Question

1, a sin(B-C) + b sin(C-A) + c sin(A-B)=0 2, sin² A + sin² B + sin² C = 2+2cosAcosBcosC 3, r= 4RsinA/2 sin B/2 sin C/2

Answer 1

Câu 1:

\(a.sin\left(B-C\right)=a.sinBcosC-a.cosB.sinC\)

\(bsin\left(C-A\right)=bsinC.cosA-bcosC.sinA\)

\(csin\left(A-B\right)=csinAcosB-csinB.cosA\)

Cộng lại:

\(VT=cosA\left(bsinC-c.sinB\right)+cosB\left(c.sinA-a.sinC\right)+cosC\left(a.sinB-bsinA\right)\)

\(=cosA\left(\frac{b.c}{2R}-\frac{bc}{2R}\right)+cosB\left(\frac{ac}{2R}-\frac{ac}{2R}\right)+cosC\left(\frac{ab}{2R}-\frac{ab}{2R}\right)=0\)

Câu 2:

\(sin^2A+sin^2B+sin^2C=\frac{1}{2}-\frac{1}{2}cos2A+\frac{1}{2}-\frac{1}{2}cos2B+1-cos^2C\)

\(=2-\frac{1}{2}\left(cos2A+cos2B\right)-cosC.cosC\)

\(=2-cos\left(A+B\right)cos\left(A-B\right)+cosC.cos\left(A+B\right)\)

\(=2+cosC.cos\left(A-B\right)+cosC.cos\left(A+B\right)\)

\(=2+cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)

\(=2+2cosA.cosB.cosC\)

Answer 2

Câu 3:

Ta có \(sin^2\frac{A}{2}=\frac{1-cosA}{2}=\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}=\frac{a^2-b^2-c^2+2bc}{4bc}=\frac{a^2-\left(b-c\right)^2}{4bc}\)

\(=\frac{\left(a+b-c\right)\left(a+c-b\right)}{4bc}=\frac{\left(p-c\right)\left(p-b\right)}{bc}\Rightarrow sin\frac{A}{2}=\sqrt{\frac{\left(p-b\right)\left(p-c\right)}{bc}}\)

Tương tự ta có \(sin\frac{B}{2}=\sqrt{\frac{\left(p-a\right)\left(p-c\right)}{ac}}\) ; \(sin\frac{C}{2}=\sqrt{\frac{\left(p-a\right)\left(p-b\right)}{ab}}\)

\(\Rightarrow4Rsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=4\left(\frac{abc}{4S}\right)\sqrt{\frac{\left(p-a\right)^2\left(p-b\right)^2\left(p-c\right)^2}{a^2b^2c^2}}\)

\(=\frac{abc.\left(p-a\right)\left(p-b\right)\left(p-c\right)}{S.abc}=\frac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{S}=\frac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}}=\sqrt{\frac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{p}}=r\)