证明三角恒等式证明:三角形中(tanA/2)^2+(tanB/2)^2+(tanC/2)^2=1恒成立...
1个回答
应该学过余弦定理吧
先化简下面这个式子
tan(a/2)=sin(a/2)/cos(a/2)
=[2sin(a/2)cos(a/2)]/[2cos²(a/2)]
=sina/(1+cosa)
故tan²(a/2)=sin²a/(1+cosa)²
=(1-cos²a)/(1+cosa)²
=(1-cosa)/(1+cosa)
故
(tanA/2)^2=(1-cosA)/(1+cosA)
=[1-(b²+c²-a²)/(2bc)]/[1+(b²+c²-a²)/(2bc)]
=[2bc-(b²+c²-a²)]/[2bc+(b²+c²-a²)]
=(2bc-b²-c²+a²)/(b²+c²+2bc-a²)
=[a²-(b-c)²]/[(b+c)²-a²]
=[(a+b-c)*(a-b+c)]/[(a+b+c)*(b+c-a)]
同理,有
(tanB/2)^2=[(a+b-c)*(-a+b+c)]/[(a+b+c)*(a+c-b)]
(tanC/2)^2=[(-a+b+c)*(a-b+c)]/[(a+b+c)*(b+a-c)]
故(tanA/2)^2+(tanB/2)^2+(tanC/2)^2
=[(a+b-c)*(a-b+c)]/[(a+b+c)*(b+c-a)]+[(a+b-c)*(-a+b+c)]/[(a+b+c)*(a+c-b)]+[(-a+b+c)*(a-b+c)]/[(a+b+c)*(b+a-c)]
=1
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