nhường cho I don't know :))

Nó bt đếch gì mà nhường :) ?

1.

\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)

\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)

\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)

\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)

\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)

\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)

\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)

\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)

2.

\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)

\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x\)

\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)

\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)

\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)

\(=tan^2a.cot^2b-2\)

Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)

Theo Viet: \(\left\{{}\begin{matrix}tana+tanb=p\\tana.tanb=q\end{matrix}\right.\)

\(\Rightarrow tan\left(a+b\right)=\frac{tana+tanb}{1-tana.tanb}=\frac{p}{1-q}\)

\(\Rightarrow A=cos^2\left(a+b\right)+psin\left(a+b\right)+q.sin^2\left(a+b\right)\)

\(=\frac{1}{cos^2\left(a+b\right)}\left(1+p.\frac{sin\left(a+b\right)}{cos\left(a+b\right)}+q.\frac{sin^2\left(a+b\right)}{cos^2\left(a+b\right)}\right)\)

\(=\left[1+tan^2\left(a+b\right)\right]\left[1+p.tan\left(a+b\right)+q.tan^2\left(a+b\right)\right]\)

\(=\left[1+\frac{p^2}{\left(1-q\right)^2}\right]\left[1+\frac{p^2}{1-q}+\frac{p^2q}{\left(1-q\right)^2}\right]\)

\(=\left[1+\frac{p^2}{\left(1-q\right)^2}\right]\left[1+\frac{p^2}{\left(1-q\right)^2}\right]=\left[1+\frac{p^2}{\left(1-q\right)^2}\right]^2\)

Theo Viet ta có \(\left\{{}\begin{matrix}tana+tanb=p\\tana.tanb=q\end{matrix}\right.\)

\(\Rightarrow tan\left(a+b\right)=\frac{tana+tanb}{1-tana.tanb}=\frac{p}{1-q}\)

\(A=cos^2\left(a+b\right)\left[1+p.tan\left(a+b\right)+q.tan^2\left(a+b\right)\right]\)

\(A=\frac{1}{1+tan^2\left(a+b\right)}\left[1+\frac{p^2}{1-q}+\frac{q.p^2}{\left(1-q\right)^2}\right]\)

\(A=\frac{\left(1-q\right)^2}{p^2+\left(1-q\right)^2}\left(1+\frac{p^2}{\left(1-q^2\right)}\right)\)

\(A=\frac{\left(1-q^2\right)}{p^2+\left(1-q\right)^2}.\left(\frac{p^2+\left(1-q\right)^2}{\left(1-q\right)^2}\right)=1\)