已知函数f(x)=cos^2x-2sinxcosx-sin^2x求f(x)最小正周期.当x属于[0,π/2]时求f(x)最小值及取得最小值时
问题描述:
已知函数f(x)=cos^2x-2sinxcosx-sin^2x求f(x)最小正周期.当x属于[0,π/2]时求f(x)最小值及取得最小值时
x的集合
答
f(x)
=cos^2x-2sinxcosx-sin^2x
=cos2x-sin2x
=√2((1/√2)cos2x - (1/√2)sin2x)
=√2( sin(π/4-2x) )
最小正周期 = π
min f(x) at x= 3π/8
min f(x) = -√2第二个问题是求集合