已知sin(x+π/3)cos(x-π/3)+cos(x-π/3)sin(x-π/3)=-(2根号2/3)且π
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![已知sin(x+π/3)cos(x-π/3)+cos(x-π/3)sin(x-π/3)=-(2根号2/3)且π](/uploads/image/z/3003191-71-1.jpg?t=%E5%B7%B2%E7%9F%A5sin%28x%2B%CF%80%2F3%29cos%28x-%CF%80%2F3%29%2Bcos%28x-%CF%80%2F3%29sin%28x-%CF%80%2F3%29%3D-%282%E6%A0%B9%E5%8F%B72%2F3%29%E4%B8%94%CF%80)
已知sin(x+π/3)cos(x-π/3)+cos(x-π/3)sin(x-π/3)=-(2根号2/3)且π
已知sin(x+π/3)cos(x-π/3)+cos(x-π/3)sin(x-π/3)=-(2根号2/3)且π<2x<3π/2,求(1)tanx的值
(2)(sin2x-2cos^2x)/(1-tanx)
已知sin(x+π/3)cos(x-π/3)+cos(x-π/3)sin(x-π/3)=-(2根号2/3)且π
是sin(x+π/3)cos(x-π/3)+cos(x+π/3)sin(x-π/3)=-(2√2/3)吧?
如果是,则上式可化为sin2x=-2√2/3,
∵π<2x<3π/2,∴cos2x=-1/3,
(1) tanx=sin2x/(1+cos2x)=-√2.
(2) 2cos²x=1+cos2x=2/3,
∴原式=(-2√2/3-2/3)/(1-√2)=(6+4√2)/3.