设θ∈(0,2π),点p(sinθ,cos^2-sin^2)在第三象限,则角θ的范围是
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![设θ∈(0,2π),点p(sinθ,cos^2-sin^2)在第三象限,则角θ的范围是](/uploads/image/z/5208140-20-0.jpg?t=%E8%AE%BE%CE%B8%E2%88%88%280%2C2%CF%80%29%2C%E7%82%B9p%28sin%CE%B8%2Ccos%5E2-sin%5E2%29%E5%9C%A8%E7%AC%AC%E4%B8%89%E8%B1%A1%E9%99%90%2C%E5%88%99%E8%A7%92%CE%B8%E7%9A%84%E8%8C%83%E5%9B%B4%E6%98%AF)
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设θ∈(0,2π),点p(sinθ,cos^2-sin^2)在第三象限,则角θ的范围是
设θ∈(0,2π),点p(sinθ,cos^2-sin^2)在第三象限,则角θ的范围是
设θ∈(0,2π),点p(sinθ,cos^2-sin^2)在第三象限,则角θ的范围是
sinθ
27的4次方根=27^(1/4)=3^(3/4)
f(3)=3^(3/4)
所以f(x)=x^(3/4)
所以f-1(x)=x^(4/3)
2xy(x2-y2)/(x2+y2)(x2-y2)
=(2x3y-2xy3)/(x2+y2)(x2-y2)
x(x2+y2)/(x2+y2)(x2-y2)
=(x3+xy2)/(x2+y2)(x2-y2)
∵点P(sinx,cos2x)在第三象限,∴sinx<0,且cos2x<0.由0<x<2π.可知,(一)当sinx<0时,π<x<2π.===>2π<2x<4π.结合cos2x<0.可知5π/2<2x<7π/2.===>5π/4<x<7π/4.综上可知,5π/4<x<7π/4.即x∈(5π/4,7π/4)