\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\)
=>\(\dfrac{1}{cos^2\alpha}=1+\left(\dfrac{5}{12}\right)^2=1+\dfrac{25}{144}=\dfrac{169}{144}\)
=>\(cos^2\alpha=\dfrac{144}{169}\)
mà \(cos\alpha< 0\left(\Omega< \alpha< \dfrac{3}{2}\Omega\right)\)
nên \(cos\alpha=-\dfrac{12}{13}\)
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)
=>\(sin\alpha=-\dfrac{12}{13}\cdot\dfrac{5}{12}=-\dfrac{5}{13}\)
\(\Omega< \alpha< \dfrac{3}{2}\Omega\)
=>\(\dfrac{\Omega}{2}< \dfrac{\alpha}{2}< \dfrac{3}{4}\Omega\)
=>\(cos\left(\dfrac{\alpha}{2}\right)< 0;sin\left(\dfrac{\alpha}{2}\right)>0\)
\(cos\alpha=cos\left(2\cdot\dfrac{\alpha}{2}\right)=2\cdot cos^2\left(\dfrac{\alpha}{2}\right)-1\)
=>\(2\cdot cos^2\left(\dfrac{\alpha}{2}\right)-1=-\dfrac{12}{13}\)
=>\(2\cdot cos^2\left(\dfrac{\alpha}{2}\right)=1-\dfrac{12}{13}=\dfrac{1}{13}\)
=>\(cos^2\left(\dfrac{\alpha}{2}\right)=\dfrac{1}{26}\)
mà \(cos\left(\dfrac{\alpha}{2}\right)< 0\)
nên \(cos\left(\dfrac{\alpha}{2}\right)=-\dfrac{1}{\sqrt{26}}\)
\(sin^2\left(\dfrac{\alpha}{2}\right)+cos^2\left(\dfrac{\alpha}{2}\right)=1\)
=>\(sin^2\left(\dfrac{\alpha}{2}\right)=1-\left(\dfrac{1}{26}\right)=\dfrac{25}{26}\)
mà \(sin\left(\dfrac{\alpha}{2}\right)>0\)
nên \(sin\left(\dfrac{\alpha}{2}\right)=\dfrac{5}{\sqrt{26}}\)
\(cos\left(\dfrac{3}{2}\alpha\right)=cos\left(\alpha+\dfrac{1}{2}\alpha\right)\)
\(=cos\alpha\cdot cos\left(\dfrac{1}{2}\alpha\right)-sin\alpha\cdot sin\left(\dfrac{1}{2}\alpha\right)\)
\(=-\dfrac{12}{13}\cdot\dfrac{-1}{\sqrt{26}}-\dfrac{5}{\sqrt{26}}\cdot\dfrac{-5}{13}=\dfrac{12}{13\sqrt{26}}+\dfrac{25}{13\sqrt{26}}=\dfrac{37}{13\sqrt{26}}\)
