y=cosx平方乘以sinx平方的周期是多少 要过程
\u51fd\u6570y=cosx\u7684\u5e73\u65b9\u4e58\u4ee5sinx\u7684\u5e73\u65b9\u7684\u6700\u5c0f\u6b63\u5468\u671f\u662f\u591a\u5c11\uff1fy=(cosx)(sinx)=(sinxcosx)=((1/2)sin2x) =(1/4)(sin2x)=(1/4)(1-cos4x)/2=1/8-(1/8)cos4x \u2234\u6700\u5c0f\u6b63\u5468\u671f\u4e3aT=2\u03c0/4=\u03c0/2
y=(cosx+sinx)^2
= (cosx)^2+(sinx)^2+2cosxsinx
= 1+sin2x
\u5468\u671f pi
T=2π/4=π/2
y=cosx平方乘以sinx平方的周期是π/2
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