Вычислите cosa, если ctga= -корень из 3 и 3π/2<a<2π.

\[ctg\ a = - \sqrt{3};\ \ \ \]

\[\frac{3\pi}{2} < a < 2\pi \Longrightarrow \sin a < 0;\ \ \]

\[\cos a > 0\]

\[\sin a = \sqrt{\frac{1}{\text{ct}g^{2}a + 1}} =\]

\[= \sqrt{\frac{1}{\left( - \sqrt{3} \right)^{2} + 1}} = \sqrt{\frac{1}{3 + 1}} =\]

\[= \sqrt{\frac{1}{4}} = \left| \frac{1}{2} \right| = - \frac{1}{2}\]

\[\cos a = \sqrt{1 - \sin^{2}a} =\]

\[{= \sqrt{1 - \left( - \frac{1}{2} \right)^{2}} = \sqrt{1 - \frac{1}{4}} = }{= \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\text{\ \ }}\]