Finding Tan(α+β) Using Trigonometric Identities

by Alex Johnson 48 views

Let's dive into the fascinating world of trigonometry! In this article, we'll tackle a classic problem: finding the value of tan⁡(α+β)\tan(\alpha + \beta) given some information about cos⁡(α)\cos(\alpha) and cos⁡(β)\cos(\beta), along with the quadrants where the angles α\alpha and β\beta lie. We'll leverage our knowledge of trigonometric identities, quadrant rules, and sum formulas to solve this problem step by step. So, buckle up and get ready to explore the beauty of trigonometric relationships!

Understanding the Problem

Before we jump into the calculations, let's break down the problem statement. We're given that cos⁡(α)=116\cos(\alpha) = \frac{\sqrt{11}}{6} and 3π2<α<2π\frac{3\pi}{2} < \alpha < 2\pi. This tells us two crucial things: the cosine of angle α\alpha is 116\frac{\sqrt{11}}{6}, and α\alpha lies in the fourth quadrant. Remember, in the fourth quadrant, cosine is positive, sine is negative, and tangent is negative. Similarly, we're given that cos⁡(β)=−457\cos(\beta) = -\frac{\sqrt{45}}{7} and β\beta is in the second quadrant. This means the cosine of angle β\beta is −457-\frac{\sqrt{45}}{7}, and β\beta lies in the second quadrant. In the second quadrant, sine is positive, cosine is negative, and tangent is negative. Our mission is to find tan⁡(α+β)\tan(\alpha + \beta) using the sum formula for tangent, which is a key trigonometric identity. To do this effectively, we need to first determine the values of sin⁡(α)\sin(\alpha), tan⁡(α)\tan(\alpha), sin⁡(β)\sin(\beta), and tan⁡(β)\tan(\beta). These values will serve as the building blocks for our final calculation. Trigonometric identities are essential tools in solving these kinds of problems, and understanding their applications is crucial for mastering trigonometry.

Finding sin(α) and tan(α)

Now, let's find sin⁡(α)\sin(\alpha) and tan⁡(α)\tan(\alpha). We know that cos⁡(α)=116\cos(\alpha) = \frac{\sqrt{11}}{6} and α\alpha is in the fourth quadrant. We can use the Pythagorean identity, which states that sin⁡2(α)+cos⁡2(α)=1\sin^2(\alpha) + \cos^2(\alpha) = 1, to find sin⁡(α)\sin(\alpha). Plugging in the value of cos⁡(α)\cos(\alpha), we get:

sin⁡2(α)+(116)2=1\sin^2(\alpha) + \left(\frac{\sqrt{11}}{6}\right)^2 = 1

sin⁡2(α)+1136=1\sin^2(\alpha) + \frac{11}{36} = 1

sin⁡2(α)=1−1136=2536\sin^2(\alpha) = 1 - \frac{11}{36} = \frac{25}{36}

Taking the square root of both sides, we get sin⁡(α)=±56\sin(\alpha) = \pm\frac{5}{6}. Since α\alpha is in the fourth quadrant, where sine is negative, we have sin⁡(α)=−56\sin(\alpha) = -\frac{5}{6}. Next, we can find tan⁡(α)\tan(\alpha) using the identity tan⁡(α)=sin⁡(α)cos⁡(α)\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}. Substituting the values we found:

tan⁡(α)=−56116=−511\tan(\alpha) = \frac{-\frac{5}{6}}{\frac{\sqrt{11}}{6}} = -\frac{5}{\sqrt{11}}

To rationalize the denominator, we multiply the numerator and denominator by 11\sqrt{11}:

tan⁡(α)=−51111\tan(\alpha) = -\frac{5\sqrt{11}}{11}

So, we've found that sin⁡(α)=−56\sin(\alpha) = -\frac{5}{6} and tan⁡(α)=−51111\tan(\alpha) = -\frac{5\sqrt{11}}{11}. These values are crucial for the next steps in solving the problem. Pythagorean identities are fundamental in trigonometry, allowing us to relate sine and cosine, which in turn helps us find other trigonometric functions. The careful application of these identities, along with quadrant rules, ensures we get the correct signs for our trigonometric values.

Finding sin(β) and tan(β)

Now, let's shift our focus to angle β\beta. We know that cos⁡(β)=−457\cos(\beta) = -\frac{\sqrt{45}}{7} and β\beta is in the second quadrant. Again, we can use the Pythagorean identity sin⁡2(β)+cos⁡2(β)=1\sin^2(\beta) + \cos^2(\beta) = 1 to find sin⁡(β)\sin(\beta). Plugging in the value of cos⁡(β)\cos(\beta), we get:

sin⁡2(β)+(−457)2=1\sin^2(\beta) + \left(-\frac{\sqrt{45}}{7}\right)^2 = 1

sin⁡2(β)+4549=1\sin^2(\beta) + \frac{45}{49} = 1

sin⁡2(β)=1−4549=449\sin^2(\beta) = 1 - \frac{45}{49} = \frac{4}{49}

Taking the square root of both sides, we get sin⁡(β)=±27\sin(\beta) = \pm\frac{2}{7}. Since β\beta is in the second quadrant, where sine is positive, we have sin⁡(β)=27\sin(\beta) = \frac{2}{7}. Next, we find tan⁡(β)\tan(\beta) using the identity tan⁡(β)=sin⁡(β)cos⁡(β)\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}. Substituting the values we found:

tan⁡(β)=27−457=−245\tan(\beta) = \frac{\frac{2}{7}}{-\frac{\sqrt{45}}{7}} = -\frac{2}{\sqrt{45}}

We can simplify 45\sqrt{45} as 9⋅5=35\sqrt{9 \cdot 5} = 3\sqrt{5}, so:

tan⁡(β)=−235\tan(\beta) = -\frac{2}{3\sqrt{5}}

To rationalize the denominator, we multiply the numerator and denominator by 5\sqrt{5}:

tan⁡(β)=−2515\tan(\beta) = -\frac{2\sqrt{5}}{15}

So, we've found that sin⁡(β)=27\sin(\beta) = \frac{2}{7} and tan⁡(β)=−2515\tan(\beta) = -\frac{2\sqrt{5}}{15}. These values, along with the values for α\alpha, will be used in the tangent sum formula. Understanding quadrant rules is just as important as knowing the trigonometric identities themselves. By knowing the sign conventions for each quadrant, we can correctly determine the sign of our trigonometric functions, ensuring accurate results.

Applying the Sum Formula for Tangent

Now comes the exciting part: applying the sum formula for tangent! The formula states:

tan⁡(α+β)=tan⁡(α)+tan⁡(β)1−tan⁡(α)tan⁡(β)\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}

We have already calculated tan⁡(α)=−51111\tan(\alpha) = -\frac{5\sqrt{11}}{11} and tan⁡(β)=−2515\tan(\beta) = -\frac{2\sqrt{5}}{15}. Let's plug these values into the formula:

tan⁡(α+β)=−51111−25151−(−51111)(−2515)\tan(\alpha + \beta) = \frac{-\frac{5\sqrt{11}}{11} - \frac{2\sqrt{5}}{15}}{1 - \left(-\frac{5\sqrt{11}}{11}\right)\left(-\frac{2\sqrt{5}}{15}\right)}

To simplify this expression, we first find a common denominator for the terms in the numerator and denominator. The common denominator for 11 and 15 is 165. So, we rewrite the numerator:

−51111−2515=−7511165−225165=−7511−225165-\frac{5\sqrt{11}}{11} - \frac{2\sqrt{5}}{15} = -\frac{75\sqrt{11}}{165} - \frac{22\sqrt{5}}{165} = \frac{-75\sqrt{11} - 22\sqrt{5}}{165}

Next, we simplify the denominator:

1−(−51111)(−2515)=1−1055165=165−10551651 - \left(-\frac{5\sqrt{11}}{11}\right)\left(-\frac{2\sqrt{5}}{15}\right) = 1 - \frac{10\sqrt{55}}{165} = \frac{165 - 10\sqrt{55}}{165}

Now, we can rewrite the entire expression:

tan⁡(α+β)=−7511−225165165−1055165=−7511−225165−1055\tan(\alpha + \beta) = \frac{\frac{-75\sqrt{11} - 22\sqrt{5}}{165}}{\frac{165 - 10\sqrt{55}}{165}} = \frac{-75\sqrt{11} - 22\sqrt{5}}{165 - 10\sqrt{55}}

This is the exact value of tan⁡(α+β)\tan(\alpha + \beta). While we could attempt to rationalize the denominator further, this form is perfectly acceptable and represents the solution. The sum and difference formulas are powerful tools that allow us to find trigonometric functions of combined angles, expanding our ability to solve complex trigonometric problems.

Conclusion

In this article, we successfully found the value of tan⁡(α+β)\tan(\alpha + \beta) given the values of cos⁡(α)\cos(\alpha) and cos⁡(β)\cos(\beta), along with the quadrants in which the angles lie. We used the Pythagorean identity to find sin⁡(α)\sin(\alpha) and sin⁡(β)\sin(\beta), then calculated tan⁡(α)\tan(\alpha) and tan⁡(β)\tan(\beta). Finally, we applied the sum formula for tangent to arrive at our solution: tan⁡(α+β)=−7511−225165−1055\tan(\alpha + \beta) = \frac{-75\sqrt{11} - 22\sqrt{5}}{165 - 10\sqrt{55}}. This problem showcases the importance of understanding trigonometric identities, quadrant rules, and sum formulas in solving trigonometric problems. By mastering these concepts, you'll be well-equipped to tackle a wide range of trigonometric challenges.

For further exploration of trigonometric identities and formulas, you can visit Khan Academy's Trigonometry section. This resource offers comprehensive lessons, practice exercises, and videos to help you deepen your understanding of trigonometry. Remember, practice makes perfect, so keep exploring and honing your skills!