Question

函数f(x)=cos^4x-2sinxcosx-sin^4x,x属于[0,π/2],求f(x)的最大值,最小值

函数f(x)=cos^4x-2sinxcosx-sin^4x,x属于[0,π/2],求f(x)的最大值,最小值,求f(x)的单调区间。
2。若π/4小于等于v=x小于等于π/3，求函数y={sin(x+π/6)}/cosx的值域。

Answer 1

f(x)=(cosx)^4-(sinx)^4-2sinxcosx
=[(cosx)^2-(sinx)^2]*[(cosx)^2+(sinx)^2]-2sinxcosx
=(cosx)^2-(sinx)^2-2sinxcosx
=cos2x-sin2x
=√2cos(2x+π/4)
x∈[0,π/2]
则2x+π/4∈[π/4,5π/4]

最大值1 此时2x+π/4=π/4 x=0
最小值-√2 此时2x+π/4=π x=3π/8