化简f(x)=sin(2x+6/π)-sin(2x-6/π)-cos2x+1

6/π?通常是π/6啊,下面我当π/6来解吧：
f(x)=sin(2x+π/6)-sin(2x-π/6)-cos2x+1
=(sin2xcosπ/6+cos2xsinπ/6)-(sin2xcosπ/6-cos2xsinπ/6)-cos2x+1
=2cos2xsinπ/6-cos2x+1
=cos2x-cos2x+1
=1抱歉，一激动就把题目打得穿越了。补充了新的题目，不嫌弃的请重新回答一下，有理必采纳。对任意x∈R，恒有f(x)=1，是常数函数没有最小正周期【任何正实数都是它的正周期】对任意a∈R，x=a都是它的对称轴对任意a∈R，点P(a，0)都是它的对称中心没有严格单调递增区间。抱歉题目还是打错了，应该是：函数f(x)=sin(2x+π/6)+sin(2x-π/6)-cos2x+1，求f(x)的最小正周期、对称轴、对称中心、单调增区间。(x)=sin(2x+π/6)+sin(2x-π/6)-cos2x+1=(sin2xcosπ/6+cos2xsinπ/6)+(sin2xcosπ/6-cos2xsinπ/6)-cos2x+1=2(sin2x)cosπ/6-2cos2xsinπ/6+1=2sin(2x-π/6)+1最小正周期T=2π/2=πsin(2x-π/6)=1或-1，即2x-π/6=kπ+π/2，即x=kπ/2+π/3是对称轴，【k是整数，下同】sin(2x-π/6)=0，即2x-π/6=kπ，即x=kπ/2+π/12，点P(kπ/2+π/12，1)是对称中心sin(2x-π/6)=1，即2x-π/6=2kπ+π/2，x=kπ+π/3，T=π，单调递增区间为[kπ-2π/3，kπ+π/3]函数的最大值3，最小值-1，即值域[-1，3]