\(A=2\cdot sin\left(\dfrac{13\Omega}{3}\right)+cos\left(\dfrac{16\Omega}{6}\right)-3\cdot tan\left(\dfrac{5\Omega}{4}\right)\)
\(=2\cdot sin\left(4\Omega+\dfrac{\Omega}{3}\right)+cos\left(2\Omega+\dfrac{2}{3}\Omega\right)-3\cdot tan\left(\Omega+\dfrac{\Omega}{4}\right)\)
\(=2\cdot sin\left(\dfrac{\Omega}{3}\right)+cos\left(\dfrac{2\Omega}{3}\right)-3\cdot tan\left(\dfrac{\Omega}{4}\right)\)
\(=2\cdot\dfrac{\sqrt{3}}{2}+\dfrac{-1}{2}-3=\dfrac{2\sqrt{3}-1}{2}-3=\dfrac{2\sqrt{3}-7}{2}\)
\(B=cos\left(-\dfrac{\Omega}{6}\right)-2\cdot sin\left(\dfrac{2\Omega}{3}\right)+4\cdot sin\left(\dfrac{\Omega}{5}\right)\cdot sin\left(\Omega\right)\)
\(=cos\left(\dfrac{\Omega}{6}\right)-2\cdot sin\left(\Omega-\dfrac{1}{3}\Omega\right)+4\cdot sin\left(\dfrac{\Omega}{5}\right)\cdot0\)
\(=\dfrac{1}{2}-2\cdot sin\left(\dfrac{1}{3}\Omega\right)=\dfrac{1}{2}-2\cdot\dfrac{1}{2}=\dfrac{1}{2}-1=-\dfrac{1}{2}\)
\(C=2\cdot sin390^0-3\cdot tan225^0+cot120^0\)
\(=2\cdot sin\left(360^0+30^0\right)-3\cdot tan\left(180^0+45^0\right)+cot\left(180^0-120^0\right)\)
\(=2\cdot sin30^0-3\cdot tan45^0+\left(-1\right)\cdot cot120^0\)
\(=2\cdot\dfrac{1}{2}-3\cdot1+\left(-1\right)\cdot\dfrac{-\sqrt{3}}{3}=1-3+\dfrac{\sqrt{3}}{3}=-2+\dfrac{\sqrt{3}}{3}=\dfrac{-6+\sqrt{3}}{3}\)
\(D=\dfrac{sin130^0-cos220^0}{cos50^0\cdot cot320^0}\)
\(=\dfrac{sin130^0-cos\left(130^0+90^0\right)}{cos50^0\cdot cot\left(270^0+50^0\right)}\)
\(=\dfrac{sin130^0-cos130^0\cdot cos90^0+sin130^0\cdot sin90^0}{cos50^0\cdot cot320^0}\)
\(=\dfrac{sin130^0+sin130^0}{cos50^0\cdot cot\left(360^0-40^0\right)}=\dfrac{2\cdot sin130^0}{cos50^0\cdot cot\left(-40^0\right)}\)
\(=\dfrac{2\cdot sin\left(180^0-50^0\right)}{cos50^0\cdot\left(-1\right)\cdot cot40^0}=-\dfrac{2\cdot sin50^0}{cos50^0\cdot tan50^0}=-2\)
mọi người giúp em với em cảm ơn nhiều lắmmmmm
