sin[α+(k+1)л]+sin[α-(k+1)л]/sin(α+kл)*cos(α-kл)=
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sin[α+(k+1)л]+sin[α-(k+1)л]/sin(α+kл)*cos(α-kл)=sin[α+(k+1)л]+sin[α-(k+1)л]/sin(α+kл)*cos(α-kл)=sin[
sin[α+(k+1)л]+sin[α-(k+1)л]/sin(α+kл)*cos(α-kл)=
sin[α+(k+1)л]+sin[α-(k+1)л]/sin(α+kл)*cos(α-kл)=
sin[α+(k+1)л]+sin[α-(k+1)л]/sin(α+kл)*cos(α-kл)=
sin[α+(k+1)л]+sin[α-(k+1)л]/sin(α+kл)*cos(α-kл)
k偶数
=[sin[α+л]+sin[α-л] ]/sinα*cosα
=2sin(α+л)/(sinacosa)=2sin(-a)/(sinacosa)=-2/cosa
k奇数
=[sinα+sinα]/sin(α+л)*cos(α-л)
=2sina/[sin(-a)*cos(л-α)]
=2/cosa
sin[α+(k+1)л]+sin[α-(k+1)л]/sin(α+kл)*cos(α-kл)=
sin(kπ+α)化简
化简 {sin[α+(k+1)π]+sin[α-(k+1)π]}/[sin(α+kπ)*cos(α-kπ)],k∈Z
化简[sin(kπ-α)*cos(kπ+α)]/{sin[(k+1)π+α]*cos[(k+1)π-α]}
化简sin(kπ-α)cos(kπ+α)/sin[(k+1)π+α]cos[(k+1)π+α]
已知sin^4α+cos^4α=1,求:sin^kα+cos^kα(k∈Z).
已知(2sin^2α+2sinαcosα)/(1+tanα)=k(0
求证:(sin(kπ-α)cos(kπ+α))/(sin((k+1)π+α)cos((k+1)π+α))=-1,k∈Z
sin(kπ-α)*cos〔(k-1)π-α〕/sin〔(k+1)π+α〕*cos(kπ+α) ,k属于Z
cosαcos[(2k+1)π]-sinαsin[(2k+1)π]为什么等于-cosα
已知 2sin(α)+2sin(α)cos(α)/1+tan(α)=k.试用k表示sin(α)-cos(α)的值.
已知(2sin^2 α+2sinα*cosα)/(1+tanα)=k 试用k表示sinα-cosα的值
(2sinαcosα+2sin^2(α))/(1+tanα)=k,用k表示sinα-cosα
化简:cos[(k+1)π-α]*sin(kπ-α)/cos(kπ+α)*sin[(k+1)π+α]拜托了各位
①利用公式sin(π-θ)=sinθ和sin(∏+θ)=-sinθ证明:sin(-θ)=-sinθ②证明tanθsinθ∕tanθ-sinθ=1+cosθ∕sinθ③已知sinα-2cosα+1=0,α≠kπ+π∕2,k∈z求:tan(3π-α)和1∕sin2α-sinαcosα+1的值
sinα
sinα
(1-cos-sin)(1-sin+cos)/sin^2α-sinα