第3个回答 2015-10-02
一、公式法
1、
cos(2x)=1-2sin^2x
cos45°=1-sin^2(45°/2)
√2/2=1-sin^2(45°/2)
sin(45°/2)=(2-√2)/4
2、
sin(2x)=2sinx*cosx
sin(45°)=2sin(45°/2)*cos(45°/2)
(√2/2)^2=4sin^2(45°/2)*cos^2(45°/2)=4sin^2(45°/2)*[1-sin^2(45°/2)]
sin(45°/2)=(2-√2)/4
3、
tg(2x)=2tgx/(1-tg^2x)
2tg(45°/2)/[1+tg^2(45°/2)]=tg45°=1
tg^2(45°/2)+2tg(45°/2)-1=0
tg(45°/2)=√2-1
sin(45°/2)/cos(45°/2)=√2-1
[sin(45°/2)/cos(45°/2)]^2=(√2-1)^2
sin^2(45°/2)/(1-sin^2(45°/2)=3-2√2
sin(45°/2)=(2-√2)/4
二、几何法
RT△ABC
A(-1,0),B(0,1),C(0,0),∠BAC=∠ABC=45°
AC=BC=1,AB=√2
过A作∠BAC的平分线交BC于D,∠DAC=∠DAE=45°/2,AE=AC=1
设d>0,CD=d,D(0,d)
过D作DE垂直AB交AB于E,AE=AC=1
k(AB)=1,k(DE)=-1
AB:y=x+1......(1)
DE:y=-x+d.....(2)
(1),(2):
x=(d-1)/2,y=(d+1)/2
[(d-1)/2+1]^2+[(d+1)/2]^2=AE^2=1
CD=d=√2-1,AC=1
tg(45°/2)=√2-1
sin(45°/2)=(2-√2)/4