Вычислите sin2a и cos2a, если sina=-5/13 и π<a<3π/2.

\[\sin a = - \frac{5}{13};\ \ \]

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

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

\[= \sqrt{1 - \left( - \frac{5}{13} \right)^{2}} = \sqrt{1 - \frac{25}{169}} =\]

\[= \sqrt{\frac{144}{169}} = \left| \frac{12}{13} \right| = - \frac{12}{13}\ \]

\[\sin{2a} = 2 \cdot \sin a \cdot \cos a =\]

\[= 2 \cdot \left( - \frac{5}{13} \right) \cdot \left( - \frac{12}{13} \right) = \frac{120}{169}\]

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

\[= \left( - \frac{12}{13} \right)^{2} - \left( - \frac{5}{13} \right)^{2} =\]

\[= \frac{144}{169} - \frac{25}{169} = \frac{119}{169}\]