Ta có \(cosa=\sqrt{1-sin^2a}=\pm\dfrac{5}{13}\)
Do 3pi/2 < a < 2pi => cosa = 5/13
\(cos2a=cos^2a-sin^2a=\left(\dfrac{5}{13}\right)^2-\left(-\dfrac{12}{13}\right)^2=-\dfrac{119}{169}\)
\(sin2a=2sina.cosa=\dfrac{2.5}{13}.\left(-\dfrac{12}{13}\right)=-\dfrac{120}{169}\)
\(cos\left(\dfrac{\pi}{3}-a\right)=cos\left(\dfrac{\pi}{3}\right).cosa+sin\left(\dfrac{\pi}{3}\right).sina=\dfrac{-12+5\sqrt{3}}{26}\)
\(\dfrac{3}{2}\Omega< \alpha< 2\Omega\)
=>\(cos\alpha>0\)
=>\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\left(-\dfrac{12}{13}\right)^2}=\dfrac{5}{13}\)
\(cos2\alpha=2\cdot cos^2\alpha-1=2\cdot\left(\dfrac{5}{13}\right)^2-1\)
\(=2\cdot\dfrac{25}{169}-1=\dfrac{50}{169}-1=-\dfrac{119}{169}\)
\(sin2\alpha=2\cdot sin\alpha\cdot cos\alpha=2\cdot\dfrac{5}{13}\cdot\dfrac{-12}{13}=\dfrac{-120}{169}\)
\(cos\left(\dfrac{\Omega}{3}-\alpha\right)=cos\left(\dfrac{\Omega}{3}\right)\cdot cos\alpha+sin\left(\dfrac{\Omega}{3}\right)\cdot sin\alpha\)
\(=\dfrac{1}{2}\cdot\dfrac{5}{13}+\dfrac{-12}{13}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{5-12\sqrt{3}}{26}\)
