It is known that the function f (x) is an even function on R, satisfying f (x) = - f (x + 1). When x ∈ [20112012], f (x) = x-2013, then () A. f(sinπ3)＞f(cosπ3)B. f（sin2）＞f（cos2）C. f(sinπ5)＜f(cosπ5)D. f（sin1）＜f（cos1）

It is known that the function f (x) is an even function on R, satisfying f (x) = - f (x + 1). When x ∈ [20112012], f (x) = x-2013, then () A. f(sinπ3)＞f(cosπ3)B. f（sin2）＞f（cos2）C. f(sinπ5)＜f(cosπ5)D. f（sin1）＜f（cos1）

From F (x) = - f (x + 1), we can get f (x) = f (x + 2), so the period of the function is 2. When x ∈ [20112012], the image shape is the same as that when x ∈ [- 1,0], but the left and right positions are different. Because when x ∈ [20112012], f (x) = x-2003, which is an increasing function, so f (x) is in [- 1,0]