证明: 令
C
C
C为
Δ
A
B
C
\Delta ABC
ΔABC最大的内角 则
C
∈
[
π
3
,
π
2
)
C\in[\frac \pi 3,\frac \pi 2)
C∈[3π,2π) 则
sin
C
∈
[
3
2
,
1
)
,
cos
C
∈
[
1
2
,
1
)
\sin C\in[\frac {\sqrt 3} 2,1),\cos C\in[\frac 1 2,1)
sinC∈[23
,1),cosC∈[21,1),根据
sin
\sin
sin和
cos
\cos
cos在这一部分的单调性可知,
sin
C
>
cos
C
\sin C>\cos C
sinC>cosC
sin
A
+
sin
B
−
(
cos
A
+
cos
B
)
\sin A+\sin B-(\cos A+\cos B)
sinA+sinB−(cosA+cosB)
=
2
sin
A
+
B
2
cos
A
−
B
2
−
2
cos
A
+
B
2
cos
A
−
B
2
=2\sin\frac{A+B}2\cos\frac{A-B}2-2\cos\frac{A+B}2\cos\frac{A-B}2
=2sin2A+Bcos2A−B−2cos2A+Bcos2A−B
=
2
cos
A
−
B
2
(
sin
A
+
B
2
−
cos
A
+
B
2
)
=2\cos\frac{A-B}2(\sin\frac{A+B}2-\cos\frac{A+B}2)
=2cos2A−B(sin2A+B−cos2A+B)
∵
A
+
B
=
π
−
C
∈
(
π
2
,
2
π
3
]
\because A+B=\pi-C\in(\frac \pi 2,\frac {2\pi}{3}]
∵A+B=π−C∈(2π,32π]
∴
A
+
B
2
∈
(
π
4
,
π
3
]
\therefore \frac{A+B}2\in(\frac \pi 4,\frac \pi 3]
∴2A+B∈(4π,3π]
∴
cos
A
+
B
2
≥
1
2
>
0
\therefore \cos \frac{A+B}{2}\geq \frac 1 2>0
∴cos2A+B≥21>0 即
sin
A
+
B
2
>
cos
A
+
B
2
\sin \frac{A+B}2>\cos \frac{A+B}2
sin2A+B>cos2A+B 又
cos
A
−
B
2
>
0
\cos \frac{A-B}2>0
cos2A−B>0
∴
sin
A
+
sin
B
>
cos
A
+
cos
B
\therefore \sin A+\sin B>\cos A+\cos B
∴sinA+sinB>cosA+cosB 综上
sin
A
+
sin
B
+
s
i
n
C
>
cos
A
+
cos
B
+
c
o
s
C
\sin A+\sin B+sinC>\cos A+\cos B+cosC
sinA+sinB+sinC>cosA+cosB+cosC
|