已知tanx/2=根号5,求(1+sinx-cosx)/(1+sinx+cosx)的值!

问题描述:

已知tanx/2=根号5,求(1+sinx-cosx)/(1+sinx+cosx)的值!

(1+sinx-cosx)/(1+sinx+cosx)=(1+2sinx/2cosx/2-1+2sin^2x/2)/(1+2sinx/2cosx/2+2cos^x/2-1)=(2sinx/2cosx/2+2sin^2x/2)/(2sinx/2cosx/2+2cos^x/2)=(tanx/2+tan^2x/2)/(tanx/2+1)=tanx/2=根号5

sinx = 2tan(x/2)/(1+tanx/2tanx/2) = 2*sqrt(5)/(1+5) = sqrt(5)/3cosx = (1-tanx/2tanx/2)/(1+tanx/2tanx/2) = -2/3(1+sinx-cosx)/(1+sinx+cosx)= (1+ sqrt(5)/3+2/3)/(1+ sqrt(5)/3-2/3)= sqrt(5)...