1-cos2θ

易溥泥679求证1 - cos2θ/1+cos2θ=tan^2θ
厉储勇17244106118 ______ (1-cos2θ)/(1+cos2θ)=2sin²θ/(2cos²θ)=tan²θ

易溥泥679sinθ=根号下1 - cos^2θ,还是正负根号下1 - cos^2θ?根号下1+2sin(π - a)cos(π+a)的结果一定是正的吧 -
厉储勇17244106118 ______[答案] 取正负取决于角θ的范围啊,看它在哪一象限中阿、一二象限取正,三四象限取负. 根号下1+2sin(π-a)cos(π+a)=根号下1+2sin a (-cos a)=|sin a-cos a |.当sin a >cosa时答案等于sin a-cos a,当sin a

易溥泥6791+cos2θ+2sin²θ=2 1 - cos2θ/1+cos2θ=tan²θ -
厉储勇17244106118 ______[答案] 1+cos2θ+2sin²θ=1+cos2θ+1-(1-2sin²θ)=2+cos2θ-cos2θ=2 (1-cos2θ)/(1+cos2θ)=(1-(1-2sin²θ))/(1+(2cos²θ-1))=2sin²θ/(2cos²θ)=tan²θ

易溥泥679证明1 - 2cos^2θ/tanθ - cotθ=sinθcosθ -
厉储勇17244106118 ______[答案] 左边=(1-cos^2(θ))/(sinθ/cosθ-cosθ/sinθ)=(sin^2(θ)-cos^2(θ))/((sin^2(θ)-cos^2(θ)/(sinθcosθ))=1/(1/(sinθcosθ))=sinθcosθ

易溥泥679已知a是第三象限角,化简(cosθ/√1 - cos^2θ)+(sinθ√1 - sin^2θ/1 - cos^θ) -
厉储勇17244106118 ______[答案] ∵sinθ cosθ=1∴﹙sinθ cosθ﹚=1∴sinθ*cosθ=5/9 sin2θ=﹙2sinθcosθ﹚=4sinθcosθ=8/9 ∴sin2θ=﹙2根号2﹚/3

易溥泥679ρ=1 - cosθ化为直角坐标方程 -
厉储勇17244106118 ______[答案] ρ=1-cosθ两边乘以ρ得到ρ^2=ρ(1-cosθ=)=ρ-ρcosθ 由ρ^2=x^2+y^2,x=ρcosθ得到x^2+y^2=根号下(x^2+y^2)-x 化简得:x^4+y^4+2(x^2)y^2+2x^3+2xy^2-y^2=0

易溥泥679求sin^2θcosθ的最大值(θ为锐角) -
厉储勇17244106118 ______[答案] y=sin^2θcosθ=(1-cos^2θ)cosθ=cosθ-cos^3θ对y求导y'=-sinθ-[3cos^2θ*(-sinθ)]=-sinθ+3sinθcos^2θ令y'=0sinθ(3cos^2θ-1)=0sinθ=0 或者cos^2θ=1/3θ=kπ 或者 θ=arccos(±√3/3)θ=kπ 函数无极值θ=...

易溥泥679求两道高三数学题的解已知向量a=(1+cos2θ ,sin2θ) b=(1 - cos2θ sin2θ)(θ是第二象限的角) 求 a+b的绝对植 的取值范围数列An的前n项和为Sn 常数r不等于0和1 ... -
厉储勇17244106118 ______[答案] 1. a+b=(2,2sin2θ) |a+b|=2[(1+sin2θ*sin2θ)1/2] 90

易溥泥6791 - cosθ/(2sinθ/2) -
厉储勇17244106118 ______[答案] 1-cosθ=2*(sinθ/2)^2,因此原式的结果是sinθ/2

易溥泥679化简√(1 - cosθ) -
厉储勇17244106118 ______ cosθ=1-2[sin(θ/2)]^21-cosθ=2[sin(θ/2)]^2 √(1-cosθ)=√2 *|sin(θ/2)|