Finding Sin(105°) Using The Half-Angle Formula With Complete Simplification
In the realm of trigonometry, half-angle formulas stand as powerful tools for evaluating trigonometric functions of angles that are half the size of known angles. These formulas are particularly useful when dealing with angles that are not standard angles (e.g., 30°, 45°, 60°, 90°) but can be expressed as half of a standard angle. In this article, we will delve into the application of the half-angle formula to determine the value of sin(105°), simplifying the answer to its most concise form. This exploration will not only enhance our understanding of trigonometric identities but also showcase their practical utility in solving mathematical problems. This method allows us to find exact values for trigonometric functions of angles that are not on the unit circle, expanding our ability to solve a wider range of problems. The half-angle formulas are derived from the double-angle formulas, which are fundamental trigonometric identities. By manipulating these double-angle formulas, we can express trigonometric functions of half an angle in terms of trigonometric functions of the full angle. This transformation is crucial for evaluating sines, cosines, and tangents of angles like 105°, which do not have readily available values. Mastering these formulas is essential for anyone studying trigonometry and its applications in fields like physics, engineering, and computer graphics. The ability to break down complex problems into simpler parts, using trigonometric identities, is a hallmark of mathematical proficiency. In the following sections, we will walk through the process step-by-step, ensuring that the logic and application of the half-angle formula are clear and understandable. The use of half-angle formulas not only provides a method for calculation but also reinforces the interconnectedness of trigonometric concepts. By understanding the relationships between different angles and their trigonometric functions, we gain a deeper appreciation for the elegance and power of mathematics.
Understanding Half-Angle Formulas
The half-angle formulas are a set of trigonometric identities that express trigonometric functions of an angle that is half of another angle. These formulas are derived from the double-angle formulas and are essential for finding exact values of trigonometric functions for angles that are not standard angles on the unit circle. The half-angle formulas for sine, cosine, and tangent are:
- sin(θ/2) = ±√[(1 - cos θ)/2]
- cos(θ/2) = ±√[(1 + cos θ)/2]
- tan(θ/2) = ±√[(1 - cos θ)/(1 + cos θ)] = (sin θ)/(1 + cos θ) = (1 - cos θ)/(sin θ)
The "±" sign in the sine and cosine formulas indicates that the sign of the result depends on the quadrant in which θ/2 lies. To determine the correct sign, we need to consider the quadrant of the angle θ/2 and the sign of the trigonometric function in that quadrant. For example, if θ/2 is in the first or second quadrant, sin(θ/2) is positive, while if it is in the third or fourth quadrant, sin(θ/2) is negative. Similarly, cos(θ/2) is positive in the first and fourth quadrants and negative in the second and third quadrants. The tangent formula, on the other hand, has multiple forms, each of which can be useful in different situations. The first form, ±√[(1 - cos θ)/(1 + cos θ)], involves a square root and a "±" sign, similar to the sine and cosine formulas. The second and third forms, (sin θ)/(1 + cos θ) and (1 - cos θ)/(sin θ), are often more convenient to use because they do not involve square roots or sign ambiguities. Understanding the derivation and application of these formulas is crucial for solving a wide range of trigonometric problems. These formulas are not just theoretical constructs; they have practical applications in various fields, including physics, engineering, and computer graphics. For instance, in physics, they can be used to analyze wave phenomena and oscillations. In engineering, they are essential for designing structures and systems that involve angular relationships. In computer graphics, they are used to perform rotations and transformations of objects in 3D space. The versatility of half-angle formulas makes them a valuable tool in any mathematician's arsenal. By mastering these formulas, students and professionals alike can tackle complex problems with greater confidence and efficiency. Moreover, the process of deriving and applying these formulas enhances one's understanding of the underlying principles of trigonometry. This deeper understanding, in turn, fosters a greater appreciation for the elegance and power of mathematics.
Applying the Half-Angle Formula to sin(105°)
To find the value of sin(105°), we can use the half-angle formula for sine. First, we need to recognize that 105° is half of 210°. Therefore, we can set θ/2 = 105°, which means θ = 210°. Now, we can apply the half-angle formula:
sin(105°) = sin(210°/2) = ±√[(1 - cos 210°)/2]
Next, we need to find the value of cos 210°. The angle 210° lies in the third quadrant, where cosine is negative. The reference angle for 210° is 210° - 180° = 30°. Therefore, cos 210° = -cos 30° = -√3/2. Substituting this value into the half-angle formula, we get:
sin(105°) = ±√[(1 - (-√3/2))/2]
Simplifying the expression inside the square root:
sin(105°) = ±√[(1 + √3/2)/2] = ±√[(2 + √3)/4]
sin(105°) = ±√(2 + √3)/2
Since 105° lies in the second quadrant, where sine is positive, we choose the positive square root:
sin(105°) = √(2 + √3)/2
This is the exact value of sin(105°). It is an irrational number, which means it cannot be expressed as a simple fraction. However, the expression √(2 + √3)/2 is a simplified form that clearly shows the value of sin(105°). This process demonstrates the power of the half-angle formula in finding exact values for trigonometric functions of angles that are not on the unit circle. The ability to manipulate trigonometric identities and apply them to specific problems is a crucial skill in mathematics. By breaking down the problem into smaller steps and carefully applying the appropriate formulas, we can arrive at the correct solution. Moreover, this example highlights the importance of understanding the signs of trigonometric functions in different quadrants. The quadrant in which an angle lies determines the sign of its sine, cosine, and tangent, which is essential for choosing the correct sign in the half-angle formulas. In summary, applying the half-angle formula to find sin(105°) involves recognizing the relationship between 105° and 210°, finding the value of cos 210°, substituting it into the formula, simplifying the expression, and choosing the correct sign based on the quadrant of 105°. This process not only provides the exact value of sin(105°) but also reinforces our understanding of trigonometric identities and their applications.
Simplifying the Answer Completely
We have found that sin(105°) = √(2 + √3)/2. To simplify this answer completely, we can try to express the nested square root √(2 + √3) in a simpler form. This often involves finding two numbers, a and b, such that √(2 + √3) = √a + √b. Squaring both sides, we get:
2 + √3 = (√a + √b)² = a + b + 2√(ab)
Equating the rational and irrational parts, we have:
a + b = 2
2√(ab) = √3, which implies ab = 3/4
Now we have a system of two equations with two variables:
a + b = 2
ab = 3/4
We can solve this system by substitution or elimination. Let's use substitution. From the first equation, we have b = 2 - a. Substituting this into the second equation, we get:
a(2 - a) = 3/4
2a - a² = 3/4
Multiplying by 4 to eliminate the fraction:
8a - 4a² = 3
Rearranging the equation:
4a² - 8a + 3 = 0
This is a quadratic equation in a. We can solve it using the quadratic formula or by factoring. Factoring the quadratic, we get:
(2a - 1)(2a - 3) = 0
So, the solutions for a are a = 1/2 and a = 3/2. If a = 1/2, then b = 2 - 1/2 = 3/2. If a = 3/2, then b = 2 - 3/2 = 1/2. In either case, we have a = 1/2 and b = 3/2 or vice versa. Therefore,
√(2 + √3) = √(1/2) + √(3/2) = √2/2 + √6/2 = (√2 + √6)/2
Now we can substitute this back into our expression for sin(105°):
sin(105°) = √(2 + √3)/2 = [(√2 + √6)/2]/2 = (√2 + √6)/4
This is the completely simplified form of sin(105°). It expresses the value in terms of simple square roots and fractions, which is often considered the most elegant form. The process of simplifying nested square roots can be challenging, but it is a valuable skill in mathematics. It often involves recognizing patterns and applying algebraic techniques to solve equations. In this case, we used a combination of substitution and factoring to find the values of a and b that allowed us to express √(2 + √3) in a simpler form. This example demonstrates the importance of perseverance and attention to detail in mathematical problem-solving. The final answer, (√2 + √6)/4, is not only simpler but also provides a clearer understanding of the value of sin(105°). It is a testament to the power of mathematical techniques in transforming complex expressions into more manageable and insightful forms. By mastering these techniques, students and professionals can tackle a wide range of mathematical problems with greater confidence and efficiency.
Conclusion
In this article, we successfully used the half-angle formula to find the exact value of sin(105°). We started by understanding the half-angle formulas and their derivation from the double-angle formulas. Then, we applied the formula sin(θ/2) = ±√[(1 - cos θ)/2] with θ = 210° to find sin(105°) = √(2 + √3)/2. Finally, we simplified the nested square root to obtain the completely simplified answer sin(105°) = (√2 + √6)/4. This process demonstrates the power and utility of trigonometric identities in solving mathematical problems. The half-angle formulas are essential tools in trigonometry, allowing us to find exact values for trigonometric functions of angles that are not standard angles on the unit circle. By mastering these formulas and the techniques for simplifying nested square roots, we can tackle a wide range of trigonometric problems with greater confidence and efficiency. The example of finding sin(105°) highlights the importance of breaking down complex problems into smaller, more manageable steps. It also underscores the value of perseverance and attention to detail in mathematical problem-solving. The ability to manipulate trigonometric identities and apply them to specific problems is a crucial skill in mathematics, and this article has provided a clear and comprehensive demonstration of how to do so. Furthermore, the simplification process showcased the elegance and power of algebraic techniques in transforming complex expressions into simpler, more insightful forms. The final answer, (√2 + √6)/4, not only provides the exact value of sin(105°) but also enhances our understanding of the relationships between different trigonometric functions and angles. In conclusion, the journey of finding and simplifying sin(105°) using the half-angle formula has been a valuable exercise in mathematical problem-solving. It has reinforced our understanding of trigonometric identities, algebraic techniques, and the importance of perseverance in the pursuit of mathematical knowledge. This knowledge is not only applicable to academic pursuits but also to various fields such as physics, engineering, and computer science, where trigonometry plays a crucial role.
