answer:We start by recognizing that the given expression is in the form of the cosine of a sum of two angles. So, we can apply the cosine of a sum identity, which is cos(a + b) = cos(a)cos(b) - sin(a)sin(b). Applying this identity to the given expression, we get: cos frac{pi}{4} cos frac{pi}{12} - sin frac{pi}{4} sin frac{pi}{12} = cos left( frac{pi}{4} + frac{pi}{12} right) = cos frac{pi}{3} Now, we know that cos frac{pi}{3} = frac{1}{2}. So, the value of the given expression is boxed{frac{1}{2}}. Therefore, the correct answer is option A.

answer:Let us consider the sequence {a_n} defined by: [ a_1 = r text{ and } a_2 = s, ] where r and s are positive odd integers. For n > 0, a_{n+1} is the greatest odd divisor of a_n + a_{n-1}. 1. **Analysis of the recurrence relation**: Since a_n and a_{n-1} are both odd, their sum a_n + a_{n-1} is even. Let d be the greatest odd divisor of an even number a_n + a_{n-1}. This implies that: [ a_{n+1} leq frac{a_n + a_{n-1}}{2}. ] Given that d | a_{n} + a_{n-1}, and d is an odd divisor, it is also true that d < a_n + a_{n-1} unless a_n + a_{n-1} itself is an odd number. 2. **Decreasing nature**: The sequence {a_n + a_{n+1}} is eventually decreasing because: [ a_{n+2} leq frac{a_{n+1} + a_n}{2}, ] means {a_n + a_{n+1}} is redundantly bounded and decreasing. As {a_n} consists of positive integers, this sequence must stabilize to a constant after some term N. 3. **Constancy of the Sequence**: Assume for some N, the sequence becomes constant: a_{n+1} = a_n: [ a_{N+1} = text{gd}(a_N + a_{N-1}) implies a_{N} = text{gd}(a_N + a_N) ] Since the greatest odd divisor of an odd number times 2 is itself, thus: [ a_N = text{gd}(2a_N) = a_N. ] 4. **Constant Value Calculation**: The limiting value of the sequence {a_n} must be gcd(r, s) since the properties of gcd are preserved: - Let (g) be the gcd(r, s). If gcd remains constant through the operation, the sequence a_{n} converges to g for gcd(a_{n}, a_{n-1})= gcd(a_{n-1}, a_{n+1}) = g. 5. **Conclusion**: The sequence {a_n} eventually becomes constant and equal to gcd(r, s), since all terms a_{n+1} from some point onwards will be: [ a_{n+1} = gcd(r, s). ] (boxed{gcd(r, s)})

answer:**Analysis** This problem tests the operation of real numbers, including the calculation of cube roots, zero exponent, absolute values, and arithmetic square roots. To solve it, one should closely follow the rules of these calculations. **Solution** The original expression can be simplified as =-2+1+1=0. Therefore, the answer is boxed{0}.

answer:First, express the sum using the formula for the sum of a geometric series: [ 1 + 3 + 3^2 + cdots + 3^{1004} = frac{3^{1005} - 1}{3-1} = frac{3^{1005} - 1}{2} ] Next, since varphi(500) = 200, by Euler's theorem, 3^{200} equiv 1 pmod{500}. Therefore, 3^{1005} equiv 3^{200 cdot 5 + 5} equiv 3^5 pmod{500}. Calculating 3^5: [ 3^5 = 243 ] So, [ 3^{1005} equiv 243 pmod{500} ] Thus, [ frac{3^{1005} - 1}{2} equiv frac{243 - 1}{2} equiv frac{242}{2} equiv 121 pmod{500} ] So the remainder when 1 + 3 + 3^2 + cdots + 3^{1004} is divided by 500 is boxed{121}.