1/2-sin十二分之十九兀＋ tan22.5度/(1-tan22.5度的平方)

Question

如题所述

Answer 1

[sin(19π/12)]^2
=(1/2)[1 - cos(19π/6)]
=(1/2)[1 - cos(4π-19π/6)]
=(1/2)[1 - cos(5π/6)]
=(1/2)[1 + cos(π/6)]
=(1/2) +(1/2)(√3/2)
tan45°= 2tan22.5°/[1 -(tan22.5°)^2]
1=2tan22.5°/[1 -(tan22.5°)^2]
tan22.5°/[1 -(tan22.5°)^2]=1/2
ie
1/2 -[sin(19π/12)]^2 +tan22.5°/[1 -(tan22.5°)^2]
=1/2 -[(1/2) +(1/2)(√3/2)] +1/2
=1/2 -√3/4本回答被网友采纳