\(A=cos3a+2cos\left(\pi-3a\right)sin^2\left(\dfrac{\pi}{4}-1,5a\right)\)
\(=cos3a-2cos3a\dfrac{1-cos\left(\dfrac{\pi}{2}-3a\right)}{2}\)
\(=cos3a-cos3a\left(1-sin3a\right)\)
\(=cos3a-cos3a+cos3asin3a=\dfrac{1}{2}sin6a\)
\(=\dfrac{1}{2}sin\left(6\dfrac{5\pi}{6}\right)=\dfrac{1}{2}sin\left(4\pi+\pi\right)=\dfrac{1}{2}sin\pi=0\)
Vì a=\(\dfrac{5\pi}{6}\) nên: \(3a=\dfrac{5\pi}{2}\) => \(\cos3a=0\)
\(\pi-3a=\pi-\dfrac{5\pi}{2}=\dfrac{-3\pi}{2}\)
=> \(\cos\left(\pi-3a\right)=0\)
ta có : \(cos\left(\Pi-3a\right)=-cosa\)
\(sin^2\left(\dfrac{\Pi}{4}-1,5a\right)=\dfrac{1-cos\left(\dfrac{\Pi}{2}-3a\right)}{2}=\dfrac{1-cos3a}{2}\)
\(\Rightarrow cos3a+2cos\left(\Pi-3a\right)sin^2\left(\dfrac{\Pi}{4}-1,5a\right)=cos3a-2cos3a\left(\dfrac{1-cos3a}{2}\right)\) =\(cos^23a=cos^23.\dfrac{5\Pi}{6}=cos^2\dfrac{5\Pi}{2}=cos^2\dfrac{\Pi}{2}=0\)
