题目

f(x)=3/2根号3sinx*cosx+cos^2(x-π/6)-3/2sin^2x+1/4 若f(X)=4/5(0＜θ＜π/2)求sin2θ​
f(x)=3/2根号3sinx*cosx+cos^2(x-π/6)-3/2sin^2x+1/4 若f(X)=4/5(0＜θ＜π/2)求sin2θ

提问时间：2021-03-31

答案

f(x)=3/2√3sinx*cosx+cos^2(x-π/6)-3/2sin^2x+1/4
=3√3/4sin2x+1/2(1+cos(2x-π/3)-3/4(1-cos2x)+1/4
=3√3/4sin2x+1/2cos(2x-π/3)+3/4cos2x
=3√3/4sin2x+1/4cos2x+√3/4sin2x+3/4cos2x
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2sin(2x+π/6)
f(θ)=2sin(2θ+π/6)=4/5
sin(2θ+π/6)=2/5
0＜θ＜π/2,π/6＜2θ+π/6＜7π/6
cos(2θ+π/6)=√21/5或cos(2θ+π/6)=-√21/5
sin2θ=sin[(2θ+π/6)-π/6]=√3/2sin(2θ+π/6)-1/2cos(2θ+π/6)
=√3/2*2/5-1/2*√21/5=(2√3-√21)/10或
sin2θ=√3/2*2/5+1/2*√21/5=(2√3+√21)/10
即sin2θ=(2√3-√21)/10或sin2θ=(2√3+√21)/10