一道特殊的三角最值问题

题目

在 ( riangle ABC) 中，(sqrt{2}sin A+sin Bsin C) 的最大值是 ((qquad))

(A. 2+dfrac{1}{2})

(B. 2)

(C. sqrt{3})

(D. sqrt{5})

解析

[egin{array}{rl}sqrt{2}sin A+sin Bsin C&=sqrt{2}sin A-dfrac{cos(B+C)-cos(B-C)}{2}\&=sqrt{2}sin A+dfrac{cos(B-C)-cos(B+C)}{2}\&leqsqrt{2}sin A+dfrac{1+cos A}{2}\&=sqrt{2}sin A+dfrac{1}{2}cos A+dfrac{1}{2}\&leqsqrt{2+dfrac{1}{4}}+dfrac{1}{2}=2end{array} ]

当且仅当 (sin B=sin C=dfrac{sqrt{6}}{3},sin A=dfrac{2sqrt{2}}{3}) 时，等号成立.

答案 (B)