1+sinA+cosA除以sinA/2+cosA/2值
来源:百度知道 编辑:UC知道 时间:2024/09/21 06:47:13
需要详细过程~~~!!!!
解:
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)
谢谢
2cosA/2
分母=(sinA/2)^2+(cosA/2)^2+2*sinA/2*cosA/2+(cosA/2)^2-(sinA/2)^2
看一下就知道再除以分子=sinA/2+cosA/2+cosA/2-sinA/2=2*cosA/2
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