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(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
求証:[tan(2π-α)cos(3π/2-α)cos(6π-α)]/[sin(α 3π/2)cos(α 3π/2)]=-tanα
0條評論