1.设等差数列{an}满足:[sin^2(a3)-cos^2(a3)+cos^2(a3)*cos^2(a6)-sin^2(a3)*sin^2(a6)]/sin(a4+a5)=1,公差d∈(-1,0).若当且仅当n=9时,数列{an}的前n项和Sn取得最大值,则首项a1的取值范围是()A.(7π/6,4π/3) B.(4π/3,3π
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![1.设等差数列{an}满足:[sin^2(a3)-cos^2(a3)+cos^2(a3)*cos^2(a6)-sin^2(a3)*sin^2(a6)]/sin(a4+a5)=1,公差d∈(-1,0).若当且仅当n=9时,数列{an}的前n项和Sn取得最大值,则首项a1的取值范围是()A.(7π/6,4π/3) B.(4π/3,3π](/uploads/image/z/8901600-24-0.jpg?t=1.%E8%AE%BE%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3%3A%5Bsin%5E2%28a3%29-cos%5E2%28a3%29%2Bcos%5E2%28a3%29%2Acos%5E2%28a6%29-sin%5E2%28a3%29%2Asin%5E2%28a6%29%5D%2Fsin%28a4%2Ba5%29%3D1%2C%E5%85%AC%E5%B7%AEd%E2%88%88%EF%BC%88-1%2C0%EF%BC%89.%E8%8B%A5%E5%BD%93%E4%B8%94%E4%BB%85%E5%BD%93n%3D9%E6%97%B6%2C%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E5%8F%96%E5%BE%97%E6%9C%80%E5%A4%A7%E5%80%BC%2C%E5%88%99%E9%A6%96%E9%A1%B9a1%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF%EF%BC%88%EF%BC%89A.%287%CF%80%2F6%2C4%CF%80%2F3%29+B.%284%CF%80%2F3%2C3%CF%80)
1.设等差数列{an}满足:[sin^2(a3)-cos^2(a3)+cos^2(a3)*cos^2(a6)-sin^2(a3)*sin^2(a6)]/sin(a4+a5)=1,公差d∈(-1,0).若当且仅当n=9时,数列{an}的前n项和Sn取得最大值,则首项a1的取值范围是()A.(7π/6,4π/3) B.(4π/3,3π
1.设等差数列{an}满足:[sin^2(a3)-cos^2(a3)+cos^2(a3)*cos^2(a6)-sin^2(a3)*sin^2(a6)]/sin(a4+a5)=1,公差d∈(-1,0).若当且仅当n=9时,数列{an}的前n项和Sn取得最大值,则首项a1的取值范围是()
A.(7π/6,4π/3) B.(4π/3,3π/2) C.[7π/6,4π/3] D.[4π/3,3π/2]
2.在△ABC中,角A、B、C的对边分别是a、b、c,则a>b是cosA
1.设等差数列{an}满足:[sin^2(a3)-cos^2(a3)+cos^2(a3)*cos^2(a6)-sin^2(a3)*sin^2(a6)]/sin(a4+a5)=1,公差d∈(-1,0).若当且仅当n=9时,数列{an}的前n项和Sn取得最大值,则首项a1的取值范围是()A.(7π/6,4π/3) B.(4π/3,3π
由题意可知,概述列为(1)2(2)4(3)8……
(1)a3=[1+(cos π/2)^2]*a1+(sin π/2)^2=a1+1=2
a4=[1+(cos π)^2]*a2+(sin π)^2=2*a2=4
(2)Sk=a1+a3+……+a2k-1=1+2+……+k=1/2*(1+k)*k=1/2(k^2+k)
(3) n=2n-1(奇数)时 an=(n+1)/2
n=2n(偶数)时 an=2^(n/2)
可先做第三小问,判断出an,再算第二小问.