设函数f(x)=sinx*(cosx-根号3sinx)(1)求函数f(x)在[0,π]上的单调递增区间,(2)若△ABC的内角A,B,C的对应边分别为a,b,c,且f(B)=0,a,b,根号3c成公差大于0的等差数列,求sinA/sinC的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 02:01:15
![设函数f(x)=sinx*(cosx-根号3sinx)(1)求函数f(x)在[0,π]上的单调递增区间,(2)若△ABC的内角A,B,C的对应边分别为a,b,c,且f(B)=0,a,b,根号3c成公差大于0的等差数列,求sinA/sinC的值](/uploads/image/z/5288229-45-9.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dsinx%2A%EF%BC%88cosx-%E6%A0%B9%E5%8F%B73sinx%EF%BC%89%EF%BC%881%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E5%9C%A8%5B0%2C%CF%80%5D%E4%B8%8A%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%2C%EF%BC%882%EF%BC%89%E8%8B%A5%E2%96%B3ABC%E7%9A%84%E5%86%85%E8%A7%92A%2CB%2CC%E7%9A%84%E5%AF%B9%E5%BA%94%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%2C%E4%B8%94f%EF%BC%88B%EF%BC%89%3D0%2Ca%2Cb%2C%E6%A0%B9%E5%8F%B73c%E6%88%90%E5%85%AC%E5%B7%AE%E5%A4%A7%E4%BA%8E0%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%B1%82sinA%2FsinC%E7%9A%84%E5%80%BC)
设函数f(x)=sinx*(cosx-根号3sinx)(1)求函数f(x)在[0,π]上的单调递增区间,(2)若△ABC的内角A,B,C的对应边分别为a,b,c,且f(B)=0,a,b,根号3c成公差大于0的等差数列,求sinA/sinC的值
设函数f(x)=sinx*(cosx-根号3sinx)
(1)求函数f(x)在[0,π]上的单调递增区间,
(2)若△ABC的内角A,B,C的对应边分别为a,b,c,且f(B)=0,a,b,根号3c成公差大于0的等差数列,求sinA/sinC的值
设函数f(x)=sinx*(cosx-根号3sinx)(1)求函数f(x)在[0,π]上的单调递增区间,(2)若△ABC的内角A,B,C的对应边分别为a,b,c,且f(B)=0,a,b,根号3c成公差大于0的等差数列,求sinA/sinC的值
(1)
f(x)=sinxcosx-√3sin²x
=(1/2)sin2x-(√3/2)(1-cox2x)
=(1/2)sin2x+(√3/2)cox2x-√3/2
=sin(2x+π/3)-√3/2
0≤x≤π
π/3≤2x+π/3≤2π+π/3
令2x+π/3=π/2==>x=π/12,为函数的一个最高点;
令2x+π/3=2π/2==>x=7π/12,为函数的一个最低点
单调增区间为:[0,π/12] ; [7π/12,π]
单调减区间为:[π/12 ,7π/12]
(2)
f(x)=sin(2x+π/3)-√3/2
sin(2B+π/3)=√3/2
2B+π/3=4π/3
B=π/2
a+√3c=2b
(a+√3c)²=4b ²=4a²+4c²
3a²+c²-2√3ac=0
(√3a-c)²=0
√3a-c =0
√3a=c
a/c=√3/3
sinA/sinC=√3/3
f(x)=sinx*(cosx-√3sinx)
=sinxcosx-√3sin²x.
=1/2sin2x-√3/2(1-cos2x)
=1/2sin2x+√3/2cos2x-√3/2
=sin(2x+π/3)-√3/2
若0≤x≤π 则 π/3≤2x+π/3≤7π/3
∴π/2≤ 2x+π/3≤3...
全部展开
f(x)=sinx*(cosx-√3sinx)
=sinxcosx-√3sin²x.
=1/2sin2x-√3/2(1-cos2x)
=1/2sin2x+√3/2cos2x-√3/2
=sin(2x+π/3)-√3/2
若0≤x≤π 则 π/3≤2x+π/3≤7π/3
∴π/2≤ 2x+π/3≤3π/2 正弦值递减
∴π/12≤x≤7π/12
∴f(x)在[0,派]上的单调递增区
间是[π/12,7π/12]
2
f(B)=0 ==> sin(2B+π/3)=√3/2
∵2B+π/3∈(π/3,7π/3)
∴2B+π/3=2π/3 ∴B=π/6
∵a,b,√3c成等差数列,公差大于0
∴2b=a+√3c, a ∴4b²=a²+3c²+2√3ac (*)
根据余弦定理
b²=a²+c²-2accosπ/6
∴ b²=a²+c²-√3ac 代入(*)
∴4(a²+c²-√3ac )=a²+3c²+2√3ac
∴3a²+c²-6√3ac=0
∴3(a/c)²-6√3(a/c)+1=0
解得:(∵a/c<1)
a/c=(3√3-2√6)/3
∴sinA/sinC=a/c=(3√3-2√6)/3
收起
f(x)=sinxcosx-√3sin^2x=1/2sin2x-√3/2cos2x+√3/2=sin(2x-π/3)+√3/2
剩下的自己算