化简√1-sin2θ
化简√1-sin2θ θ∈(3π/2,2π)
θ∈(3π/2,2π) sinθ0 √1-sin2θ =√[(sinθ)^2-2sinθcosθ+(cosθ)^2] =√(sinθ-cosθ)^2 =|sinθ-cosθ| =cosθ-sinθ
答:解:∵1-sin2=1-2sin1*cos1=(cos1)^2-2sin1*cos1+(sin1)^2 =(cos1-sin1)^2 ∵1≈53.7度.∴cos...详情>>
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