已知tanA=3,计算:(4sinA-2cosA)/(5cosA+3sinA)?(sinA-cosA)^2?
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已知tanA=3,计算:(4sinA-2cosA)/(5cosA+3sinA)?(sinA-cosA)^2?
已知tanA=3,计算:(4sinA-2cosA)/(5cosA+3sinA)?(sinA-cosA)^2?
已知tanA=3,计算:(4sinA-2cosA)/(5cosA+3sinA)?(sinA-cosA)^2?
(4sinA-2cosA)/(5cosA+3sinA) =(4tanA-2)/(5+3tanA) =(4*3-2)/(5+3*3) =5/7 (sinA-cosA)^2 =1-2sinAcosA =1-sin2A =1-(2tanA/(1+tanA)) =1-2*3/(1+9) =2/5