Question

三角形ABC中,求证(sinA)^2+(sinB)^2+(cosC)^2+2sinAcosBcos(A+B)=1 谢谢！

Answer 1

证明:
(sinA)^2+(sinB)^2+(cosC)^2
=(1-cos2A)/2+(1-cos2B)/2+(1+cos2C)/2
=3/2-(1/2)(cos2A+cos2B-cos2C)
=3/2-(1/2)[2cos(A+B)cos(A-B)-2(cosC)^2-1]
=1-[cos(A+B)cos(A-B)-(cosC)^2]
=1-[-cos(A-B)cosC-(cosC)^2)
=1+cosC[cosC+cos(A-B)]
=1+cosC[-cos(A+B)+cos(A-B)]
=1+2cosC*sinAsinB
=1-2cos(A+B)sinAsinB

∴ 你的题目给的有问题，应该是
∴ (sinA)^2+(sinB)^2+(cosC)^2+2sinAsinBcos(A+B)
=1-2cos(A+B)sinAsinB+2sinAsinBcos(A+B)
=1
∴ 所证等式成立。

Answer 2

题目有错误，cosB应该是sinB。
证：
cosC=-cos[180°-(A+B)]=-cos(A+B)
由正弦定理，令a/sinA=b/sinB=c/sinC=t，则
a=tsinA，b=tsinB，c=tsinC
由余弦定理得
c²=a²+b²-2abcosC
t²sin²C=t²sin²A+t²sin²B-2(tsinA)(tsinB)cosC
sin²C=sin²A+sin²B-2sinAsinBcosC
sin²A+sin²B+2sinAsinBcos(A+B)=1-cos²C
sin²A+sin²B+cos²C+2sinAsinBcos(A+B)=1

Answer 3

(sinA)^2+(sinB)^2+(cosC)^2
=(1-cos2A)/2+(1-cos2B)/2+(1+cos2C)/2
=3/2-(1/2)(cos2A+cos2B-cos2C)
=3/2-(1/2)[2cos(A+B)cos(A-B)-2(cosC)^2-1]
=1-[cos(A+B)cos(A-B)-(cosC)^2]
=1-[-cos(A-B)cosC-(cosC)^2)
=1+cosC[cosC+cos(A-B)]
=1+cosC[-cos(A+B)+cos(A-B)]
=1+2cosC*sinAsinB
=1-2cos(A+B)sinAsinB
∴ 你的题目给的有问题，应该是
∴ (sinA)^2+(sinB)^2+(cosC)^2+2sinAsinBcos(A+B)
=1-2cos(A+B)sinAsinB+2sinAsinBcos(A+B)
=1
∴ 所证等式成立。