Question

已知函数f(x)=cos(2x-π/3）+2sin(x-π/4）sin(x+π/4)

Answer 1

解析：1)利用倍角和辅助角化为一角一函，f(x)=sin4x+cos4x=√2*sin(4x+π/4)
所以最小正周期t=2π/4=π/2.
2)∵0
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Answer 2

f(x)=cos(2x-π/3）+2sin(x-π/4）sin(x+π/4)
=cos2xcosπ/3+sin2xsinπ/3+2(sinxcosπ/4-cosxsinπ/4)(sinxcosπ/4+cosxsinπ/4)
可以这样打开，不用记太多公式，但计算复杂。