Question

1-sinx=（π/2-1）*cosx如何得出tan（x/2）=(π-2)/2

Answer 1

(1-sinx)/cosx=π/2-1
((sin(x/2))^2+(cos(x/2))^2-2sin(x/2)cos(x/2)) / ((cos(x/2))^2-(sin(x/2))^2)
=(cos(x/2)-sin(x/2))^2 / ((cos(x/2))^2-(sin(x/2))^2)
=(cos(x/2)-sin(x/2)) / (cos(x/2)+sin(x/2))
=(1-tan(x/2))/(1+tan(x/2)) =π/2-1
所以 tan(x/2)=(4-π)/π

Answer 2

tanx=2tan(x/2)/(1-(tanx/2)^2)
cosx=2cos(x/2)^2-1=2/(1+tan(x/2)^2)-1=( 1-tan(x/2)^2)/(1+tanx/2)^2
sinx=2tan(x/2)/(1+(tan(x/2)^2)

1-sinx=[1-tan(x/2)]^2/(1+(tanx/2)^2)
(π/2-1)cosx=(π/2-1)[(1-tan(x/2)/(1+tan(x/2)^2)]
[1-tan(x/2)]^2=(π/2-1)(1-tan(x/2)
1-tan(x/2)=π/2-1
2-π/2=tan(x/2)