Question

срочно Алгебру 5 задания​

Answer 1

Ответ:

$sina=\dfrac{4}{5}\\\\\dfrac{\pi }{2}<a<\pi \ \ \ \Rightarrow \ \ \ cosa<0\ \ \Rightarrow \ \ \ cosa=-\sqrt{1-sin^2a}\\\\cosa=-\sqrt{1-\dfrac{16}{25}}=-\sqrt{\dfrac{9}{25}}=-\dfrac{3}{5}\\\\\\sin2a=2\cdot sina\cdot cosa=2\cdot \dfrac{4}{5}\cdot (-\dfrac{3}{5})=-\dfrac{24}{25}\\\\cos2a=cos^2a-sin^2a=\dfrac{9}{25}-\dfrac{16}{25}=-\dfrac{7}{25}$