若tan(π+α)=m,3π/2<α<2π,则sin(π-α)=
若tanα=m 则sin(-5π-α)cos(3x+α)=
已知sinα=m-3/m+5,cosα=4-2m/m+5,若α∈(π/2,π),则tanα/2的值
已知tanα=2,sinα+cosα<0,求[sin(2π-α)*sin(π+α)*cos(-π+α)]/[sin(3π
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(1)确定tan(-3)/cos8×tan5的符号;(2)已知α∈(0,π),且sinα+cosα=m(0<m<1)
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sin(π-α)=-2/3 α∈(-π/2,0) 则tanα等于
1.已知cos(π/6-α)=1/3,求sin(2/3π-α) 2.求证tan^2α-sin^2α=tan^2α-sin
若f(α)=[sin(π-α)cos(2π-α)tan(-α+π)]/[-tan(-α-π)sin(-π-α)]化简
已知f(α)=sin(π-α)cos(2π-α)tan(-α+3π/2)tan(-α-π) /sin(-π-α).(1)
α是第三象限角,f(α)={sin(α-π/2)cos(3π/2+α)tan(π-α)}/{tan(-α-π)sin(-
②若cos(已知α是第三象限角f(α)=sin(π-α)cos(2π-α)tan(-α-π)/tan(-α)sin(-π