已知A、B、C是锐角,求证:cosA+cosB+cosC=1+4sinA/2sinB/2sinC/2的充要条件是A+B+C=π.
问题描述:
已知A、B、C是锐角,求证:cosA+cosB+cosC=1+4sin
sinA 2
sinB 2
的充要条件是A+B+C=π. C 2
答
cosA+cosB+cosC=1+4sin
sinA 2
sinB 2
C 2
⇔2cos
cosA+B 2
=2sin2A−B 2
+2sinC 2
(−cosC 2
+cosA+B 2
)A−B 2
⇔cos
cosA+B 2
=sin2A−B 2
−sinC 2
cosC 2
+sinA+B 2
cosC 2
A−B 2
⇔0=sin
(sinC 2
−cosC 2
)+(sinA+B 2
−cosC 2
)cosA+B 2
A−B 2
⇔0=(sin
−cosC 2
)(sinA+B 2
+cosC 2
)A−B 2
∵A、B、C是锐角,∴sin
+cosC 2
>0A−B 2
所以上式⇔0=sin
−cosC 2
A+B 2
⇔
+C 2
=A+B 2
⇔A+B+C=ππ 2