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)]
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![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)]](/uploads/image/z/3710823-15-3.jpg?t=1.%E5%B7%B2%E7%9F%A5cos%28%CF%80%2F6-%CE%B1%29%3D1%2F3%2C%E6%B1%82sin%282%2F3%CF%80-%CE%B1%29+2.%E6%B1%82%E8%AF%81tan%5E2%CE%B1-sin%5E2%CE%B1%3Dtan%5E2%CE%B1-sin%5E2%CE%B13.%E8%8B%A5sinx%3D%28m-3%29%2F%28m%2B5%29%2Ccosx%3D%284-2m%29%2F%28m%2B5%29%2Cx%E5%B1%9E%E4%BA%8E%28%CF%80%2F2%2C%CF%80%29%2C%E6%B1%82tanx4.%E5%8C%96%E7%AE%80%5Btan%28%CF%80-%CE%B1%29%2Asin%5E2%28%CE%B1%2B%CF%80%2F2%29%2Acos%282%CF%80-%CE%B1%29%5D%2F%5Bcos%5E3%28-%CE%B1-%CF%80%29%2Atan%28%CE%B1-2%CF%80%29%5D)
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^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)]
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[(2/3)π-α]
sin(2π/3-α)=sin[π/2+(π/6-α)]=cos(π/6-α)=1/3
3.若sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5),x属于(π/2,π),求tanx
tanx=sinx/cosx=[(m-3)/(m+5)]/[(4-2m)/(m+5)]=(m-3)/(4-2m)
4.化简[tan(π-α)*sin^2(α+π/2)*cos(2π-α)]/[cos^3(-α-π)*tan(α-2π)]
原式=(-tanαsin²αcosα)/(cos³αtanα)=sin²α/cos²α=tan²α
第2题不用证,两边本来就相等.