（2014•安徽）设函数f（x）（x∈R）满足f（x+π）=f（x）+sinx．当0≤x＜π时，f（x）=0，则f（23π6）=（）A．12B．32C．0D．-12

Question

（2014•安徽）设函数f（x）（x∈R）满足f（x+π）=f（x）+sinx．当0≤x＜π时，f（x）=0，则f（(23π)/(6)）=（　　）

A．(1)/(2)
B．(√(3))/(2)
C．0
D．-(1)/(2)

Answer 1

∵函数f（x）（x∈R）满足f（x+π）=f（x）+sinx．当0≤x＜π时，f（x）=0，
∴f（(23π)/(6)）=f（π+(17π)/(6)）
=f（(17π)/(6)）+sin(17π)/(6)
=f（(11π)/(6)）+sin(11π)/(6)+sin(17π)/(6)
=f（(5π)/(6)）+sin(5π)/(6)+sin(11π)/(6)+sin(17π)/(6)
=sin(5π)/(6)+sin(11π)/(6)+sin(17π)/(6)
=(1)/(2)−(1)/(2)+(1)/(2)
=(1)/(2)．
故选：A．