Question

求解18题

Answer 1

　向量AC=[2sin(a-π/4)-1.cos(a-π/4)] , 向量BC=[2sin(a-π/4).cos(a-π/4)-1]
l向量ACl=l向量bCl
[2sin(a-π/4)-1]^2+[cos(a-π/4)]^2=[2sin(a-π/4)]^2+[cos(a-π/4)-1] ^2
4[sin(a-π/4)]^2-4sin(a-π/4)+1+[cos(a-π/4)]^2=4[sin(a-π/4)]^2+[cos(a-π/4)]^2
-2cos(a-π/4)+1
2sin(a-π/4)=cos(a-π/4)
tan(a-π/4)=1/2
2)a-π/4在(0,π/2)
tan(a-π/4)=sin(a-π/4)/cos(a-π/4)=1/2
cos(a-π/4)=2sin(a-π/4)
[cos(a-π/4)]^2+[sin(a-π/4)]^2=1
5sin(a-π/4)^2=1
sin(a-π/4)=√5/5
cos(a-π/4)=2√5/5
cosa=cos[(a-π/4)+π/4]=√2/2[cos(a-π/4)-sin(a-π/4)]
=√2/2[2√5/5-√5/5]
=√10/10
cosa=√10/10

追问

3Q～～