1.已知cos(π/6-α)=1/3,求sin(2/3π-α) 2.求證tan^2α-sin^2α=tan^2α-sin^2α
1.已知cos(π/6-α)=1/3,求sin(2/3π-α) 2.求證tan^2α-sin^2α=tan^2α-sin^2α
3.若sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5),x屬于(π/2,π),求tanx
4.化簡[tan(π-α)*sin^2(α+π/2)*cos(2π-α)]/[cos^3(-α-π)*tan(α-2π)]
3.若sinx=(m-3)/(m+5),cosx=(4-2m)/(m+5),x屬于(π/2,π),求tanx
4.化簡[tan(π-α)*sin^2(α+π/2)*cos(2π-α)]/[cos^3(-α-π)*tan(α-2π)]
數(shù)學人氣：515 ℃時間：2020-05-19 21:25:48
優(yōu)質解答
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屬于(π/2,π),求tanxtanx=sinx/cosx=[(m-3)/(m+5)]/[(4-2m)/(m+5)]=(m-3)/...打錯了..是tan^2α-sin^2α=tan^2α*sin^2α抱歉證明 tan²α-sin²α=tan²αsin²α證明：左邊＝sin²α/cos²α-sin²α=sin²α(1-cos²α)/cos²α=sin²αsin²α/cos²α=tan²αsin²α=右邊
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