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)=1/33.若sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5),x属于(π/2,π),求tanxtanx=sinx/cosx=[(m-3)/(m+5)]/[(4-2m)/(m+5)]=(m-3)/...打错了..是tan^2α-sin^2α=tan^2α*sin^2α抱歉证明 tan²α-sin²α=tan²αsin²α证明：左边＝sin²α/cos²α-sin²α=sin²α(1-cos²α)/cos²α=sin²αsin²α/cos²α=tan²αsin²α=右边