已知m向量=(sinx,2cosx),n向量=(2sinx,根号3sinx),函数f(x)=mn(1).求f(x)的最小正周期及减区间;(2).若f(α/2+π/4)=-3/5.α属于(2π/3,7π/6).求sinα的值
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![已知m向量=(sinx,2cosx),n向量=(2sinx,根号3sinx),函数f(x)=mn(1).求f(x)的最小正周期及减区间;(2).若f(α/2+π/4)=-3/5.α属于(2π/3,7π/6).求sinα的值](/uploads/image/z/1627508-20-8.jpg?t=%E5%B7%B2%E7%9F%A5m%E5%90%91%E9%87%8F%3D%28sinx%2C2cosx%29%2Cn%E5%90%91%E9%87%8F%3D%282sinx%2C%E6%A0%B9%E5%8F%B73sinx%29%2C%E5%87%BD%E6%95%B0f%28x%29%3Dmn%281%29.%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E5%8F%8A%E5%87%8F%E5%8C%BA%E9%97%B4%EF%BC%9B%EF%BC%882%EF%BC%89.%E8%8B%A5f%EF%BC%88%CE%B1%2F2%2B%CF%80%2F4%EF%BC%89%3D-3%2F5.%CE%B1%E5%B1%9E%E4%BA%8E%EF%BC%882%CF%80%2F3%2C7%CF%80%2F6%EF%BC%89.%E6%B1%82sin%CE%B1%E7%9A%84%E5%80%BC)
已知m向量=(sinx,2cosx),n向量=(2sinx,根号3sinx),函数f(x)=mn(1).求f(x)的最小正周期及减区间;(2).若f(α/2+π/4)=-3/5.α属于(2π/3,7π/6).求sinα的值
已知m向量=(sinx,2cosx),n向量=(2sinx,根号3sinx),函数f(x)=mn
(1).求f(x)的最小正周期及减区间;
(2).若f(α/2+π/4)=-3/5.α属于(2π/3,7π/6).求sinα的值
已知m向量=(sinx,2cosx),n向量=(2sinx,根号3sinx),函数f(x)=mn(1).求f(x)的最小正周期及减区间;(2).若f(α/2+π/4)=-3/5.α属于(2π/3,7π/6).求sinα的值
f(x)=向量m.向量n
f(x)=2sin^2x+2√3sinxcosx.
=1-cos2x+√3sin2x.
∴f(x)=2sin(2x-π/6)+1.
(1)函数f(x)的最小正周期T:T=2π/2,∴T=π.
令2kπ+π/2≤2x-π/6≤2kπ+3π/2,经过简单运算,得:
kπ+π/3≤x≤kπ+5π/6 k∈Z.---所求函数f(x)的单调递减区间.
(2)由f(x)=2sin(2x-π/6)+1,得:
f(α/2+π/4)=2sin{[2*(α/2)+π/4]-π/6}+1=-3/5 ,α∈(2π/3,7π/6).
2sin(α+π/4-π/6)+1=-3/5.
sin(α+π/12)=-4/5.
sinα=sin(α+π/12-π/12).
=sin(α+π/12)cosπ/12-cos(α+π/12)sinπ/12.
=sin(α+π/12)cos(π/3-π/4)-cos(α+π/12)sin(π/3-π/4).
=(-4/5)*√2/4(√3+1)-(-3/5)(√2/4)(√3-1).
∴sinα=(-√2/20)*(√3+7),(sinα≈ -0.6174).
f(x)=m*n=2(sinx)^2+2√3sinxcosx=1-cos(2x)+√3sin(2x)=1+2sin(2x-π/6) 。
(1)最小正周期为 T=2π/2=π ,
由 2kπ+π/2<=2x-π/6<=2kπ+3π/2 得 kπ+π/3<=x<=kπ+5π/6 ,
因此减区间为 [kπ+π/3,kπ+5π/6] ,k∈Z 。
(2)f(α/2+π/4)...
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f(x)=m*n=2(sinx)^2+2√3sinxcosx=1-cos(2x)+√3sin(2x)=1+2sin(2x-π/6) 。
(1)最小正周期为 T=2π/2=π ,
由 2kπ+π/2<=2x-π/6<=2kπ+3π/2 得 kπ+π/3<=x<=kπ+5π/6 ,
因此减区间为 [kπ+π/3,kπ+5π/6] ,k∈Z 。
(2)f(α/2+π/4)=1+2sin(α+π/3) = -3/5 ,因此 sin(α+π/3) = -4/5 ,
根据 α 范围可得 cos(α+π/3) = -√[1-(-4/5)^2] = -3/5 ,
所以 sinα=sin[(α+π/3)-π/3]
=sin(α+π/3)cos(π/3)-cos(α+π/3)sin(π/3)
=(-4/5)*(1/2)-(-3/5)*(√3/2)
=(3√3-4)/10 。
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