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a) ta có : \(cos^2\left(a-b\right)-sin^2\left(a+b\right)\)
\(=\left(cosa.cosb+sina.sinb\right)^2-\left(sina.cosb+sinb.cosa\right)^2\)
\(=cos^2a.cos^2b+sin^2a.sin^2b-sin^2a.cos^2b-sin^2b.cos^2a\)
\(=cos^2a.cos^2b-sin^2a.cos^2b+sin^2a.sin^2b-sin^2b.cos^2a\)\(=cos^2b\left(cos^2a-sin^2a\right)-sin^2b\left(cos^2a-sin^2a\right)\)
\(=\left(cos^2b-sin^2b\right)\left(cos^2a-sin^2a\right)=cos2a.cos2b\left(đpcm\right)\)
Ta có: A = \(sin\dfrac{A}{2}+sin\dfrac{B}{2}+sin\dfrac{C}{2}=cos\dfrac{B+C}{2}+2sin\dfrac{B+C}{4}cos\dfrac{B-C}{4}\)
\(\Leftrightarrow A-2sin\dfrac{B+C}{4}cos\dfrac{B-C}{4}-cos^2\dfrac{B+C}{4}+sin^2\dfrac{B+C}{4}=0\)\(\Leftrightarrow A-2sin\dfrac{B+C}{4}cos\dfrac{B-C}{4}+2sin^2\dfrac{B+C}{4}-1=0\)
Δ' = \(cos^2\dfrac{B-C}{4}-2\left(A-1\right)\ge0\)
\(\Rightarrow A-1\le\dfrac{1}{2}\Leftrightarrow A\le\dfrac{3}{2}\)
Ta có:
\(r^2+p^2+4Rr=\left(\dfrac{S}{p}\right)^2+p^2+\dfrac{abc}{S}.\dfrac{S}{p}\)
\(=\dfrac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{p}+p^2+\dfrac{abc}{p}\)
\(=\dfrac{p^3+\left(ab+bc+ac\right)p-p^2\left(a+b+c\right)-abc+p^3+abc}{p}\)
\(=ab+bc+ca\)
Do đó:
\(\dfrac{ab+bc+ca}{4R^2}=\dfrac{r^2+p^2+4Rr}{4R^2}\)
\(\Leftrightarrow sinAsinB+sinBsinC+sinCsinA=\dfrac{r^2+p^2+4Rr}{4R^2}\)\(\left(đpcm\right)\)
bạn giải thích chi tiết đoạn này hộ mình được ko ạ
p^3+(ab+bc+ac)p−p^2(a+b+c)−abc+p^3+abc/p
=ab+bc+ca
a, Áp dụng BĐT Cosi:
\(\sqrt{\left(p-a\right)\left(p-b\right)}\le\dfrac{p-a+p-b}{2}=\dfrac{c}{2}\)
\(\sqrt{\left(p-b\right)\left(p-c\right)}\le\dfrac{p-b+p-c}{2}=\dfrac{a}{2}\)
\(\sqrt{\left(p-c\right)\left(p-a\right)}\le\dfrac{p-c+p-a}{2}=\dfrac{b}{2}\)
\(\Rightarrow\left(p-a\right)\left(p-b\right)\left(p-c\right)\le\dfrac{1}{8}abc\)
oh my gut : O