就 專(zhuān)欄 cv19195894 一位朋友 所述證明 具體過(guò)程 饗以諸君
S
=
1/2absinC
=
sinC
sin(A-B)
ab
/
2sin(A-B)
=
sin(A+B)
sin(A-B)
ab
/
2sin(A-B)
=
2sin((A+B)/2)cos((A+B)/2)
2sin((A-B)/2)cos((A-B)/2)
ab
/
2sin(A-B)
=
2sin((A+B)/2)cos((A-B)/2)
2cos((A+B)/2)sin((A-B)/2)
ab
/
2sin(A-B)
=
(sinA+sinB)(sinA-sinB)
ab
/
2sin(A-B)
=
(sinA+sinB)·2R
(sinA-sinB)·2R
a/(2R)·b/(2R)
/
2sin(A-B)
=
(a+b)(a-b)
sinAsinB
/
2sin(A-B)
=
(a2-b2)
sinAsinB
/
2sin(A-B)
得證
標(biāo)簽：
