Question

已知函数f（x）=cosx

证明（1）1/2[f²（π/4－x)＋f²(π/4＋x)]≥√f²（π/4－x）*f²（π/4＋x）

（2）4f（π/3－x）*f（x）*f（π/3+x）=f（3x）

Answer 1

1、1/2[f²（π/4－x)＋f²(π/4＋x)]
=[cos²(π/4－x)＋cos²(π/4＋x)]/2
=[((cos(π/2-2x)+1)/2+((cos(π/2+2x)+1)/2]/2
=[sin2x+1+(-sin2x+1)]/4
=2/4=1/2
f²（π/4－x）*f²（π/4＋x）=cos²(π/4－x)*cos²(π/4＋x)
=((cos(π/2-2x)+1)/2*((cos(π/2+2x)+1)/2=(1+sin2x)(1-sin2x)/4
=(1-sin²2x)/4=cos²2x/4<=1/4
所以 1/2[f²（π/4－x)＋f²(π/4＋x)]≥√f²（π/4－x）*f²（π/4＋x）
2、4f（π/3－x）*f（x）*f（π/3+x）
=4cos(π/3-x)cosxcos(π/3+x)
=4cosx(cos2x+cos2π/3)/2
=2cosxcos2x+2cos2π/3cosx
=(cos3x+cosx)+2*(-1/2)cosx
=cos3x
=f(3x)
证毕