Question

12£®ÒÑÖªÏòÁ¿$\overrightarrow m=£¨\sqrt{2}cos\frac{x}{4}£¬2cos\frac{x}{4}£©$£¬$\overrightarrow n=£¨\sqrt{2}cos\frac{x}{4}£¬\sqrt{3}sin\frac{x}{4}£©$£¬Éè$f£¨x£©=\overrightarrow m•\overrightarrow n$£®
£¨¢ñ£©Èôf£¨¦Á£©=2£¬Çó$cos£¨¦Á+\frac{¦Ð}{3}£©$µÄÖµ£»
£¨¢ò£©ÔÚ¡÷ABCÖУ¬½ÇA£¬B£¬CµÄ¶Ô±ß·Ö±ðÊÇa£¬b£¬c£¬ÇÒÂú×㣨2a-b£©cosC=ccosB£¬Çóf£¨A£©µÄÈ¡Öµ·¶Î§£®

Answer 1

·ÖÎö £¨¢ñ£©¸ù¾Ý$f£¨x£©=\overrightarrow m•\overrightarrow n$£®ÀûÓÃÏòÁ¿µÄÊýÁ¿»ýµÄÔËÓÃÇó½âf£¨x£©µÄ½âÎöʽ£¬f£¨¦Á£©=2£¬ÕÒ³öÓë$cos£¨¦Á+\frac{¦Ð}{3}£©$µÄ¹Øϵ¼´¿ÉµÃ½â£®
£¨¢ò£©ÀûÓÃÕýÏÒ¶¨Àí»¯¼ò£¬Çó½âC½ÇµÄ´óС£®½áºÏÈý½Çº¯ÊýµÄÐÔÖÊÇó½â¼´¿É£®

½â´ð ½â£º£¨¢ñ£©ÏòÁ¿$\overrightarrow m=£¨\sqrt{2}cos\frac{x}{4}£¬2cos\frac{x}{4}£©$£¬$\overrightarrow n=£¨\sqrt{2}cos\frac{x}{4}£¬\sqrt{3}sin\frac{x}{4}£©$£¬
¡ß$f£¨x£©=\overrightarrow m•\overrightarrow n$
ÄÇô£º$f£¨x£©=2{cos^2}\frac{x}{4}+2\sqrt{3}sin\frac{x}{4}cos\frac{x}{4}$=$\sqrt{3}sin\frac{x}{2}+cos\frac{x}{2}+1$=$2sin£¨\frac{x}{2}+\frac{¦Ð}{6}£©+1$£®
¡ßf£¨¦Á£©=2£¬¼´$sin£¨\frac{¦Á}{2}+\frac{¦Ð}{6}£©$=$\frac{1}{2}$£¬
¡à$cos£¨¦Á+\frac{¦Ð}{3}£©=1-2{sin^2}£¨\frac{¦Á}{2}+\frac{¦Ð}{6}£©=\frac{1}{2}$£®
£¨¢ò£©¡ß£¨2a-b£©cosC=ccosB£¬
¡à£¨2sinA-sinB£©cosC=sinCcosB£¬
⇒2sinAcosC=sinBcosC+cosBsinC=sin£¨B+C£©£¬
¡à2sinAcosC=sinA£¬
¡ßsinA¡Ù0£¬
¡à$cosC=\frac{1}{2}$£¬¡à$C=\frac{¦Ð}{3}$£®
¡à$0£¼A£¼\frac{2¦Ð}{3}$£¬
$\frac{¦Ð}{6}£¼\frac{A}{2}+\frac{¦Ð}{6}£¼\frac{¦Ð}{2}$£¬
¡à$\frac{1}{2}£¼sin£¨\frac{A}{2}+\frac{¦Ð}{6}£©£¼1$£¬
¡ß$f£¨A£©=2sin£¨\frac{A}{2}+\frac{¦Ð}{6}£©+1$£¬
¡àf£¨A£©µÄÈ¡Öµ·¶Î§Îª£¨2£¬3£©£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²é¶ÔÈý½Çº¯ÊýµÄ»¯¼òÄÜÁ¦ºÍÈý½Çº¯ÊýµÄͼÏóºÍÐÔÖʵÄÔËÓã¬ÀûÓÃÈý½Çº¯Êý¹«Ê½½«º¯Êý½øÐл¯¼òÊǽâ¾ö±¾ÌâµÄ¹Ø¼ü£®ÊôÓÚÖеµÌ⣮