Question

cos(arctg(1/3)+arcctg(-корень из 3))

Answer 1

Пусть ${\rm arctg}\,\frac{1}{3}=\alpha$, тогда ${\rm tg}\, \alpha=\frac{1}{3},~~~ 0<\alpha<\frac{\pi}{2}$

$\cos^2\alpha=\dfrac{1}{1+{\rm tg}^2\alpha}=\dfrac{1}{1+\frac{1}{9}}=\dfrac{9}{10}\\ \\ \\ \cos\alpha=\dfrac{3}{\sqrt{10}};~~~~ \sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{1}{\sqrt{10}}$

$\cos(\alpha+{\rm arcctg}(-\sqrt{3}))=\cos(\alpha+\frac{5\pi}{6})=\cos\alpha\cos\frac{5\pi}{6}-\sin\alpha\sin\frac{5\pi}{6}=\\ \\ \\ =\dfrac{3}{\sqrt{10}}\cdot\bigg(-\dfrac{\sqrt{3}}{2}\bigg)-\dfrac{1}{\sqrt{10}}\cdot\dfrac{1}{2}=-\dfrac{3\sqrt{3}+1}{2\sqrt{10}}$