1+sinA+cosA除以sinA/2+cosA/2值
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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
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
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