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Proving trigonometric function: cos ^ 2 (a) - cos (2a) * cos (4a) = sin ^ 2 (3a) Verification: cos ^ 2 (a) - cos (2a) * cos (4a) = sin ^ 2 (3a)
because
cos²a-cos2acos4a-sin²3a
=（1+cos2a）/2-cos2acos4a-（1-cos6a）/2
=cos2a/2-cos2acos4a+cos6a/2
=cos2a/2-cos2acos4a+（cos2acos4a-sin2asin4a）/2
=cos2a/2-（cos2acos4a+sin2asin4a）
=cos2a/2-cos（2a-4a）/2
=cos2a/2-cos（2a）/2
=0
therefore
cos^2a-cos2acos4a=sin^2 3a
The sum of the first n terms of the arithmetic sequence {an}, {BN} is represented by Sn and TN respectively. If Sn / TN = n / (n + 3), then the value of A4 / B5 is
Let the first term and tolerance of the arithmetic sequence an be a, D1 respectively, and let the first term and common ratio of the arithmetic sequence BN be B, d2sn = (2a + (n-1) D1) n / 2tn = (2B + (n-1) D2) n / 2Sn / TN = n / (n + 3) = > D1 = D2, D1 = 2A, B = 4A respectively, so A4 / B5 = (a + 3D1) / (B + 4D)
If the sum of the first n terms of two arithmetic sequences {an}, {BN} is an and BN respectively, and anbn = 7n + 14N + 27 (n ∈ n +), then the value of a11b11 is ()
A. 74B. 32C. 43D. 7871
∵ sequences {an}, {BN} are arithmetic sequences, and the sum of the first n terms is an and BN respectively. According to the properties of arithmetic sequences, A21 = (a1 + A21) × 212 = 21a11, B21 = (B1 + B21) × 212 = 21b11, ∵ anbn = 7n + 14N + 27 (n ∈ n +), ∵ a11b11 = 21a1121b11 = a21b21 = 7 × 21 + 14 × 21 + 27 = 4
If the sum of the first n terms of two arithmetic sequences {an}, {BN} is an and BN respectively, and anbn = 7n + 14N + 27 (n ∈ n +), then the value of a11b11 is ()
A. 74B. 32C. 43D. 7871
∵ sequence {an}, {BN} are arithmetic sequence, and the sum of the first n terms are an and BN respectively. According to the properties of arithmetic sequence, A21 = (a1 + A21) × 212 = 21a11, B21 = (B1 + B21) × 212 = 21b11, ∵ anbn = 7n + 14N + 27 (n ∈ n +), ∵ a11b11 = 21a1121b11 = a21b21 = 7 × 21 + 14 × 21 + 27 = 43
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