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\(A=\frac{1}{2}-\frac{1}{2}cos6a+\frac{1}{2}-\frac{1}{2}cos\left(6a-\frac{2\pi}{3}\right)+\frac{1}{2}-\frac{1}{2}cos\left(6a+\frac{2\pi}{3}\right)\)
\(=\frac{3}{2}-\frac{1}{2}cos6a-\frac{1}{2}\left[cos\left(6a-\frac{2\pi}{3}\right)+cos\left(6a+\frac{2\pi}{3}\right)\right]\)
\(=\frac{3}{2}-\frac{1}{2}cos6a-cos6a.cos\frac{2\pi}{3}\)
\(=\frac{3}{2}-\frac{1}{2}cos6a+\frac{1}{2}cos6a=\frac{3}{2}\)
\(cot\alpha=3\Leftrightarrow\dfrac{cos\alpha}{sin\alpha}=3\Leftrightarrow cos\alpha=3sin\alpha\)
Khi đó:
\(\dfrac{3sin\alpha-2cos\alpha}{12sin^3\alpha+4cos^3\alpha}=\dfrac{3sin\alpha-6sin\alpha}{12sin^3\alpha+108sin^3\alpha}=-\dfrac{3sin\alpha}{120sin^3\alpha}=-\dfrac{1}{40sin^2\alpha}\)
\(A=cosx+cos\left(n+y\right)+cos\left(x+2y\right)+...+cos\left(x+ny\right)=\left(n+1\right)cosn\)
\(\dfrac{sinx+sin3x+sin5x+...+sin\left(2n-1\right)x}{cosx+cos3x+cos5x+...+cos\left(2n-1\right)x}\)
\(=tan\left(nx\right)\)
\(sinx+sin2x+sin3x+...+sinnx\)
\(=\dfrac{sin\dfrac{nx}{2}sin\dfrac{\left(n+1\right)x}{2}}{sin\dfrac{x}{2}}\)
\(cosx+cos2x+cos3x+cosnx\)
\(=\dfrac{sin\dfrac{nx}{2}cos\dfrac{\left(n+1\right)x}{2}}{sin\dfrac{x}{2}}\)