Finding The Exact Value Of Sin 75 Degrees Using Half-Angle Formulas

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Introduction

In the realm of trigonometry, half-angle formulas provide a powerful tool for determining the sine, cosine, and tangent of angles that are half the size of known angles. This is especially useful when dealing with angles that are not standard values found on the unit circle, such as 30°, 45°, 60°, and their multiples. In this comprehensive exploration, we will delve into the application of half-angle formulas to find the exact value of sin 75°. This exercise not only demonstrates the utility of these formulas but also reinforces the fundamental principles of trigonometric identities and their manipulation. Let's embark on a journey to unravel the intricacies of half-angle formulas and their application in solving trigonometric problems. Our focus will be on the sine half-angle formula, which is a cornerstone in this process, and we will meticulously walk through each step to ensure a clear understanding of the solution.

Understanding Half-Angle Formulas

Before we tackle the problem at hand, it is crucial to establish a firm grasp of half-angle formulas. These formulas are derived from the power-reducing formulas and offer a direct relationship between trigonometric functions of an angle and those of half that angle. Specifically, the half-angle formulas for sine, cosine, and tangent are given by:

  • sin(x/2) = ±√[(1 - cos x) / 2]
  • cos(x/2) = ±√[(1 + cos x) / 2]
  • tan(x/2) = ±√[(1 - cos x) / (1 + cos x)] = sin x / (1 + cos x) = (1 - cos x) / sin x

The ± sign in front of the square root indicates that the sign of the result depends on the quadrant in which the angle x/2 lies. For instance, if x/2 is in the first or second quadrant, sin(x/2) is positive, whereas if it is in the third or fourth quadrant, sin(x/2) is negative. Similarly, the sign of cos(x/2) depends on whether x/2 is in the first or fourth quadrant (positive) or in the second or third quadrant (negative). Understanding these sign conventions is paramount to applying the half-angle formulas correctly. In the context of finding sin 75°, we will focus on the sine half-angle formula and carefully determine the appropriate sign based on the quadrant in which 75° lies.

Applying the Half-Angle Formula to sin 75°

Now, let's put our knowledge into action and find the exact value of sin 75°. The key to using the half-angle formula lies in recognizing that 75° is half of 150°. Thus, we can express sin 75° as sin(150°/2). Our strategic move is to leverage the half-angle formula for sine, which states that sin(x/2) = ±√[(1 - cos x) / 2]. In our scenario, x is 150°, and we need to determine the cosine of 150°. Recall that 150° is in the second quadrant, where cosine values are negative. The reference angle for 150° is 30°, and we know that cos 30° = √3/2. Therefore, cos 150° = -√3/2. Now, we can substitute this value into the half-angle formula:

sin 75° = sin(150°/2) = ±√[(1 - cos 150°) / 2] = ±√[(1 - (-√3/2)) / 2] = ±√[(1 + √3/2) / 2]

Simplifying the expression inside the square root, we get:

sin 75° = ±√[(2 + √3) / 4] = ±√(2 + √3) / 2

Since 75° is in the first quadrant, where sine values are positive, we choose the positive square root. Therefore, the exact value of sin 75° is √(2 + √3) / 2. This result showcases the power and elegance of half-angle formulas in determining trigonometric values for angles that are not directly available on the unit circle. In the next section, we will further refine our understanding by verifying this result and exploring alternative approaches to solving this problem.

Detailed Steps to Calculate sin 75° Using Half-Angle Formula

To find the exact value of sin 75° using the half-angle formula, we will break down the process into clear, manageable steps. This systematic approach will not only help us arrive at the correct answer but also reinforce the underlying principles and techniques involved in trigonometric calculations. Let's embark on this step-by-step journey:

  1. Recognize the Relationship: The first step is to recognize that 75° is half of 150°. This is crucial because it allows us to utilize the half-angle formula for sine, which relates the sine of an angle to the cosine of twice that angle. Thus, we can express sin 75° as sin(150°/2).
  2. State the Half-Angle Formula: Recall the half-angle formula for sine: sin(x/2) = ±√[(1 - cos x) / 2]. This formula is the cornerstone of our calculation, and it provides the direct link between sin 75° and cos 150°.
  3. Determine cos 150°: To apply the formula, we need to find the value of cos 150°. Since 150° is in the second quadrant, its cosine value will be negative. The reference angle for 150° is 180° - 150° = 30°. We know that cos 30° = √3/2, so cos 150° = -√3/2. This step involves recalling the unit circle and the properties of trigonometric functions in different quadrants.
  4. Substitute into the Formula: Now, substitute cos 150° = -√3/2 into the half-angle formula: sin 75° = ±√[(1 - (-√3/2)) / 2]. This step is a direct application of the formula, replacing the variable with the specific value we have determined.
  5. Simplify the Expression: Next, simplify the expression inside the square root: sin 75° = ±√[(1 + √3/2) / 2] = ±√[(2 + √3) / 4]. This involves basic algebraic manipulation to combine the terms and express them in a more manageable form.
  6. Take the Square Root: Evaluate the square root: sin 75° = ±√(2 + √3) / 2. This step involves understanding how to handle square roots of fractions and simplifying them accordingly.
  7. Determine the Sign: Since 75° is in the first quadrant, where sine values are positive, we choose the positive square root. This is a critical step in ensuring the correct sign of the final answer, and it relies on our understanding of the unit circle and the properties of trigonometric functions.
  8. Final Answer: Therefore, the exact value of sin 75° is √(2 + √3) / 2. This is the culmination of our step-by-step calculation, and it represents the exact trigonometric value we sought.

By meticulously following these steps, we have successfully found the exact value of sin 75° using the half-angle formula. This process not only provides the answer but also reinforces the importance of understanding trigonometric identities, quadrant rules, and algebraic manipulation. In the subsequent sections, we will explore alternative methods for calculating sin 75° and discuss the significance of this result in various contexts.

Verifying the Result and Alternative Approaches

Having found the exact value of sin 75° using the half-angle formula, it is prudent to verify our result and explore alternative approaches to solving the same problem. This not only reinforces our understanding but also provides a broader perspective on trigonometric problem-solving. Let's delve into a couple of methods to verify our answer and gain deeper insights.

One way to verify our result is by using the sum-to-product formulas. We can express 75° as the sum of two angles whose sine and cosine values are known, such as 45° and 30°. The sine of a sum formula is given by:

sin(A + B) = sin A cos B + cos A sin B

Applying this formula to sin 75°, we have:

sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

We know that sin 45° = √2/2, cos 30° = √3/2, cos 45° = √2/2, and sin 30° = 1/2. Substituting these values, we get:

sin 75° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2) / 4

Now, let's examine our result from the half-angle formula, √(2 + √3) / 2, and see if it is equivalent to (√6 + √2) / 4. To do this, we can square both expressions and see if they yield the same value. Squaring √(2 + √3) / 2, we get:

[√(2 + √3) / 2]² = (2 + √3) / 4

Squaring (√6 + √2) / 4, we get:

[(√6 + √2) / 4]² = (6 + 2√12 + 2) / 16 = (8 + 4√3) / 16 = (2 + √3) / 4

Since both expressions yield the same value when squared, we can conclude that √(2 + √3) / 2 and (√6 + √2) / 4 are equivalent. This verification step confirms the correctness of our result obtained using the half-angle formula.

Another approach to verifying our result is by using a calculator to approximate the decimal values of both expressions and comparing them. Using a calculator, we find that:

sin 75° ≈ 0.9659 √(2 + √3) / 2 ≈ 0.9659 (√6 + √2) / 4 ≈ 0.9659

These approximations further validate our result and demonstrate the consistency of our calculations. By exploring these alternative approaches and verifying our result, we have strengthened our understanding of trigonometric principles and problem-solving techniques.

Significance and Applications of Half-Angle Formulas

The half-angle formulas are not just mathematical curiosities; they hold significant importance and have a wide array of applications in various fields. Understanding their significance and applications enriches our appreciation for these trigonometric tools and their role in solving real-world problems. Let's explore some key aspects of their significance and applications.

One of the primary significances of half-angle formulas lies in their ability to determine exact trigonometric values for angles that are not standard values on the unit circle. As we have seen with sin 75°, the half-angle formula allows us to find its exact value, which would be challenging to determine otherwise. This capability is crucial in various mathematical and scientific contexts where precise trigonometric values are required.

In the realm of calculus, half-angle formulas are invaluable for simplifying integrals involving trigonometric functions. They can transform complex integrals into more manageable forms, making them easier to solve. This is particularly useful in areas such as physics and engineering, where trigonometric integrals frequently arise.

Engineering extensively utilizes trigonometric functions and identities, including half-angle formulas, in various applications. For instance, in electrical engineering, they are used in the analysis of AC circuits. In mechanical engineering, they are applied in the study of oscillations and vibrations. The precision offered by these formulas is essential for accurate modeling and design in these fields.

Physics also benefits significantly from the use of half-angle formulas. They are employed in areas such as optics, where the behavior of light waves is analyzed, and in mechanics, where the motion of objects is studied. The ability to express trigonometric functions in different forms allows physicists to simplify equations and gain deeper insights into physical phenomena.

Beyond these specific applications, half-angle formulas contribute to a deeper understanding of trigonometric relationships and identities. They serve as a bridge between different trigonometric functions and provide a powerful tool for manipulating and simplifying trigonometric expressions. This understanding is fundamental for advanced mathematical studies and problem-solving.

In conclusion, the half-angle formulas are not merely theoretical constructs; they are practical tools with far-reaching applications. Their ability to determine exact trigonometric values, simplify integrals, and aid in various engineering and physics applications underscores their significance in mathematics and science. By mastering these formulas, we equip ourselves with a powerful toolset for tackling a wide range of problems and gaining a deeper appreciation for the interconnectedness of mathematical concepts.

Conclusion

In this comprehensive exploration, we have successfully navigated the application of half-angle formulas to find the exact value of sin 75°. Our journey began with a thorough understanding of the half-angle formulas themselves, focusing on the sine half-angle formula as our primary tool. We meticulously broke down the problem into manageable steps, starting from recognizing the relationship between 75° and 150°, applying the formula, simplifying the expression, and determining the correct sign based on the quadrant. This systematic approach not only led us to the correct answer but also reinforced the importance of understanding trigonometric identities, quadrant rules, and algebraic manipulation. Furthermore, we verified our result using alternative methods, such as the sum-to-product formulas and calculator approximations, which strengthened our confidence in the solution and provided a broader perspective on trigonometric problem-solving. Finally, we delved into the significance and applications of half-angle formulas, highlighting their role in determining exact trigonometric values, simplifying integrals in calculus, and aiding in various engineering and physics applications. The ability to express trigonometric functions in different forms allows scientists and engineers to simplify equations and gain deeper insights into physical phenomena. In essence, the half-angle formulas are not just mathematical tools; they are powerful instruments that bridge the gap between theoretical concepts and practical applications, fostering a deeper understanding of the world around us. As we conclude this exploration, we recognize that the mastery of trigonometric identities and formulas is an essential skill for anyone venturing into advanced mathematics, science, and engineering. The journey of finding the exact value of sin 75° serves as a testament to the power and elegance of these mathematical tools and their ability to unravel the intricacies of the trigonometric world.