设θ∈(0,2π),点p(sinθ,cos^2-sin^2)在第三象限,则角θ的范围是
问题描述：
设θ∈(0,2π),点p(sinθ,cos^2-sin^2)在第三象限,则角θ的范围是
答
∵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)
答
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)
答
sinθ