函数f(x)=cos1/4xcos(π/2-x/4)cos(π-x/2)在区间(0.+∞)内全部的极值点按从小到大排成数列an,求通项

Question

函数f(x)=cos1/4xcos(π/2-x/4)cos(π-x/2)在区间(0.+∞)内全部的极值点按从小到大排成数列an,求通项

Answer 1

f(x)=cos(x/4)cos(π/2-x/4)cos(π-x/2)
f'(x)=(1/2)cos(x/4)cos(π/2-x/4)sin(π-x/2)+ (1/4)cos(x/4)cos(π-x/2) sin(π/2-x/4)
-(1/4)cos(π/2-x/4)cos(π-x/2)sin(x/4)
= (1/2)cos(x/4)sin(x/4)sin(x/2)- (1/4)cos(x/4)cos(x/2) cos(x/4) +(1/4)sin(x/4)cos(x/2)sin(x/4)
= (1/2)cos(x/4)sin(x/4)sin(x/2) -(1/4) cos(x/2)[cos(x/4)cos(x/4) -sin(x/4)sin(x/4)]
= (1/2)cos(x/4)sin(x/4)sin(x/2) -(1/4) cos(x/2)cos(x/2)
= (1/4)[sin(x/2)]^2-(1/4)[cos(x/2)]^2
= (-1/4)(cosx)
f'(x)=0
cosx=0
x= π/2,3π/2,5π/2,.
an= (2n-1)π/2