Trigonometric Expression: Find The Exact Value

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Hey math enthusiasts! Today, we're diving into a fascinating trigonometric expression and figuring out its exact value without using a calculator. This kind of problem is a great exercise in understanding trigonometric identities and inverse functions. We'll break down the expression step by step, using some clever tricks to simplify it. So, grab your pencils and let's get started!

Understanding the Problem: The Trigonometric Challenge

Our mission, should you choose to accept it, is to find the exact value of the following expression: sin⁑(sinβ‘βˆ’1(14)+cosβ‘βˆ’1(βˆ’34))\sin \left(\sin^{-1}\left(\frac{1}{4}\right) + \cos^{-1}\left(-\frac{3}{4}\right)\right). At first glance, it might seem a bit intimidating, right? But don't worry, we'll conquer this beast together! The expression involves inverse trigonometric functions (sinβ‘βˆ’1\sin^{-1} and cosβ‘βˆ’1\cos^{-1}) and the sine function. The key to solving this problem lies in recognizing the relationships between these functions and applying trigonometric identities effectively. The goal here isn't just to get an answer; it's to understand how we get the answer, building a solid foundation in trigonometry. This problem is a beautiful blend of algebra and geometry, so get ready to flex those brain muscles. Let's start by understanding what each part of the expression means. sinβ‘βˆ’1(14)\sin^{-1}\left(\frac{1}{4}\right) represents the angle whose sine is 14\frac{1}{4}, and cosβ‘βˆ’1(βˆ’34)\cos^{-1}\left(-\frac{3}{4}\right) represents the angle whose cosine is βˆ’34-\frac{3}{4}. We're essentially adding these two angles and then finding the sine of the resulting angle. Sounds fun, doesn't it? Let's break this down into smaller, more manageable parts. We'll begin by focusing on the inverse trigonometric functions and see how we can express them in a way that is easier to work with. Remember, the goal is to find the exact value, which means no decimals and no calculators! This is where our knowledge of trigonometric identities will truly shine. We will utilize the sum and difference identities, along with some clever substitutions, to find our solution.

Breaking Down the Components

First, let's look at sinβ‘βˆ’1(14)\sin^{-1}\left(\frac{1}{4}\right). Let's call this angle A. So, sin⁑(A)=14\sin(A) = \frac{1}{4}. Since the sine function is positive in the first and second quadrants, and the range of sinβ‘βˆ’1\sin^{-1} is [βˆ’Ο€2,Ο€2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], angle A lies in the first quadrant. This gives us a good starting point. Now, let's consider cosβ‘βˆ’1(βˆ’34)\cos^{-1}\left(-\frac{3}{4}\right). Let's call this angle B. So, cos⁑(B)=βˆ’34\cos(B) = -\frac{3}{4}. The cosine function is negative in the second and third quadrants, but the range of cosβ‘βˆ’1\cos^{-1} is [0,Ο€]\left[0, \pi\right], which means angle B lies in the second quadrant. This is important because it tells us the angle's position on the unit circle. Knowing the quadrants of angles A and B helps us use the Pythagorean identity and other trigonometric identities correctly. To proceed, we'll need to find sin⁑(B)\sin(B) and cos⁑(A)\cos(A). Since we know sin⁑(A)\sin(A) and cos⁑(B)\cos(B), we can use the Pythagorean identity: sin⁑2(x)+cos⁑2(x)=1\sin^2(x) + \cos^2(x) = 1. This is our secret weapon! We can use this to find the cosine of A and the sine of B. This will allow us to simplify the original expression.

Unveiling the Solution: Step-by-Step Approach

Now, let's dive into the solution! We know sin⁑(A)=14\sin(A) = \frac{1}{4}. Using the Pythagorean identity, we get cos⁑2(A)=1βˆ’sin⁑2(A)=1βˆ’(14)2=1βˆ’116=1516\cos^2(A) = 1 - \sin^2(A) = 1 - \left(\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16}. Therefore, cos⁑(A)=1516=154\cos(A) = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} (since A is in the first quadrant, cosine is positive). Next, we know cos⁑(B)=βˆ’34\cos(B) = -\frac{3}{4}. Again, using the Pythagorean identity, we get sin⁑2(B)=1βˆ’cos⁑2(B)=1βˆ’(βˆ’34)2=1βˆ’916=716\sin^2(B) = 1 - \cos^2(B) = 1 - \left(-\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16}. Since B is in the second quadrant, sin⁑(B)\sin(B) is positive, so sin⁑(B)=716=74\sin(B) = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}. Now that we have sin⁑(A)\sin(A), cos⁑(A)\cos(A), sin⁑(B)\sin(B), and cos⁑(B)\cos(B), we can use the sum identity for sine: sin⁑(A+B)=sin⁑(A)cos⁑(B)+cos⁑(A)sin⁑(B)\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B). Substituting the values we found, we get: sin⁑(A+B)=(14)(βˆ’34)+(154)(74)=βˆ’316+10516=105βˆ’316\sin(A + B) = \left(\frac{1}{4}\right)\left(-\frac{3}{4}\right) + \left(\frac{\sqrt{15}}{4}\right)\left(\frac{\sqrt{7}}{4}\right) = -\frac{3}{16} + \frac{\sqrt{105}}{16} = \frac{\sqrt{105} - 3}{16}. And there you have it, guys! The exact value of the trigonometric expression is 105βˆ’316\frac{\sqrt{105} - 3}{16}. We did it! This solution demonstrates the power of trigonometric identities and how they can be used to solve complex problems. Isn’t it amazing how we broke down a complicated-looking expression into something we could solve using simple principles?

Detailed Calculation Breakdown

Let's meticulously detail each step to ensure clarity. We begin with sinβ‘βˆ’1(14)=A\sin^{-1}\left(\frac{1}{4}\right) = A and cosβ‘βˆ’1(βˆ’34)=B\cos^{-1}\left(-\frac{3}{4}\right) = B. This implies sin⁑(A)=14\sin(A) = \frac{1}{4} and cos⁑(B)=βˆ’34\cos(B) = -\frac{3}{4}. Our aim is to find sin⁑(A+B)\sin(A + B). We use the sum identity: sin⁑(A+B)=sin⁑(A)cos⁑(B)+cos⁑(A)sin⁑(B)\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B). We already know sin⁑(A)\sin(A) and cos⁑(B)\cos(B), so we need to find cos⁑(A)\cos(A) and sin⁑(B)\sin(B). For angle A: cos⁑2(A)=1βˆ’sin⁑2(A)=1βˆ’(14)2=1516\cos^2(A) = 1 - \sin^2(A) = 1 - \left(\frac{1}{4}\right)^2 = \frac{15}{16}, and since A is in the first quadrant, cos⁑(A)=154\cos(A) = \frac{\sqrt{15}}{4}. For angle B: sin⁑2(B)=1βˆ’cos⁑2(B)=1βˆ’(βˆ’34)2=716\sin^2(B) = 1 - \cos^2(B) = 1 - \left(-\frac{3}{4}\right)^2 = \frac{7}{16}, and since B is in the second quadrant, sin⁑(B)=74\sin(B) = \frac{\sqrt{7}}{4}. Now, substitute these values into the sum identity: sin⁑(A+B)=(14)(βˆ’34)+(154)(74)=βˆ’316+10516=105βˆ’316\sin(A + B) = \left(\frac{1}{4}\right)\left(-\frac{3}{4}\right) + \left(\frac{\sqrt{15}}{4}\right)\left(\frac{\sqrt{7}}{4}\right) = -\frac{3}{16} + \frac{\sqrt{105}}{16} = \frac{\sqrt{105} - 3}{16}. This detailed breakdown highlights the logical flow of the solution and ensures that every step is clear and easy to follow. This approach helps in building a strong foundation and understanding of trigonometric concepts. The meticulous steps ensure that no detail is missed, making the process easily replicable for similar problems.

Practical Applications and Further Exploration

Understanding and solving trigonometric expressions like this isn't just a mathematical exercise; it has real-world applications. Trigonometry is used in physics, engineering, computer graphics, and many other fields. The ability to manipulate and simplify trigonometric expressions is a fundamental skill.

Where Trigonometry Shines

In physics, trigonometry is crucial for analyzing motion, forces, and waves. Engineers use it to design structures, analyze circuits, and model systems. In computer graphics, trigonometry is used to create realistic 3D models and animations. Even in navigation, trigonometry is essential for calculating distances and directions. So, the skills you develop while solving these types of problems have far-reaching implications. Furthermore, this problem can be extended to explore other trigonometric identities and functions. For example, you could explore tangent functions or different combinations of angles. You could also try similar problems with different values, or even create your own! The possibilities are endless. Keep practicing, and you'll find that these trigonometric concepts become second nature. You can experiment with different problems, changing the values or the trigonometric functions involved. This exploration will deepen your understanding and build your confidence in tackling more complex problems. Remember, the journey of learning is just as important as the destination.

Conclusion: Mastering the Trigonometric Expression

Congratulations, guys! We've successfully found the exact value of the trigonometric expression without the use of a calculator. This exercise not only strengthens your understanding of inverse trigonometric functions and identities but also reinforces problem-solving skills. Remember that practice is key, so keep exploring and experimenting with different problems to deepen your understanding. This problem serves as a great example of how mathematical tools can be used to break down complex expressions into manageable parts. Keep up the great work, and happy calculating!

Key Takeaways

  • Understanding Inverse Functions: Knowing the ranges and properties of inverse trigonometric functions is crucial. This helps determine the quadrants of the angles.
  • Pythagorean Identity: This fundamental identity (sin⁑2(x)+cos⁑2(x)=1\sin^2(x) + \cos^2(x) = 1) is a powerful tool for finding missing trigonometric values.
  • Sum and Difference Identities: These identities (like sin⁑(A+B)=sin⁑(A)cos⁑(B)+cos⁑(A)sin⁑(B)\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)) are essential for simplifying expressions.
  • Step-by-Step Approach: Breaking down the problem into smaller, manageable steps makes it easier to solve.

Keep practicing, and you'll become a trigonometry pro in no time! Keep exploring, keep questioning, and most importantly, keep enjoying the beautiful world of mathematics!