函数f(x)=sin^2(x+pi/12)+cos^2(x-pi/12)的最大值

函数f(x)=sin^2(x+pi/12)+cos^2(x-pi/12)的最大值

f(x)=sin²(x+π/12)+cos²(x-π/12)
=1-cos²(x+π/12)+cos²(x-π/12)
=1+[cos(x-π/12)+cos(x+π/12)][cos(x-π/12)-cos(x+π/12)]
=1+(cosx cosπ/12 + sinx sinπ/12 + cosx cosπ/12 -sinx sinπ/12)(cosx cosπ/12 + sinx sinπ/12 - cosx cosπ/12 + sinx sinπ/12)
=1 + 2 cosx cosπ/12 × 2 sinx sinπ/12
=1 + 2sinxcosx×2sinπ/12cosπ/12
=1+sin2x×sinπ/6
=1/2·sin2x+1
因为sin2x的最大值为1,
所以f(x)的最大值为1/2+1=3/2.
由于百度上有些数学符号打得不清楚,所以小弟的解题过程有点繁杂,看起来有点乱,请见谅!不过本题主要就是根据“余弦加法公式”“正弦二倍角公式”来化简.