Question

高一三角函数题

Answer 1

解：
f(x)＝2√3sinxcosx＋2cos²x－1
＝√3(2sinxcosx)＋(2cos²x－1)
＝√3sin2x＋cos2x
＝2sin(2x＋π/6)
∵f(x0)＝6/5
∴2sin(2x0＋π/6)＝6/5
sin(2x0＋π/6)＝3/5
由x0∈[π/4，π/2]，得：2x0＋π/6∈[2π/3，7π/6]
∴cos(2x0＋π/6)＝－√[1－sin²(2x0＋π/6)]＝－4/5
∴cos2x0＝cos[(2x0＋π/6)－π/6]＝cos(2x0＋π/6)cosπ/6＋sin(2x0＋π/6)sinπ/6＝(3－4√3)/10

Answer 2

f(X)=√3sin2x+cos2x=2sin(2x+30度),因为f(x)=6/5,所以sin(2x+30度),=3/5,所以√3sin2x+cos2x=3/5,因为(sin2x)2+(cos2x)2=1,所以 cos2x=(3－4√3)/10