Trigonometric Identities Expressing Cos 2A, Finding Sin 2A, Sin 2θ, Cos 2θ And Proving Cos 15°
In trigonometry, double angle formulas are fundamental identities that relate trigonometric functions of double angles to those of single angles. The cosine double angle formula has three common forms, and we aim to express cos 2A specifically in terms of sin A. The primary identity we will use is:
cos 2A = cos²A - sin²A
Our goal is to eliminate the cos²A term and express the entire identity using sin A. We can use the Pythagorean identity, which states:
sin²A + cos²A = 1
From this, we can express cos²A as:
cos²A = 1 - sin²A
Now, substitute this expression for cos²A back into the double angle formula:
cos 2A = (1 - sin²A) - sin²A
Simplify the equation by combining like terms:
cos 2A = 1 - 2sin²A
Therefore, we have successfully expressed cos 2A in terms of sin A. This form of the double angle formula is particularly useful when you know the value of sin A and need to find cos 2A without calculating cos A first. This identity finds frequent use in solving trigonometric equations and simplifying expressions. For instance, in problems involving integration or differentiation of trigonometric functions, converting cos 2A into this form can simplify the process significantly. Furthermore, in geometric applications, such as calculating areas or angles in triangles, this identity can provide a direct pathway to the solution when the sine of an angle is known. Understanding and applying this identity efficiently is a crucial skill in trigonometry and related fields.
This transformation highlights the interconnectedness of trigonometric functions and showcases how fundamental identities can be manipulated to suit specific problem-solving needs. By mastering such manipulations, one can develop a deeper understanding of trigonometric relationships and their applications in various mathematical and scientific contexts. The ability to express trigonometric functions in different forms is not just a theoretical exercise; it is a practical tool that enhances problem-solving capabilities across a wide range of disciplines.
To find sin 2A, we will use the double angle formula for sine, which is given by:
sin 2A = 2 sin A cos A
We are given that sin A = 3/4. To use the formula, we also need to find cos A. We can use the Pythagorean identity:
sin²A + cos²A = 1
Substitute the given value of sin A:
(3/4)² + cos²A = 1
9/16 + cos²A = 1
Now, solve for cos²A:
cos²A = 1 - 9/16
cos²A = 16/16 - 9/16
cos²A = 7/16
Take the square root of both sides to find cos A. Since we don't have information about the quadrant of angle A, we'll consider both positive and negative roots. However, for this example, let's assume A is in the first quadrant where cos A is positive:
cos A = √(7/16)
cos A = √7 / 4
Now that we have both sin A and cos A, we can substitute these values into the double angle formula:
sin 2A = 2 * (3/4) * (√7 / 4)
sin 2A = (6/4) * (√7 / 4)
sin 2A = (3/2) * (√7 / 4)
sin 2A = (3√7) / 8
Therefore, if sin A = 3/4, then sin 2A = (3√7) / 8. This result demonstrates the application of the double angle formula in conjunction with the Pythagorean identity to solve trigonometric problems. Understanding these relationships and their applications is crucial in various fields, including physics and engineering, where trigonometric functions are frequently used to model periodic phenomena. The ability to manipulate these identities and solve for unknown values is a fundamental skill in mathematical analysis and problem-solving.
This process highlights the importance of understanding the connections between different trigonometric identities and how they can be used together to solve problems. Mastering these techniques is essential for advanced studies in mathematics and related disciplines. Furthermore, this example serves as a practical illustration of how trigonometric functions and identities can be applied in real-world scenarios, reinforcing their relevance and utility.
This problem presents a unique challenge because the given values of cos θ and sin θ do not belong to the same angle. If cos θ = 12/13, then sin θ should be 5/13 (or -5/13), based on the Pythagorean identity. Similarly, if sin θ = 4/5, then cos θ should be 3/5 (or -3/5). The inconsistency suggests that there might be two different angles involved, let's call them θ and φ, such that cos θ = 12/13 and sin φ = 4/5. We'll proceed with this assumption and find sin 2θ and cos 2φ.
First, let's find sin 2θ using the double angle formula:
sin 2θ = 2 sin θ cos θ
We need to find sin θ when cos θ = 12/13. Using the Pythagorean identity:
sin²θ + cos²θ = 1
sin²θ + (12/13)² = 1
sin²θ + 144/169 = 1
sin²θ = 1 - 144/169
sin²θ = 25/169
sin θ = ± 5/13
Assuming θ is in the first quadrant, sin θ = 5/13. Now we can find sin 2θ:
sin 2θ = 2 * (5/13) * (12/13)
sin 2θ = 120/169
Next, let's find cos 2φ using the double angle formula:
cos 2φ = cos²φ - sin²φ
We have sin φ = 4/5. We need to find cos φ using the Pythagorean identity:
sin²φ + cos²φ = 1
(4/5)² + cos²φ = 1
16/25 + cos²φ = 1
cos²φ = 1 - 16/25
cos²φ = 9/25
cos φ = ± 3/5
Assuming φ is in the first quadrant, cos φ = 3/5. Now we can find cos 2φ:
cos 2φ = (3/5)² - (4/5)²
cos 2φ = 9/25 - 16/25
cos 2φ = -7/25
Therefore, sin 2θ = 120/169 and cos 2φ = -7/25. This solution highlights the importance of verifying the consistency of given trigonometric values and the flexibility required to adapt problem-solving approaches when inconsistencies arise. The application of trigonometric identities and the Pythagorean theorem remains central to the solution, but the recognition of the potential for multiple angles adds a layer of complexity that must be carefully addressed.
This example underscores the need for a thorough understanding of trigonometric principles and the ability to apply them in various contexts. The careful consideration of angles and the use of appropriate identities are crucial for accurate problem-solving in trigonometry and related fields. Furthermore, this problem illustrates the importance of critical thinking and attention to detail when working with trigonometric functions and identities.
To prove that cos 15° = (√3 + 1)/(2√2), we can use the half-angle formula for cosine. The half-angle formula is derived from the double angle formula and allows us to find trigonometric functions of half an angle when we know the trigonometric function of the full angle. The half-angle formula for cosine is:
cos(A/2) = ±√[(1 + cos A) / 2]
In this case, we want to find cos 15°, which is cos(30°/2). So, A = 30°. We are given that cos 30° = √3/2. Plugging this into the half-angle formula, we get:
cos 15° = cos(30°/2) = ±√[(1 + cos 30°) / 2]
Substitute the value of cos 30°:
cos 15° = ±√[(1 + √3/2) / 2]
To simplify, we first deal with the fraction inside the square root. Multiply the numerator and denominator by 2 to eliminate the fraction within the fraction:
cos 15° = ±√[(2 + √3) / 4]
Since 15° is in the first quadrant, cosine is positive, so we take the positive root:
cos 15° = √[(2 + √3) / 4]
Now, we can take the square root of the denominator:
cos 15° = √(2 + √3) / 2
To further simplify, we want to express the numerator in a form that eliminates the nested square root. We can do this by trying to express 2 + √3 as a perfect square. Suppose √(2 + √3) = a + b, where a and b are rational numbers. Squaring both sides, we get:
2 + √3 = (a + b)²
2 + √3 = a² + 2ab + b²
Equating the rational and irrational parts, we have:
a² + b² = 2
2ab = √3
From the second equation, ab = √3/2. Let's try to find a and b such that a = p/√2 and b = q/√2 for some integers p and q. Then:
(p/√2)(q/√2) = √3/2
pq/2 = √3/2
pq = √3
This approach doesn't seem to simplify easily. Instead, let's try to manipulate the expression √(2 + √3) directly. We want to find a form (x + y)² such that (x + y)² = 2 + √3. Multiplying the numerator and denominator of the original expression inside the square root by 2, we get:
cos 15° = √(2 + √3) / 2 = √(4 + 2√3) / (2√2)
Now, we can rewrite 4 + 2√3 as (√3 + 1)²:
4 + 2√3 = (√3)² + 2(√3)(1) + 1² = (√3 + 1)²
So, cos 15° becomes:
cos 15° = √[(√3 + 1)²] / (2√2)
cos 15° = (√3 + 1) / (2√2)
Thus, we have proven that cos 15° = (√3 + 1) / (2√2). This proof demonstrates the application of the half-angle formula and algebraic manipulation to find trigonometric values of specific angles. The ability to manipulate expressions and recognize patterns is crucial in solving trigonometric problems and in various areas of mathematics and physics.
This detailed solution highlights the importance of strategic problem-solving techniques and the ability to connect different mathematical concepts. The use of the half-angle formula, combined with algebraic manipulation and pattern recognition, is a powerful approach to solving complex trigonometric problems. Furthermore, this example underscores the importance of understanding the underlying principles of trigonometry and the ability to apply them in diverse contexts.