在三角形abc中,角A,B,C的对边分别为a,b,c,求证:a*2-b*2/c*2=sin(A-B
在三角形abc中,角A,B,C的对边分别为a,b,c,求证:a*2-b*2/c*2=sin(A-B)/sinC...
在三角形abc中,角A,B,C的对边分别为a,b,c,求证:a*2-b*2/c*2=sin(A-B)/sinC
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(a^2-b^2)/c^2=(a+b/c)(a-b/c)
根据正弦定理:
(a+b/c)(a-b/c)
=(sinA+sinB/sinC)(sinA-sinB/sinC)
分别处理,用和化为积公式:
sinA+sinB/sinC=2sin(A+B/2)cos(A-B/2)/sin(A+B)
=2sin(A+B/2)cos(A-B/2)/2sin(A+B/2)cos(A+B/2)
=cos(A-B/2)/cos(A+B/2)
同理:a-b/c=sin(A-B/2)/sin(A+B/2)
所以原式=sin(A-B/2)cos(A-B/2)/sin(A+B/2)cos(A+B/2)
=sin(A-B)/sin(A+B)=sin(A-B)/sinC
根据正弦定理:
(a+b/c)(a-b/c)
=(sinA+sinB/sinC)(sinA-sinB/sinC)
分别处理,用和化为积公式:
sinA+sinB/sinC=2sin(A+B/2)cos(A-B/2)/sin(A+B)
=2sin(A+B/2)cos(A-B/2)/2sin(A+B/2)cos(A+B/2)
=cos(A-B/2)/cos(A+B/2)
同理:a-b/c=sin(A-B/2)/sin(A+B/2)
所以原式=sin(A-B/2)cos(A-B/2)/sin(A+B/2)cos(A+B/2)
=sin(A-B)/sin(A+B)=sin(A-B)/sinC
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