Question

已知函数f(x)=2sin(ωx+6/π)(ω＞0，x∈R)的最小正周期为π。若f(α)=2/3

α∈(0,π/8),求cos2α的值

Answer 1

解由函数f(x)=2sin(ωx+6/π)(ω＞0，x∈R)的最小正周期为π
知T=2π/ω=π
解得ω=2
故f(x)=2sin(2x+π/6)
由f(α)=2/3
知2sin(2a+π/6)=2/3
即sin(2a+π/6)=1/3
α∈(0,π/8)

知2a+π/6是锐角
故cos(2a+π/6)=2√2/3
故cos2a
=cos(2a+π/6-π/6)
=cos(2a+π/6)sinπ/6-sin(2a+π/6)cosπ/6
=2√2/3×1/2-1/3×√3/2
=(2√2-√3)/6

Answer 2

f(x)=2sin(ωx+π/6)
(ω＞0，x∈R)
最小正周期为π
ω=2π/π=2
f(x)=2sin(2x+π/6)
f(α)=2sin(2α+π/6)=2/3
sin(2α+π/6)=1/3
α∈(0,π/8),
2α+π/6∈(π/6，5π/12)
cos(2α+π/6)=✔【1-(1/3)²】=2✔2/3
cos2α=cos(2α+π/6-π/6)
=cos(2α+π/6)cosπ/6-sin(2α+π/6)sinπ/6
=2✔2/3×✔3/2-1/3×1/2
=(2✔6-1)/6