高中数学三角函数题（简单）
\u9ad8\u4e2d\u6570\u5b66\u4e09\u89d2\u51fd\u6570\u7b80\u5355\u9898cosx-(1/2)[cosxcos(\u03c0/3)+sinxsin(\u03c0/3)]\u5982\u4f55\u5f97\u5230-(\u221a3/2)[sinxcos(\u03c0/3)-cosxsin9\u03c0/3)]\uff1f
\u89e3:cosx-(1/2)[cosxcos(\u03c0/3)+sinxsin(\u03c0/3)]=cosx-(1/2)[(1/2)cosx+(\u221a3/2)sinx]
=cosx-(1/4)cosx-(\u221a3/4)sinx=(3/4)cosx-(\u221a3/4)sinx
=(-\u221a3/2)[(1/2)sinx-(\u221a3/2)cosx]
=(-\u221a3/2)[sinxcos(\u03c0/3)-cosxsin(\u03c0/3)]
1,\u7531\u9898\u610f\u5f97\u51fa\u56fe\uff08\u5927\u6982\u5c31\u662f\u8fd9\u4e2a\u6837\u5b50\uff0c\u53ef\u80fd\u6709\u8bef\u5dee\uff09\uff0cCD\u5782\u76f4AB
\u5219\uff1a|\u5411\u91cfAB\u51cf\u5411\u91cfAC|=|BC|\uff0c
AB=8\uff0c\u5219BD=5.5\uff0cAD=2.5
\u4e09\u89d2ACD,AC=5,AD=2.5,\u5219CD=\uff1f
\u4e09\u89d2BCD,CD\u77e5\uff0cBD=5.5\uff0cBC=\uff1f
2\uff0c\u5728\u76f4\u89d2\u4e09\u89d2\u5f62ACD\u4e2d\uff0cAC=5,AD=2.5,\u5219\u89d2BAC=k=60\u5ea6
cos(60\u5ea6+x)=4/5
\u5269\u4e0b\u7684\u4e0d\u7528\u6211\u5199\u4e86\u5427\uff01
=1/2-1/2cos2wx+√3/2sin2wx
=1/2+sin(2wx-π/6)
π=2π/2w
解之：w=1
所以：f(x)=1/2+sin(2x-π/6)
(2) f(x)+f(x+2)=2sin(π/4x+π/4)+2sin(π/4x+π/2+π/4)
=2sin(π/4x+π/4)+2cos(π/4x+π/4)
=2√2cos(π/4x)
所以：f(x)+f(x+2)最大值在x=-2/3时取得；其值为-√6/3
f(x)+f(x+2)最小值在x=-6是取得；其值为-2√2。
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