Question

急，请大佬帮忙求解

Answer 1

f(θ)=(R-2Rcos^2θ+Rcosθ)/sinθ
f'(θ)=[(4Rcosθsinθ-Rsinθ)sinθ-(R-2Rcos^2θ+Rcosθ)cosθ]/sin^2θ
=[4Rcosθsin^2θ-Rsin^2θ-Rcosθ+2Rcos^3θ-Rcos^2θ]/sin^2θ
=[2Rsinθsin2θ-R(sin^2θ+cos^2θ)+Rcosθ(2cos^2θ-1)]/sin^2θ
=R[2sinθsin2θ+cosθcos2θ-1]/sin^2θ
=R{2×1/2[cos(2θ-θ)-cos(2θ+θ)]+1/2[cos(2θ-θ)+cos(2θ+θ)]-1}/sin^2θ
=R{cosθ-cos3θ+1/2cosθ+1/2cos3θ}/sin^2θ
=R(cosθ-cos3θ)/(2sin^2θ)
=2Rsin[(θ+3θ)/2]sin[(θ-3θ)/2]/(2sin^2θ)
=-Rsin2θsinθ/sin^2θ
=-Rsin2θ/sinθ
=-2Rcosθ
f'(θ)>0
-2Rcosθ>0
2Rcosθ<0
cosθ<0 (R>0)
π/2+2kπ<θ<3π/2+2kπ
单调递增区间：(π/2+2kπ,3π/2+2kπ) (k∈Z)
单调递减区间：(-π/2+2kπ,π/2+2kπ) (k∈Z)

Answer 2

把sinx的平方换成1-cosx的平方来算算