已知函数f(x)=sin2xcosfai-2cos^2xsin(x-fai)-cos(π/2+fai)(-π/2<fai<π/2)在x=π/6时取得最大值。
求fai的值将函数y=f(x)图象上各点的横坐标扩大到原来的2倍,纵坐标不变,得到函数y=g(x)的图象,若g(α)=1/3,α∈(-π/2,0),求cosα的值...
求fai的值
将函数y=f(x)图象上各点的横坐标扩大到原来的2倍,纵坐标不变,得到函数y=g(x)的图象,若g(α)=1/3,α∈(-π/2,0),求cosα的值 展开
将函数y=f(x)图象上各点的横坐标扩大到原来的2倍,纵坐标不变,得到函数y=g(x)的图象,若g(α)=1/3,α∈(-π/2,0),求cosα的值 展开
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题出错了sin(π-φ)-否则没法算
f(x)=sin2xcosφ-2cos^2xsin(π-φ)-cos(π/2+φ)(-π/2<φ<π/2)
=sin2xcosφ-sinφ(2cos^2x-1)
=sin2xcosφ-cos2xsinφ
=sin(2x-φ)
x=π/6时取得最大值(-π/2<φ<π/2)
π/3-φ=π/2+2kπ,k是整数
φ=-π/6
g(x)=sin(x+π/6)
cos(a)=sin(a+π/6)=1/3
α∈(-π/2,0)
α+π/6∈(-π/3,π/6)
cos(α+π/6)=√(1-1/9)=2√2/3
cosα=cos((α+π/6)-α)
=cos(α+π/6)cosπ/6+sin(α+π/6)sinπ/6
=(2√6+1)/6
望采纳,有问题请追问
f(x)=sin2xcosφ-2cos^2xsin(π-φ)-cos(π/2+φ)(-π/2<φ<π/2)
=sin2xcosφ-sinφ(2cos^2x-1)
=sin2xcosφ-cos2xsinφ
=sin(2x-φ)
x=π/6时取得最大值(-π/2<φ<π/2)
π/3-φ=π/2+2kπ,k是整数
φ=-π/6
g(x)=sin(x+π/6)
cos(a)=sin(a+π/6)=1/3
α∈(-π/2,0)
α+π/6∈(-π/3,π/6)
cos(α+π/6)=√(1-1/9)=2√2/3
cosα=cos((α+π/6)-α)
=cos(α+π/6)cosπ/6+sin(α+π/6)sinπ/6
=(2√6+1)/6
望采纳,有问题请追问
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