将函数f(x)=sin ¼x•sin ¼ (x+2π)•sin ½
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/10 11:50:08
将函数f(x)=sin ¼x•sin ¼ (x+2π)•sin ½ (x+3π)-½cos²x/2
在区间(0,+∞)内的全部极值点按从小到大的顺序排成数列{an}(n∈N*)(1)求数列{an}的通项公式;(2)设bn=2nan,数列{bn}的前n项和为Tn,求Tn的表达式.
在区间(0,+∞)内的全部极值点按从小到大的顺序排成数列{an}(n∈N*)(1)求数列{an}的通项公式;(2)设bn=2nan,数列{bn}的前n项和为Tn,求Tn的表达式.
![将函数f(x)=sin ¼x•sin ¼ (x+2π)•sin ½](/uploads/image/z/16785698-50-8.jpg?t=%E5%B0%86%E5%87%BD%E6%95%B0f%28x%29%3Dsin+%26%23188%3Bx%26%238226%3Bsin+%26%23188%3B+%28x%2B2%CF%80%29%26%238226%3Bsin+%26%23189%3B)
(1)
f(x)=sin(x/4)•sin[(x+2π)/4]•sin [(x+3π)/2]-(1/2)cos²(x/2)
=sin(x/4)• [cos(x/4)] • [-cos(x/2)] -(1/2)cos²(x/2)
= -(1/2)sin(x/2)cos(x/2)-(1/2)cos²(x/2)
=-(1/4)sinx -(1/4)(cosx+1)
f'(x) = -(1/4)cosx + (1/4)sinx =0
tanx =1
x = kπ + π/4
an = nπ + π/4
(2)
bn = 2nan
= 2n(nπ + π/4)
Tn =b1+b2+...+bn
=2π[ (1/6)n(n+1)(2n+1) + n(n+1)/8 ]
=(π/12)n(n+1)[ 4(2n+1)+3 ]
=(π/12)n(n+1)(8n+7)
f(x)=sin(x/4)•sin[(x+2π)/4]•sin [(x+3π)/2]-(1/2)cos²(x/2)
=sin(x/4)• [cos(x/4)] • [-cos(x/2)] -(1/2)cos²(x/2)
= -(1/2)sin(x/2)cos(x/2)-(1/2)cos²(x/2)
=-(1/4)sinx -(1/4)(cosx+1)
f'(x) = -(1/4)cosx + (1/4)sinx =0
tanx =1
x = kπ + π/4
an = nπ + π/4
(2)
bn = 2nan
= 2n(nπ + π/4)
Tn =b1+b2+...+bn
=2π[ (1/6)n(n+1)(2n+1) + n(n+1)/8 ]
=(π/12)n(n+1)[ 4(2n+1)+3 ]
=(π/12)n(n+1)(8n+7)
已知函数f(x)=2sin(π-x)sin(π/2-x)
高中数学:已知函数f(x)=2sin(x+π/2).sin(x+7π/3)-
已知函数f(x)=sin(π-x)sin(π2-x)+cos2x
已知函数f(x)=2sinx*sin(π/2+x)-2sin^2x+1
已知函数f(x)=2√3sin²x-sin(2x-π/3)
已知函数f(x)=2根号3sin平方x-sin(2x-π/3)
已知函数f(x)=sin(2x+π/3)
已知函数f(x)=2sin(π-x)cosx
设函数f(x)=sin(2x+φ)(-π
已知函数f(x)=sin(π/2-x)+sinx
已知函数f(x)=sinx+sin(x+π/2) ,
函数f(x)=sin(2x+a) -π