10 вариант, задание 6. Created by apietrov. algebra-ru

10 вариант, задание 6...

Codelobster

$\frac{2Sin2\alpha-Sin4\alpha}{2Sin2\alpha+Sin4\alpha}=\frac{2Sin2\alpha-2Sin2\alpha Cos2\alpha}{2Sin2\alpha+2Sin2\alpha Cos2\alpha}=\frac{2Sin2\alpha(1-Cos2\alpha)}{2Sin2\alpha(1+Cos2\alpha)}=\frac{1-Cos2\alpha}{1+Cos2\alpha}=\frac{1-(2Cos^{2}\alpha-1)}{1+(2Cos^{2}\alpha-1)}=\frac{1-2Cos^{2}\alpha +1}{1+2Cos^{2}\alpha-1}=\frac{2(1-Cos^{2}\alpha)}{2Cos^{2}\alpha}$

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

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Ответ:3

Объяснение:

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

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