已知函数f(x)=2cos(sinx-cosx)+1,x∈R
问题描述:
已知函数f(x)=2cos(sinx-cosx)+1,x∈R
①求函数的最小正周期;
②求函数在区间[π/8,3π/4]上的最值.
答
f(x)=2cosx(sinx-cosx)+1=2sinxcosx-2(cosx)^2+1 =sin2x-cos2x-1+1 =√2(√2/2*sin2x-√2/2*cos2x) =√2(sin2xcosπ/4-cos2xsinπ/4) =√2sin(2x-π/4) ①f(x)=√2sin(2x-π/4)∴函数f(x)的最小正周期:T=2π/2=π ②...