Question

在三角形ABC内，求sinA+sinB+sinC-cosA-cosB-cosC的最大值

请给详细解答~~

Answer 1

解:
因为sinA+sinB
=2sin(A+B)/2*cos(A+B)/2
≤ 2sin(A+B)/2...... ①
sinC+sinπ/3
=2sin(C+π/3)/2*cos(C-π/3)/2
≤ 2sin(C+π/3)/2 ......... ②

又因为sin(A+B)/2+sin(C+π/3)/2
=2sin(A+B+C+π/3)/4*cos(A+B-C-π/3)/4
≤2sinπ/3............... ③

由①，②，③得
sinA+sinB+sinC+sinπ/3 ≤4sinπ/3 ,
所以sinA+sinB+sinC≤3sinπ/3=3√3/2 ,
当A=B=C=π/3 时，
（sinA+sinB+sinC）max= 3√3/2 .

(sinA+sinB+sinC-cosA-cosB-cosC)max
=3√3/2-3/2