卖大理石的挂画:一道数学题
来源:百度文库 编辑:高校问答 时间:2024/05/06 02:21:01
结论:sinA+sinB+sinC>cosA+cosB+cosC
证明:∵∠A+∠B>90°
∴∠A>90°-∠B
∴sinA>sin(90°-∠B)
∴sinA>cos∠B
同理,sinB>cosC
sinC>cosA
∴sinA+sinB+sinC>cosA+cosB+cosC
a大
COSACOSBCOSC
=COSACOSB(-COS(A+B))
=-1/2(COS(A+B)+COS(A-B))COS(A+B)
=-1/2(COS(A+B)COS(A+B)+COS(A-B)COS(A+B))
=-1/2((-COSC)(-COSC)+1/2(COS2A+cos(-2B))
=-1/2(1-SINCSINC+1/2(1-2SINASINA+1-2sinbsinb))
=sinasina+sinbsinb+sincsinc
由于A,B,C都是锐角
所以SINA,SINB,SINC大于SINASINA.......且都为正
所以SINA+SINB+SINC大于COSA+COSB+COSC
COSACOSBCOSC
=COSACOSB(-COS(A+B))
=-1/2(COS(A+B)+COS(A-B))COS(A+B)
=-1/2(COS(A+B)COS(A+B)+COS(A-B)COS(A+B))
=-1/2((-COSC)(-COSC)+1/2(COS2A+cos(-2B))
=-1/2(1-SINCSINC+1/2(1-2SINASINA+1-2sinbsinb))
=sinasina+sinbsinb+sincsinc
由于A,B,C都是锐角
所以SINA,SINB,SINC大于SINASINA.......且都为正
所以SINA+SINB+SINC大于COSA+COSB+COSC
我真叫你开了!!!!!!!
结论:sinA+sinB+sinC>cosA+cosB+cosC
证明:∵∠A+∠B>90°
∴∠A>90°-∠B
∴sinA>sin(90°-∠B)
∴sinA>cos∠B
同理,sinB>cosC
sinC>cosA
∴sinA+sinB+sinC>cosA+cosB+cosC