sin²α+√3sinαcosα-2cos²α=0,α∈(π/6,5π/12),求:(1)sin(2α-π/3)的值.(2)cos2α的值··
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![sin²α+√3sinαcosα-2cos²α=0,α∈(π/6,5π/12),求:(1)sin(2α-π/3)的值.(2)cos2α的值··](/uploads/image/z/6997814-62-4.jpg?t=sin%26%23178%3B%CE%B1%2B%E2%88%9A3sin%CE%B1cos%CE%B1-2cos%26%23178%3B%CE%B1%3D0%2C%CE%B1%E2%88%88%28%CF%80%2F6%2C5%CF%80%2F12%29%2C%E6%B1%82%EF%BC%9A%281%29sin%282%CE%B1%EF%BC%8D%CF%80%EF%BC%8F3%29%E7%9A%84%E5%80%BC.%282%29cos2%CE%B1%E7%9A%84%E5%80%BC%C2%B7%C2%B7)
sin²α+√3sinαcosα-2cos²α=0,α∈(π/6,5π/12),求:(1)sin(2α-π/3)的值.(2)cos2α的值··
sin²α+√3sinαcosα-2cos²α=0,α∈(π/6,5π/12),求:(1)sin(2α-π/3)的值.(2)cos2α的值··
sin²α+√3sinαcosα-2cos²α=0,α∈(π/6,5π/12),求:(1)sin(2α-π/3)的值.(2)cos2α的值··
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
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/...
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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
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