Question

求下列函数的周期 (1)y=3cosx ,x∈ R; (2)y=sin2x,x∈ R;

Answer 1

f(x)=f(x+T)
3cosx=3cos(x+T)对于任意x恒成立
3cosx=3cosxcosT-3sinxsinT
3cosx(1-cosT)+3sinxsinT=0
cosx(1-cosT)+sinxsinT=0
1-cosT=0,sinT=0
cosT=1,T=2kpai
sinT=0,T=kpai
T=2kpai,k=1,T=2pai
(2)y=sin2x
f(x)=f(x+T)
sin2x=sin2(x+T)
sin2x=sin(2x+2T)=sin2xcos2T+cos2xsin2T
sin2x(1-cos2T)-cos2xsin2T=0
1-cos2T=0,sin2T=0
cos2T=1,2T=2kpai,T=kpai
2T=kpai,T=kpai/2
T=kpai,k=1,T=pai