an=n^2(cos-sin),S30=
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∵数列{a[n]}的通项a[n]=n^2[(cosnπ/3)^2-(sinnπ/3)^2],前n项和为S[n]∴a[n]=n^2Cos(2nπ/3)∴S[n]=1^2(-1/2)+2^2(-1/2)+
建议你把括号标注明白一些:cosθ^2n是cosθ的2n次方,还是cos(θ^2n),还是cosθ…^2*
n为奇数时sin(nπ/2)^2=1偶数时为0n为奇数时1+cos(nπ/2)^2=0偶数时为2故n为奇数时an+2=0an+1=1n为偶数时an+2=2an故an=1(n为奇数)2^(n/2)(n为
|m+n|的平方=6+4(cosβ-sinβ)=128/25,cosβ-sinβ=-11/50cos(β+π/4)=根号2/2*cosβ-根号2/2*sinβ再用半角公式cos(2分之β+8分之π)即
(1)向量m+向量n=(√2+cosα-sinα,sinα+cosα)│向量m+向量n│=√[(√2+cosα-sinα)²+(sinα+cosα)²]=√[4+4sin(π/4-
lim(n→∞)[(sin(2/n)+cos(3/n))^(-n)]=lim(n→∞)[(sin(2/n)+1)^(-n)]=e^[lim(n→∞)(-n)ln(sin(2/n)+1)](等价无穷小替
是cos(2nπ/3)的意思吧? 晕死.懒得重新编辑了. 按照那个思路每一项乘以cosA-sinA.应该有更简单的方法. PS.连加的意思.就是以
an=n(cosnπ/3-sinnπ/3)=n^2*cos(2nπ/3)(二倍角公式)cos(2π/3)=-1/2cos(4π/3)=-1/2cos(6π/3)=1所以a(3k-2)+a(3k-1)+
(1)根据a(1)=1、a(2)=2及a(n+2)=[1-(1/3)cos²(nπ/2)]a(n)+2sin²(nπ/2)可得a(3)=[1-(1/3)cos²(π/2)
f(x)=(sinωx+cosωx))×(cosωx-sinωx)+cosωx2sinωx+t=cos2ωx+sin2ωx+t=根号2sin(2ωx+四分之π)+t根据相邻对称轴间的距离求出周期就能求
利用三角函数的积化和差公式,得到an=sin(n+1)cos(n-1)/n^p=[sin(2n)+sin2]/2n^p={sin(2n)/n^p+sin2/n^p}/2可证当0再问:确实是条件收敛,我
a(n+2)=[1+cos^2(nπ/2)]an+sin^2(nπ/2)]若n为偶数a(n+2)=2an若n为奇数a(n+2)=an+1∴a1,a3,a5……形成等差数列a2,a4.26……形成等比数
(1)f(x)=mn=(cosα+sinα)(cosα-sinα)+2√3sinαcosα=cos2α+√3sin2α=2[sin2α·cosπ/6+cos2α·sinπ/6]=2sin(2α+π/6
是否求S2013∵a1=1,a2=2a(n+2)=cos²(nπ/2)an+sin²(nπ/2)∴a3=cos²(π/2)a1+sin²(π/2)=1a5=co
(1)、an+2=[1+cos(nπ/2)]an+sin(nπ/2),n∈N因为cosnπ=2cos(nπ/2)-1=1-sin(nπ/2),所以an+2=[1+cos(nπ/2)]an+sin(nπ
(1)an=n^2cos2πn/3cos2πn/3取到的值为-1/2,-1/2,1,-1/2,-1/2,1,.对于n=3k(k∈N*),a(3k-2)+a(3k-1)+a(3k)=-1/2(3k-2)
sin^2/(sin-cos)-(sin+cos)/(tan^2-1)=sin^2/(sin-cos)-(sin+cos)/[(sin^2/cos^2)-1]=sin^2/(sin-cos)-(sin
an=n^2{cos^(n*180)/3-sin^(n*180)/3}=n^2*cos(2n*180)/3=n^2cos(120n)a1=1*cos120=-1/2a2=4cos240=-2a3=9c
∵an=n[cos(nπ)/3-sin(nπ)/3]=n×cos(2π/3)n∵n取1到n时,cos(2π/3)n的取值依次为-1/2,-1/2,1,-1/2,-1/2,1,……∴S10=-1/2×(
之前你出过这种题了吧,原来让求的是前30项.也不说清楚是从a0还是a1开始,不过不要紧a0=0;之前求的是S29,S30如下cos(nπ/3)^2-sin(nπ/3)^2=1-2sin(nπ/3)^2