求函數f（ θ ）=（sin θ -1）/（cos θ -2）的最大值與最小值！

熱心網友

因為：sinθ=2tan(θ/2)/[1+{tan(θ/2)}^2] cosθ=[1-{tan(θ/2)}^2]/[1+{tan(θ/2)}^2]所以: f(θ)=[2tan(θ/2)/[1+{tan(θ/2)}^2]-1]/[[1-{tan(θ/2)}^2]/[1+{tan(θ/2)}^2]-2]整理:f(θ)=[{tan(θ/2)}^2-2tan(θ/2)+1]/[3{tan(θ/2)}^2+1]設:f(θ)=T {tan(θ/2)}^2=A則:原式為 T=(A^2-2A+1)/(3A^2+1) 3T*A^2+T=A^2-2A+1 (3T-1)A^2+2A+(T-1)=0 判別式: 2^2-4(3T-1)(T-1)=0 T(3T-4)=<0 0=< T =<(4/3)所以: f(θ)最大值為 4/3 f(θ)最小值為 0 。

熱心網友

y=(sinT-1)/(cosT-2)---sinT-ycosT=1-2y---sin(T-A)=(1-2y)/√(1+y^2) (tanA=y)|sin(T-A)|=|1-2y|/√(1+y^2}=(1-2y)^2=3y^2-4y=0=ymin=0;ymax=4/3.