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a) ta có : \(cos^2\left(a-b\right)-sin^2\left(a+b\right)\)
\(=\left(cosa.cosb+sina.sinb\right)^2-\left(sina.cosb+sinb.cosa\right)^2\)
\(=cos^2a.cos^2b+sin^2a.sin^2b-sin^2a.cos^2b-sin^2b.cos^2a\)
\(=cos^2a.cos^2b-sin^2a.cos^2b+sin^2a.sin^2b-sin^2b.cos^2a\)\(=cos^2b\left(cos^2a-sin^2a\right)-sin^2b\left(cos^2a-sin^2a\right)\)
\(=\left(cos^2b-sin^2b\right)\left(cos^2a-sin^2a\right)=cos2a.cos2b\left(đpcm\right)\)
\(b.cosC+c.cosB=b.\frac{a^2+b^2-c^2}{2ab}+c.\frac{a^2+c^2-b^2}{2ac}\)
\(=\frac{a^2+b^2-c^2}{2a}+\frac{a^2+c^2-b^2}{2a}=\frac{a^2+b^2-c^2+a^2+c^2-b^2}{2a}=\frac{2a^2}{2a}=a\)
Hai cái dưới chứng minh y hệt
\(\dfrac{b^2-a^2}{2c}=b.\dfrac{\left(b^2+c^2-a^2\right)}{2bc}-a.\dfrac{\left(a^2+c^2-b^2\right)}{2ac}\)
\(\Leftrightarrow\dfrac{b^2-a^2}{2c}=\dfrac{b^2+c^2-a^2}{2c}-\dfrac{a^2+c^2-b^2}{2c}\)
\(\Leftrightarrow b^2-a^2=\left(b^2+c^2-a^2\right)-\left(a^2+c^2-b^2\right)\)
\(\Leftrightarrow3b^2=3a^2\Leftrightarrow a=b\)
Hay tam giác cân tại C
\(cosA=2sin\frac{B+C}{2}.cos\frac{B-C}{2}-\frac{3}{2}\)
\(\Leftrightarrow cosA=2cos\frac{A}{2}.cos\frac{B-C}{2}-\frac{3}{2}\)
\(\Leftrightarrow2cos^2\frac{A}{2}-1=2cos\frac{A}{2}cos\frac{B-C}{2}-\frac{3}{2}\)
\(\Leftrightarrow4cos^2\frac{A}{2}-4cos\frac{A}{2}cos\frac{B-C}{2}+1=0\)
Mà \(\left\{{}\begin{matrix}cos\frac{A}{2}>0\\0< cos\frac{B-C}{2}\le1\end{matrix}\right.\) \(\Rightarrow cos\frac{A}{2}.cos\frac{B-C}{2}\le cos\frac{A}{2}\)
\(\Rightarrow4cos^2\frac{A}{2}-4cos\frac{A}{2}+1\le0\)
\(\Leftrightarrow\left(2cos\frac{A}{2}-1\right)^2\le0\)
BPT có nghiệm khi và chỉ khi:
\(\left\{{}\begin{matrix}2cos\frac{A}{2}-1=0\\B-C=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cos\frac{A}{2}=\frac{1}{2}\\B=C\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}A=120^0\\B=C\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}A=120^0\\B=C=30^0\end{matrix}\right.\)