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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
人气:115 ℃ 时间:2019-08-22 12:51:37
解答
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).
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