求証:[tan(2π-α)cos(3π2-α)cos(6π-α)][sin(α+3π2)cos(α+3π2)]求証:[tan(2π-α)cos(3π2-α)cos(6π-α)][sin(α+3π2)cos(α+3π2)]=-tanα

求証:[tan(2π-α)cos(3π2-α)cos(6π-α)][sin(α+3π2)cos(α+3π2)]求証:[tan(2π-α)cos(3π2-α)cos(6π-α)][sin(α+3π2)cos(α+3π2)]=-tanα,第1張

求証:[tan(2π-α)cos(3π/2-α)cos(6π-α)]/[sin(α 3π/2)cos(α 3π/2)]
求証:[tan(2π-α)cos(3π/2-α)cos(6π-α)]/[sin(α 3π/2)cos(α 3π/2)]=-tanα

生活常識_百科知識_各類知識大全»求証:[tan(2π-α)cos(3π2-α)cos(6π-α)][sin(α+3π2)cos(α+3π2)]求証:[tan(2π-α)cos(3π2-α)cos(6π-α)][sin(α+3π2)cos(α+3π2)]=-tanα

0條評論

    發表評論

    提供最優質的資源集郃

    立即查看了解詳情