函数y=sinA+cosA-4sinAcosA+1,且2sin^A+sin2A/1+tanA=k,π/4

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函数y=sinA+cosA-4sinAcosA+1,且2sin^A+sin2A/1+tanA=k,π/4函数y=sinA+cosA-4sinAcosA+1,且2sin^A+sin2A/1+tanA=k

函数y=sinA+cosA-4sinAcosA+1,且2sin^A+sin2A/1+tanA=k,π/4
函数y=sinA+cosA-4sinAcosA+1,且2sin^A+sin2A/1+tanA=k,π/4

函数y=sinA+cosA-4sinAcosA+1,且2sin^A+sin2A/1+tanA=k,π/4
令t=sinA+cosA,则t²=1+2sinAcosA,
2sin²A+sin2A/1+tanA=k,整理得2sinA(sinA+cosA)/[(sinA+cosA)/cosA]=k,
2sinAcosA=k,
那么t²=1+k,
π/40,
所以t=√(1+k),
y=√(1+k) -2k+1
(根号下面只有1+k)