高中数学求化简.f(x)=[cos(α-π/2)sin(π/2+α)tan(-α+3π/2)]/cos(-π-α)
高中数学求化简.f(x)=[cos(α-π/2)sin(π/2+α)tan(-α+3π/2)]/cos(-π-α)
高中数学三角函数化简题已知 fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin
化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(
②若cos(已知α是第三象限角f(α)=sin(π-α)cos(2π-α)tan(-α-π)/tan(-α)sin(-π
f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简
设函数f(x)=[cot(-x-π)sin(2π+x)]/[cos(-x)tan(3π-x)].(1)若f(α)=(根3
已知tan(3π+α)=2,求:1、(sinα+cosα)²;2、sinα-cosα/2sinα+cosα
已知tan(π-α)=2,求sinα-2sinαcosα-cosα/4cosα-3sinα的值
已知tan(α+π/4)=2,则(2sinα+cosα)/(3cosα-2sinα)
已知tan(α+π/4)=2,求2cosα-sinα/cosα+3sinα
已知tan(α+π/4)=3,求cosα+2sinα/cosα-sinα的值
化简{sin(π+α)^3cos(-α)cos(π+α)}/{tan(π+α)^3cos(-π-α)^2}