有

cos2B+sin2B=cos2A+sin2A

即

cos2B-sin2B+2sin2B

=

2sin2A+cos2A-sin2A

即

cos2B-sin2B-cos2A+sin2A

=

2sin2A-2sin2B

即

cos(2B)-cos(2A)

/

2

=

sin2A-sin2B

即

sin(A+B)sin(A-B)=sin2A-sin2B

即

sinCsin(A-B)=sin2A-sin2B

即

S

=

(absinC·sinCsin(A-B))

/

2（sin2A-sin2B）

即

S

=

(abc2sin(A-B))

/

2（a2-b2）

即

1=（a2-b2）2S/(abc2sin(A-B))

即

S

=

(a2-b2)2S2/(abc2sin(A-B))

即

S

=

(a2-b2)2S/(bc)·2S/(ac)/(2sin(A-B))

即

S

=

(a2-b2)sinAsinB/(2sin(A-B))

得證