Cái này chỉ cần dùng kiến thức đơn thuần lớp 10 thôi mà.
Có $4sin\dfrac{A}{2}.sin\dfrac{B}{2}.sin\dfrac{C}{2}$
$=\dfrac{sinA.sinB.sinC}{2cos\dfrac{A}{2}cos\dfrac{B}{2}cos\dfrac{C}{2}}$
$=\dfrac{sinA.sinB.sinC}{(cos\dfrac{A+B}{2}+cos\dfrac{A-B}{2})cos\dfrac{C}{2}}$
$=\dfrac{sinA.sinB.sinC}{(cos\dfrac{A+B}{2}.cos\dfrac{C}{2})+(cos\dfrac{A-B}{2}cos\dfrac{C}{2})}$
$=2\dfrac{sinA.sinB.sinC}{cos\dfrac{A+B+C}{2}+cos\dfrac{A+B-C}{2}+cos\dfrac{A-B+C}{2}+cos\dfrac{A-B-C}{2}}$
$=2\dfrac{sinA.sinB.sinC}{(cos\dfrac{A+B}{2}.cos\dfrac{C}{2})+(cos\dfrac{A-B}{2}cos\dfrac{C}{2})}$
$=2\dfrac{sinA.sinB.sinC}{cos\dfrac{180}{2}+cos\dfrac{180-2C}{2}+cos\dfrac{180-2B}{2}+cos\dfrac{2A-180}{2}}$
$=2\dfrac{sinA.sinB.sinC}{sin\dfrac{C}{2}+sin\dfrac{B}{2}+sin\dfrac{A}{2}}$
Áp dụng công thức:$sinA=\dfrac{a}{2R}$, $sinA=\dfrac{a}{2R}$, $sinA=\dfrac{a}{2R}$
$S=\dfrac{abc}{4R}=pr$
VT$=2\dfrac{\dfrac{a}{2R}\dfrac{b}{2R}\dfrac{c}{2R}}{\dfrac{a}{2R}+\dfrac{b}{2R}+\dfrac{c}{2R}}=\dfrac{abc}{4R^{3}}.\dfrac{2R}{a+b+c}$
$=\dfrac{2S}{R.2p}=\dfrac{r}{R}$ (Đ.P.C.M)
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