1.已知cos(π/6-α)=1/3,求sin(2/3π-α) 2.求证tan^2α-sin^2α=tan^2α-sin

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.