Bài 1:
a: \(\sin15^0=\sin\left(45^0-30^0\right)=\sin45^0\cdot cos30^0-cos45^0\cdot\sin30^0\)
\(=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6-\sqrt2}{4}\)
\(cos15^0=cos\left(45^0-30^0\right)=cos45^0\cdot cos30^0+\sin45^0\cdot\sin30^0\)
\(=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6+\sqrt2}{4}\)
\(\tan15^0=\frac{\sin15^0}{cos15^0}=\frac{\sqrt6-\sqrt2}{4}:\frac{\sqrt6+\sqrt2}{4}=\frac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\)
\(=\frac{\left(\sqrt6-\sqrt2\right)\left(\sqrt6-\sqrt2\right)}{\left(\sqrt6+\sqrt2\right)\left(\sqrt6-\sqrt2\right)}=\frac{8-2\sqrt{12}}{6-2}=\frac{8-4\sqrt3}{4}=2-\sqrt3\)
\(\cot15^0=\frac{1}{\tan15^0}=\frac{1}{2-\sqrt3}=2+\sqrt3\)
\(\sin75^0=cos\left(90^0-75^0\right)=cos15^0=\frac{\sqrt6+\sqrt2}{4}\)
\(cos75^0=\sin15^0=\frac{\sqrt6-\sqrt2}{4}\)
\(\tan75^0=\cot15^0=2+\sqrt3\)
\(\cot75^0=\tan15^0=2-\sqrt3\)
\(\sin105^0=\sin\left(90^0+15^0\right)=cos15^0=\frac{\sqrt6+\sqrt2}{4}\)
\(cos105^0=cos\left(90^0+15^0\right)=-\sin15^0=\frac{-\sqrt6+\sqrt2}{4}\)
\(\tan105^0=\frac{\sin105^0}{cos105^0}=\frac{cos15^0}{-\sin15^0}=-\cot15^0=-2-\sqrt3\)
\(\cot105^0=\frac{1}{\tan105^0}=\frac{1}{-2-\sqrt3}=\frac{-1}{2+\sqrt3}=-\frac{-1\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}=-2+\sqrt3\)
b: \(\sin\left(\frac{\pi}{12}\right)=\sin\left(\frac{\pi}{4}-\frac{\pi}{6}\right)=\sin\frac{\pi}{4}\cdot cos\left(\frac{\pi}{6}\right)-cos\left(\frac{\pi}{4}\right)\cdot\sin\left(\frac{\pi}{6}\right)\)
\(=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6-\sqrt2}{4}\)
\(cos\left(\frac{\pi}{12}\right)=cos\left(\frac{\pi}{4}-\frac{\pi}{6}\right)=cos\left(\frac{\pi}{4}\right)\cdot cos\left(\frac{\pi}{6}\right)+\sin\left(\frac{\pi}{4}\right)\cdot\sin\left(\frac{\pi}{6}\right)\)
\(=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6+\sqrt2}{4}\)
\(\tan\left(\frac{\pi}{12}\right)=\frac{\sin\left(\frac{\pi}{12}\right)}{cos\left(\frac{\pi}{12}\right)}=\frac{\sqrt6-\sqrt2}{4}:\frac{\sqrt6+\sqrt2}{4}=\frac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\)
\(=\frac{\left(\sqrt6-\sqrt2\right)\left(\sqrt6-\sqrt2\right)}{\left(\sqrt6+\sqrt2\right)\left(\sqrt6-\sqrt2\right)}=\frac{8-2\sqrt{12}}{6-2}=\frac{8-4\sqrt3}{4}=2-\sqrt3\)
\(\cot\left(\frac{\pi}{12}\right)=\frac{1}{\tan\left(\frac{\pi}{12}\right)}=\frac{1}{2-\sqrt3}=2+\sqrt3\)
\(\sin\left(\frac{5}{12}\pi\right)=cos\left(\frac{\pi}{2}-\frac{5}{12}\pi\right)=cos\left(\frac{\pi}{12}\right)=\frac{\sqrt6+\sqrt2}{4}\)
\(cos\left(\frac{5}{12}\pi\right)=\sin\left(\frac{\pi}{12}_{}\right)=\frac{\sqrt6-\sqrt2}{4}\)
\(\tan\left(\frac{5}{12}\pi\right)=\cot\left(\frac{\pi}{12}\right)=2+\sqrt3\)
\(\cot\left(\frac{5}{12}\pi\right)=\tan\left(\frac{\pi}{12}_{}\right)=2-\sqrt3\)
\(\sin\left(\frac{7}{12}\pi\right)=\sin\left(\frac{\pi}{2}+\frac{\pi}{12}\right)=cos\left(\frac{\pi}{12}\right)_{}=\frac{\sqrt6+\sqrt2}{4}\)
\(cos\left(\frac{7}{12}\pi\right)=cos\left(\frac{\pi}{2}+\frac{\pi}{12}\right)=-\sin\left(\frac{\pi}{12}\right)=\frac{-\sqrt6+\sqrt2}{4}\)
\(\tan\left(\frac{7}{12}\pi\right)=\frac{\sin\left(\frac{7}{12}\pi\right)}{cos\left(\frac{7}{12}\pi\right)}=\frac{cos\left(\frac{\pi}{12}\right)}{-\sin\left(\frac{\pi}{12}\right)}=-\cot15^0=-2-\sqrt3\)
\(\cot\left(\frac{7}{12}\pi\right)=\frac{1}{\tan\left(\frac{7}{12}\pi\right)}=\frac{1}{-2-\sqrt3}=\frac{-1}{2+\sqrt3}=-\frac{-1\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}=-2+\sqrt3\)
