Question

sin²α+√3sinαcosα-2cos²α=0，α∈(π/6，5π/12)，求：(1)s

sin²α+√3sinαcosα-2cos²α=0，α∈(π/6，5π/12)，求：(1)sin(2α－π／3)的值。(2)cos2α的值··

Answer 1

sin²α+√3sinαcosα-2cos²α=0
1-cos²α+√3sinαcosα-2cos²α=0
1+√3sinαcosα-3cos²α=0
√3/2*sin2α-3(1+cos2α)/2+1=0
√3/2*sin2α-3/2-3/2cos2α+1=0
√3/2*sin2α-3/2cos2α-1/2=0
√3(1/2*sin2α-√3/2*cos2α)=1/2
√3(sin2αcosπ/3-cos2αsinπ/3)=1/2
√3sin(2α-π/3)=1/2
sin(2α-π/3)=1/(2√3)
sin(2α-π/3)=√3/6

α∈(π/6，5π/12)
2α∈(π/3，5π/6)
2α-π/3∈(0，π/2)
cos(2α-π/3)=√33/6

cos2α
=cos[(2α-π/3)+π/3]
=cos(2α-π/3)cosπ/3-sin(2α-π/3)sinπ/3
=√33/6*1/2-√3/6*√3/2
=(√33-3)/12

Answer 2

sin²α+√3sinαcosα-2cos²α=0
1-3cos²α+√3sinαcosα=0
1-3（1＋cos2α）/2+√3sinαcosα=0
√3/2sin2α-3/2cos2α=1/2
√3sin2α-3cos2α=1
1/2 sin2α-√3/2cos2α=1/[2√3]
cosπ/3sin2α-sinπ/3cos2α=1/[2√3]
即
sin（2α-π/3）=√3/6
又α∈(π/6，5π/12)，
2α∈(π/3，5π/6)，
2α-π/3∈(0，π/2)，
所以
cos（2α-π/3）= √33/6
所以
cos（2α）=cos【（2α-π/3）＋π/3】
=sin（2α-π/3）cosπ/3＋cos（2α-π/3）sinπ/3
=√3/6×1/2＋√33/6×√3/2
=【√3＋3√11】/12