作业帮(根号下1-cosA^2)+根号下(1-sinA^2)=sinA-cosA,已知A属于[0,2π) 求A的取值范围a,[0,π/2] b,[π/2,π] c,[π,3π/2] d,[3π/2,2π）
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![(根号下1-cosA^2)+根号下(1-sinA^2)=sinA-cosA,已知A属于[0,2π) 求A的取值范围a,[0,π/2] b,[π/2,π] c,[π,3π/2] d,[3π/2,2π）](/uploads/image/z/3594928-40-8.jpg?t=%28%E6%A0%B9%E5%8F%B7%E4%B8%8B1-cosA%5E2%29%2B%E6%A0%B9%E5%8F%B7%E4%B8%8B%281-sinA%5E2%29%3DsinA-cosA%2C%E5%B7%B2%E7%9F%A5A%E5%B1%9E%E4%BA%8E%5B0%2C2%CF%80%29+%E6%B1%82A%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4a%2C%5B0%2C%CF%80%2F2%5D+b%2C%5B%CF%80%2F2%2C%CF%80%5D+c%2C%5B%CF%80%2C3%CF%80%2F2%5D+d%2C%5B3%CF%80%2F2%2C2%CF%80%EF%BC%89)
(根号下1-cosA^2)+根号下(1-sinA^2)=sinA-cosA,已知A属于[0,2π) 求A的取值范围a,[0,π/2] b,[π/2,π] c,[π,3π/2] d,[3π/2,2π）
(根号下1-cosA^2)+根号下(1-sinA^2)=sinA-cosA,已知A属于[0,2π) 求A的取值范围
a,[0,π/2] b,[π/2,π] c,[π,3π/2] d,[3π/2,2π）
(根号下1-cosA^2)+根号下(1-sinA^2)=sinA-cosA,已知A属于[0,2π) 求A的取值范围a,[0,π/2] b,[π/2,π] c,[π,3π/2] d,[3π/2,2π）
(根号下1-cosA^2)+根号下(1-sinA^2)=|sinA|+|cosA|=sinA-cosA
所以sinA>0,cosA