您的位置: 首页  >  作业答案  >  数学  >  1.已知cos(π/6-α)=1/3,求sin(2/3π-α) 2.求证tan^2α-sin^2α=tan^2α-sin^2α3.若sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5),x属于(π/2,π),求tanx4.化简[tan(π-α)*sin^2(α+π/2)*cos(2π-α)]/[cos^3(-α-π)*tan(α-2π)]

1.已知cos(π/6-α)=1/3,求sin(2/3π-α) 2.求证tan^2α-sin^2α=tan^2α-sin^2α3.若sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5),x属于(π/2,π),求tanx4.化简[tan(π-α)*sin^2(α+π/2)*cos(2π-α)]/[cos^3(-α-π)*tan(α-2π)]

分类: 作业答案 • 2021-12-18 16:38:31
问题描述：

1.已知cos(π/6-α)=1/3,求sin(2/3π-α) 2.求证tan^2α-sin^2α=tan^2α-sin^2α
3.若sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5),x属于(π/2,π),求tanx
4.化简[tan(π-α)*sin^2(α+π/2)*cos(2π-α)]/[cos^3(-α-π)*tan(α-2π)]

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