Question

函数y=sin^4x+cos²x的最小正周期为

求详解

Answer 1

y=sin^4 x+cos^2 x
=sin^4 x+1-sin^2 x
=(sin^2 x-1/2)²+3/4
=-1/4(cos2x)^2+3/4
=-1/8(2cos2x)^2+1/8-1/8+3/4
=-(cos4x)/8+5/8

所以周期为π/2

Answer 2

y=sin^4x+cos²x
=sin^4x+1-sin²x
=(sin²x-1/2)²+3/4
=-1/2(1-2sin²x)²+3/4
=-1/2cos²2x+3/4
=-1/4(2cos²2x)+3/4
=-1/4(2cos²2x-1)+1
=-1/4cos4x+1
所以T=~~~