Giải bài tập 1.10 trang 10 SBT toán 11 tập 1 kết nối

$cos105^{o}=cos(60^{o}+45^{o})=cos60^{o}cos45^{o}-sin60^{o}sin45^{o}$

$=\frac{1}{2}.\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{2}-\sqrt{6}}{4}$

$sin105^{o}=sin(60^{o}+45^{o})=sin60^{o}cos45^{o}+cos60^{o}sin45^{o}$

$=\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}+\frac{1}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{2}+\sqrt{6}}{4}$

$tan105^{o}=\frac{sin105^{o}}{cos105^{o}}=\frac{\sqrt{2}+\sqrt{6}}{\sqrt{2}-\sqrt{6}}$

$cot105^{o}=\frac{1}{tan105^{o}}=\frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}+\sqrt{6}}$