举报
ddnsd
解:
根号下1-cosa=√2丨sin(a/2)丨
1+sinA+cosA除以sinA/2+cosA/2
=(1+sinA+cosA)/(sinA/2+cosA/2)
={[sin(A/2)]^2+[cos(A/2)]^2+2sin(A/2)cos(A/2)+[cos(A/2)]^2-[sin(A/2)]^2}/(sinA/2+cosA/2)
={2[cos(A/2)]^2+2sin(A/2)cos(A/2)}/(sinA/2+cosA/2)
=2cos(A/2)[cos(A/2)+sin(A/2)]/(sinA/2+cosA/2)
=2cos(A/2)