f(x)=(cosx)^4-2sinxcosx-(sinx)^4
=[(cosx)^2+(sinx)^2]*[(cosx)^2-(sinx)^2]-sin2x
=cos2x-sin2x
=√2cos(2x+π/4)
所以T=2π/2=π
x∈[0,2π]
2x+π/4∈[π/4,17π/4]
所以f(x)的最大值是√2,最小值是-√2
如果不懂,请Hi我,祝学习愉快!
f(x)=(cosx)^4-2sinxcosx-(sinx)^4
=[(cosx)^2+(sinx)^2]*[(cosx)^2-(sinx)^2]-sin2x
=cos2x-sin2x
=√2cos(2x+π/4)
所以T=2π/2=π
x∈[0,2π]
2x+π/4∈[π/4,17π/4]
所以f(x)的最大值是√2,最小值是-√2
如果不懂,请Hi我,祝学习愉快!