Simplification 2 / (COS ^ 2a-sin ^ 2a)

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• 1. If 3sina + cosa = 0, then 1 / sin ^ 2A + cos ^ 2A = a.10/3 B.5 / 3 C.2 / 3 d. - 2
• 2. Proving trigonometric function: cos ^ 2 (a) - cos (2a) * cos (4a) = sin ^ 2 (3a) Verification: cos ^ 2 (a) - cos (2a) * cos (4a) = sin ^ 2 (3a)
• 3. If SN: TN = (7n + 1): (4N + 27), find the ratio of a11: B11
• 4. The sum of the first n terms of the sequence {an} and {BN} is denoted as an and BN respectively. It is known that an = - n-3 / 2,4bn-12an = 13N (n ∈ natural number) (base n after a, B, a, b) ① Find the analytic expression of B base n about N and the general term formula of the sequence {B base n}. ② let C base n = (1 / 2) ∧ 2B base n-3a base n, prove that {C base n} is an equal ratio sequence, and find the first n term and C base n of the sequence {C base n} (thank you for the detailed process
• 5. (2010 Susong County three modules) arithmetic {an} series of the former n items and Sn, S9=-18, S13=-52, geometric ratio {bn}, b5=a5, b7=a7, the value of B15 is () A. 64B. -64C. 128D. -128
• 6. Given two arithmetic sequences {an}, {BN}, the sum of the first n terms is Sn, TN respectively, and Sn / TN = 7n + 2 / N + 3, then A7 / B8= The answer is 31 / 6
• 7. In the arithmetic sequence {an}, S9 = - 36, S13 = - 104, in the arithmetic sequence {BN}, B5 = A5, B7 = A7, then B6 = () A. ±42B. 42C. ±6D. 6
• 8. In the arithmetic sequence an, the sum of the first n terms is Sn, BN = 1 / Sn, B4 = 1 / 10, s6-s3 = 15. Find the general term of BN
• 9. Let the first term A1 = a not = 1 / 4 and an + 1 = 1 / 2An n be an even number or an + 1 / 4 n be an odd number, denote BN = a (2n-1) - 1 / 4, n = 1,2,3 Let A1 = a not = 1 / 4 and an + 1 = 1 / 2An n be even or an + 1 / 4 n be odd, denote BN = a (2n-1) - 1 / 4, n = 1, 3 Find the value of A2 and A3 Judge whether {BN} is equal ratio sequence and prove it
• 10. The sum of the first n terms of the sequence {an} is Sn, and an is the median of Sn and 1. The sequence {BN} satisfies B1 = A1, B4 = S3 (1) Finding the general term formula of sequence {an}, {BN} (2) Let CN = 1 / BN * B (n + 1) and the sum of the first n terms of the sequence {CN} be TN. it is proved that 1 / 3 ≤ TN
• 11. Given Sina + SINB = 1 / 3, find the range of square 2b of sina cos! Given sin (pie-a) = 1 / 2, calculate cos (3 pie + a)
• 12. 2*sin(60°-B)*sinB=cos(60°-2B)-cos60°
• 13. Sin α = A-3 / A + 5, cos α = 4-2a / A + 5, find Tan (α / 2)
• 14. Calculate 1 × 4 / 1 + 4 × 7 / 1 + 7 × 10 / 1 + 10 × 13 / 1 +... 2008 × 2010 / 1
• 15. 11×3+13×5+15×7+… +12009×2011．
• 16. Given that f (x) belongs to R for any x, f (- x) + F (x) = 0, f (x + 1) = f (x-1), then what is f (2011) equal to
• 17. Given K ∈ n *, let F: n * → n * satisfy: for any positive integer greater than k Given that K belongs to n *, let F: n * → n * satisfy that for any positive integer n greater than k, f (n) = n-k (1) Let k = 1, then the value of one of the functions f at n = 1 is? (2) Let k = 4, and when n ≤ 4, 2 ≤ f (n) ≤ 3, then the number of different functions f is? The problem implies that for a positive integer n less than or equal to K, its function value should also be a positive integer. I can't see how it is implied in the problem Question 2: I haven't learned the principle of step-by-step counting. Is there any other method
• 18. Given that the function f (x) defined on the positive integer set satisfies the condition: F (1) = 2F (2) = - 2F (n + 2) = f (n + 1) - f (n), then the value of F (2011) is?
• 19. About the function, just answer the fourth question
• 20. It is known that the two points of the equation 2x2-3x-5 = 0 are 5 / 2, - 1, then the image of the quadratic function y = 2x2-3x-5 and the two points of intersection of the X axis