化简cos2(-α)-tan(2π+α)/sin(-α)
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化简cos2(-α)-tan(2π+α)/sin(-α)
化简cos2(-α)-tan(2π+α)/sin(-α)
化简cos2(-α)-tan(2π+α)/sin(-α)
=cos2a+tana/sina=cos2a+1/cosa=(cosacosa-sinasina+cosacosa+sinasina)/cosa=2cosa
化简cos2α/tan(π/4+α)
化简cos2(-α)-tan(2π+α)/sin(-α)
化简 2cos2α+sin2αtanα
cos2α/tan(α+π/4)=
化简[tan(π/4+α)cos2α]/[2cos^2(π/4-α)]
sin2(-α)+tan(2π+α)cos2(π+α)-sin(2π-α)cos(π+α)cos2(π+α)化简 求助各位大神!
若tan(π-α)=1/3,则cos2α/2sinαcosα+cos2α的值为
已知tan(π/4+α)=1/2 求(1)tanα的值 (2)sin2α-cos2(平方)α/1+cos2α
(sin2α-cos2α+1)/(1+tanα)=2sin2αcos2α 为什么
化简:sin2α/(1-cos2α)-1/tanα=
sin2α+2cos2α=-1 求tanα
tan(π/4+α)=k 求cos2α
化学cos2α/tan(π/4-a)如题
求证:若tan^2 θ=tan^2 α+sec^2 α.则cos2α-2cos2θ=1
已知tan(α+β)=-1,tan(α-β)=1/2,求cos2α/cos2β
已知:(tanα)^2-2(tanβ)^2=1,求证cos2β=2cos2α+1
已知(tanα)^2=2(tanβ)^2+1,求证:cos2β=2cos2α+1
已知tanαtanβ=√3/3,求证:(2-cos2α)(2-cos2β)=3