Question

求下列函数的最小正周期：（1）y=sin2xcos2x（2）y=cos平方2x-sin平方2x 怎么做啊！！

我要具体步骤

Answer 1

1>y=sin2xcos2x=sin4x/2 所以周期T=2π/4=π/22>y=cos平方2x-sin平脊册掘方樱核2x =cos4x 所以周期T=2π/4=π/姿灶2

Answer 2

(1)解：y=sin2xcos2x =(1/2)sin4x
得最小周期为π/2 单调增区间为猜橡[(kπ)/2-π/8，(kπ/2)+π/8] 单调减区间为[(kπ)/穗扮旁缺滚2+π/8，(kπ/2)+3π/8]；最大值为1/2 (2)解：呃..不好打..

Answer 3

y=sin2xcos2x=sin4x T=2pi/4

Answer 4

(1)y=1/2sin4x (2)y=cox4x