灾后重建工作方案:锐角三角形ABC中,求证sinA+sinB+sinC>2

先证明锐角三角形中sinA+sinB+sinC>1+cosA+cosB+cosC

再证明cosA+cosB+cosC>1

过程:(1)1+cosA+cosB+cosC-sinA+sinB+sinC

=2cos2(A/2)+2cos((B+C)/2)·cos((B-C)/2)-2(sin(A/2)·cos(A/2)+sin((B+C)/2)·cos((B-C)/2))

=2[cos(A/2)·(cos(A/2)-sin(A/2))+cos((B-C)/2)·(cos((B+C)/2)-sin((B+C)/2))]

=2[cos(A/2)·(cos(A/2)-sin(A/2))+cos((B-C)/2)·(sin(A/2)-cos(A/2))]

=2(cos(A/2)-sin(A/2))·(cos(A/2)-cos((B-C)/2))

=-4(cos(A/2)-sin(A/2))·sin((A+B-C)/4)·sin((A+C-B)/4)..........①

由于A,B,C∈(0,π/2),所以0<(A/2)<π/4,0<(A+B-C)/4<π/4,0<(A+C-B)/4<π/4

所以①式<0

∴sinA+sinB+sinC>1+cosA+cosB+cosC

(2)cosA+cosB+cosC=2cos((A+B)/2)cos((A-B)/2)-2cos2((A+B)/2)+1

=2cos((A+B)/2)(cos((A-B)/2)-cos((A+B)/2)+1

=4cos((A+B)/2)sin(A/2)sin(B/2)+1

=1+4sin(A/2)sin(B/2)sin(C/2)

>1

综合(1)(2)知锐角三角形中sinA+sinB+sinC>2

因为是锐角
所以：0<A<90
0<B<90
0<c<90
然后画三个角在同一单位圆中的图象一比可知
sinA+sinB+sinC>2