Question

10 вариант, задание 6

Answer 1

Ответ:3

Объяснение:

Sin4a=2sin2a * cos 2a = 4sina*cosa(cos²a-sin²a) Полное решение на фото

Answer 2

$frac{2Sin2alpha-Sin4alpha}{2Sin2alpha+Sin4alpha}=frac{2Sin2alpha-2Sin2alpha Cos2alpha}{2Sin2alpha+2Sin2alpha Cos2alpha}=frac{2Sin2alpha(1-Cos2alpha)}{2Sin2alpha(1+Cos2alpha)}=frac{1-Cos2alpha}{1+Cos2alpha}=frac{1-(2Cos^{2}alpha-1)}{1+(2Cos^{2}alpha-1)}=frac{1-2Cos^{2}alpha +1}{1+2Cos^{2}alpha-1}=frac{2(1-Cos^{2}alpha)}{2Cos^{2}alpha}$

$frac{2(1-(-frac{1}{2})^{2})}{(2*(-frac{1}{2})^{2})}=frac{2(1-frac{1}{4}) }{frac{1}{2} }=4*frac{3}{4}=3$