已知函数fx=sin(2x+π/6)-cos(2x+π/3)+cos2x,①f(π/12)的值②函数fx单调递增区间③函数fx最大值和相应x的值

(1)f（x）=sin（2x+π/6）-cos（2x+π/3）+2cos²x
= sin2x cosπ/6+cos2x sinπ/6-[ cos2x cosπ/3- sin2x sinπ/3] +2cos²x
=√3 sin2x+2cos²x=√3 sin2x+1+cos2x
=2 sin(2x+π/6)+1,
f（π/12）=2sin(2*π/12+π/6)+1=√3+1.
（2）当2x+π/6 ∈[ -π/2+2kπ ,π/2+2kπ ] 函数fx单调递增
(3)f(x)的最大值是3,
此时2x+π/6=2kπ+π/2,x=kπ+π/6,k∈Z.再教我一条好不好。。你先采纳吧！然后发过来我看看！！！已知cos(α+β)=5/13,cos(π/2+β)=-3/5,α,β均为锐角，cosα的值拜托了。。？。。已知cos(α+β)=5/13,cos(π/2+β)=-3/5,α,β均为锐角，cosα的值cos(α+β)=5/13所以sin(α+β)=12/13cos(π/2+β)=-3/5 ∴sin β=3/5cos β=4/5cos a=cos [(a+β)-β]=cos(a+β)cosβ+sin(a+β) sinβ=5/13*4/5+12/13*3/5=56/65嗯嗯嗯谢啦