Question

若sinα是方程6x=1-√x的根,试求tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α)的值

若sinα是方程6x=1-√x的根,试求tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α)的值

Answer 1

化简：tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α) = -tanα
解根号方程：6x+√x-1=0
x=1/3（舍去负根）
所以：sinα = 1/3
因α没有限制，所以α可能锐角可能钝角
若α锐角
tanα = √2/4
tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α) = -√2/4
若α钝角
tanα = -√2/4
tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α) = √2/4

Answer 2

若sinα是方程6x=1-√x的根,试求tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α)的值
6x=1-√x
1-6x=√x
36x^2-12x+1=x
36x^2-13x+1=0
(9x-1)(4x-1)=0
x=1/9,x=1/4
tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α)
=-tana*tana(-cosa)/cosa
=tan^2a=sec^2a-1=1/(1-sin^2a)-1=1/80或1/15

Answer 3

6x=1-√x的根为x=1/9, 即sinα=1/9, cosα=±4*根号5/9

tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α)=tan2α×(-cosα）/sinα=-sinα/cosα=±根号5/20