Question

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.

详细过程!^.^

Answer 1

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)
①
f(x)=√2sin(2x-π/4)
∴函数f(x)的最小正周期:
T=2π/2=π
②
∵x∈[π/8,3π/4]
∴当2x-π/4=π/2时
即:x=3π/8时
f(x)=√2

Answer 2

f(x)=2sinx*cosx+1-2cosx*cosx=sin2x-cos2x=根号2*sin(2x-π/4)
第一问：
2kπ-π/2<=2x-π/4<=2kπ+π/2
所以单调区间为：[kπ-π/8,kπ+3π/8]
第二问：
g(x)=根号2*sin(x+π/4)
其最大值=根号2，此时：x+π/4=2kπ+π/2，所以：x=2kπ+π/4