is positive, we choose the positive value:

sin⁡β=398\sin \beta = \frac{\sqrt{39}}{8}

Step 3: Apply the Cosine Difference Identity

Now that we have found $\cos \alpha$, $\sin \alpha$, $\cos \beta$, and $\sin \beta$, we can apply the cosine difference identity:

cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos (\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta

Substituting the values we found:

cos⁡(α−β)=(−35)(−58)+(−45)(398)\cos (\alpha-\beta) = \left(-\frac{3}{5}\right)\left(-\frac{5}{8}\right) + \left(-\frac{4}{5}\right)\left(\frac{\sqrt{39}}{8}\right)

cos⁡(α−β)=1540−43940\cos (\alpha-\beta) = \frac{15}{40} - \frac{4\sqrt{39}}{40}

Combining the terms:

cos⁡(α−β)=15−43940\cos (\alpha-\beta) = \frac{15 - 4\sqrt{39}}{40}

Therefore, the exact value of $\cos (\alpha-\beta)$ is $\frac{15 - 4\sqrt{39}}{40}$.

Conclusion

In this detailed exploration, we successfully determined the exact value of $\cos (\alpha-\beta)$ given $\sin \alpha = -\frac{4}{5}$ and $\cos \beta = -\frac{5}{8}$, with $\alpha$ in Quadrant III and $\beta$ in Quadrant II. Our journey involved a careful application of trigonometric identities, particularly the cosine difference identity and the Pythagorean identity. We also emphasized the importance of quadrant rules in determining the signs of trigonometric functions.

The step-by-step solution illuminated the process of finding $\cos \alpha$ and $\sin \beta$ using the Pythagorean identity and the quadrant information. Finally, we substituted these values into the cosine difference identity to arrive at the exact value of $\cos (\alpha-\beta)$, which is $ rac{15 - 4\sqrt{39}}{40}$. This problem exemplifies the power and elegance of trigonometry in solving geometric and algebraic problems. By mastering these fundamental concepts and techniques, you will be well-equipped to tackle a wide array of trigonometric challenges. Remember, practice is key to solidifying your understanding and building confidence in your problem-solving abilities. Keep exploring the fascinating world of trigonometry, and you will discover its profound applications in various fields of science, engineering, and mathematics.