解
tanA/2
=(sinA/2)/(cosA/2)
=(2sinA/2cosA/2)/(2cos²A/2)
=sinA/(2cos²A/2-1+1)
=sinA/(cosA+1)
tanA/2
=(2sin²A/2)/(2sinA/2cosA/2)
=[1-(1-2sin²A/2)]/(sinA)
=(1-cosA)/(sinA)
解
tanA/2
=(sinA/2)/(cosA/2)
=(2sinA/2cosA/2)/(2cos²A/2)
=sinA/(2cos²A/2-1+1)
=sinA/(cosA+1)
tanA/2
=(2sin²A/2)/(2sinA/2cosA/2)
=[1-(1-2sin²A/2)]/(sinA)
=(1-cosA)/(sinA)