given that x=(Sin(2/5))^(-1),find the exact value of 1.(Cosx)^2 2.(Tanx)^2

问题描述:

given that x=(Sin(2/5))^(-1),find the exact value of 1.(Cosx)^2 2.(Tanx)^2

x=(Sin(2/5))^(-1)
就是说,arcsin2/5=x,所以,sinx=2/5
根据(sinx)^2+(cosx)^2=1
解出,cosx=(根号下21)/5
或者,cosx=-(根号下21)/5
1)所以,(cosx)^2=21/25
2)tanx)^2=(sinx/cosx)^2=4/21

1.(Cosx)^2
=19/25
2.(Tanx)^2
=4/19

(sinx)^2+(cosx)^2=1
来算