answer:From the angle addition formula, we have: [ cos 75^circ = cos (60^circ + 15^circ) ] Applying the formula: [ cos (a + b) = cos a cos b - sin a sin b ] where ( a = 60^circ ) and ( b = 15^circ ). We know: [ cos 60^circ = frac{1}{2}, quad sin 60^circ = frac{sqrt{3}}{2}, quad cos 15^circ = frac{sqrt{6} + sqrt{2}}{4}, quad sin 15^circ = frac{sqrt{6} - sqrt{2}}{4} ] Substituting these values, we get: [ cos 75^circ = left(frac{1}{2}right)left(frac{sqrt{6} + sqrt{2}}{4}right) - left(frac{sqrt{3}}{2}right)left(frac{sqrt{6} - sqrt{2}}{4}right) ] [ cos 75^circ = frac{sqrt{6} + sqrt{2}}{8} - frac{sqrt{18} - sqrt{6}}{8} ] [ cos 75^circ = frac{sqrt{6} + sqrt{2}}{8} - frac{3sqrt{2} - sqrt{6}}{8} ] [ cos 75^circ = frac{sqrt{6} + sqrt{2} - 3sqrt{2} + sqrt{6}}{8} ] [ cos 75^circ = frac{2sqrt{6} - 2sqrt{2}}{8} = frac{sqrt{6} - sqrt{2}}{4} ] Thus, we have: [ boxed{frac{sqrt{6} - sqrt{2}}{4}} ]

answer:To solve the limit problem, we need to calculate: lim_{x rightarrow 2} left(frac{sin (3 pi x)}{sin (pi x)}right)^{sin^2(x-2)} Let's proceed step-by-step: 1. **Trigonometric Transformation**: Rewrite (sin(3 pi x)) using the angle sum identity: [ sin(3 pi x) = sin(2 pi x + pi x) ] Using the sum of angles formula for sine: [ sin(2 pi x + pi x) = sin(2 pi x) cos(pi x) + cos(2 pi x) sin(pi x) ] 2. **Substitute in the Limit Expression**: Substitute the expanded form back into the limit: [ lim_{x rightarrow 2} left(frac{sin(2 pi x + pi x)}{sin(pi x)}right)^{sin^2(x-2)} = lim_{x rightarrow 2} left( frac{sin (2 pi x) cos (pi x) + cos (2 pi x) sin (pi x)}{sin (pi x)} right)^{sin^2(x-2)} ] 3. **Factor Out (sin(pi x))**: Observe that (sin(pi x)) is common in the numerator, leading to: [ frac{sin(2 pi x) cos(pi x) + cos(2 pi x) sin(pi x)}{sin(pi x)} = frac{sin(pi x) (2 cos(pi x) cos(pi x)) + cos(2 pi x) sin(pi x)}{sin(pi x)} ] Simplify the fraction: [ = frac{sin(pi x) (2 cos^2(pi x) + cos(2 pi x))}{sin(pi x)} = 2 cos^2(pi x) + cos(2 pi x) ] 4. **Evaluation as (x rightarrow 2)**: First note that as (x rightarrow 2), (pi x rightarrow 2pi), [ 2cos^2(pi x) + cos(2 pi x) = 2cos^2(2pi) + cos(4pi) ] Since (cos(2pi) = cos(4pi) = 1), we have: [ 2 cos^2(2pi) + cos(4pi) = 2 cdot 1^2 + 1 = 3 ] 5. **Evaluate the Exponent**: As (x rightarrow 2), ( sin^2(x-2) rightarrow sin^2(0) = 0). 6. **Combine Terms to Find the Limit**: Therefore, the given limit can now be evaluated as follows: [ lim_{x rightarrow 2} left(3right)^{sin^2(x-2)} = 3^{0} = 1 ] # Conclusion: [ boxed{1} ]

answer:First, we proceed by using a factoring trick similar to Simon's. We can rewrite the equation by adding a constant to both sides, specifically 56, which is the product of the coefficients of a and b: ab - 8a + 7b + 56 = 627 Now, this can be factored as: (a+7)(b-8) = 627 To find pairs (a, b) that satisfy this, we look at the factors of 627. The prime factorization of 627 is 3 times 11 times 19. From this, we derive the factor pairs: - (1, 627) - (3, 209) - (11, 57) - (19, 33) Substituting back into the expressions for a and b: - From (a+7, b-8) = (1, 627), we get (a, b) = (-6, 635), which is invalid as a must be positive. - From (a+7, b-8) = (3, 209), we get (a, b) = (-4, 217), which is again invalid. - From (a+7, b-8) = (11, 57), we get (a, b) = (4, 65). - From (a+7, b-8) = (19, 33), we get (a, b) = (12, 41). We now find the absolute differences: - |4 - 65| = 61 - |12 - 41| = 29 Thus, the minimal possible value of |a - b| is boxed{29}.

answer:Solving (x^2 - 5x + 6 = 0) yields the set (A). According to (A cup B = A), it can be inferred that (B subseteq A). Considering (B = {2}), (B = {3}), and (B = emptyset) as three possible scenarios, we can find the corresponding values of (a) for each case, thus forming the set (C) composed of real numbers. This problem examines the parameter issues within set operations, specifically focusing on the condition (A cup B = A) which implies (B subseteq A). The key to solving this problem is recognizing that (B) can be a singleton set or an empty set. A common oversight is to neglect the possibility of (B = emptyset), leading to the incorrect conclusion that (C = {2, 3}). Therefore, the correct answer is boxed{C = {2, 3}}.