Question

Скажите, , как получилось такое преобразование? тригонометрия

Answer 1

$\frac{2\sin^2(\frac{\pi}{4} - \alpha)}{2\cos^2(\frac{\pi}{4} - \alpha)} = \frac{2\frac{1 - \cos(\frac{\pi}{2} - 2\alpha)}{2}}{2\frac{1 + \cos(\frac{\pi}{2} - 2\alpha)}{2}} = \frac{1 - \cos(\frac{\pi}{2} - 2\alpha)}{1 + \cos(\frac{\pi}{2} - 2\alpha)} = \frac{1 - \sin(2\alpha)}{1 + \sin(2\alpha)} = \frac{1 - 2\sin(\alpha)\cos(\alpha)}{1 + 2\sin(\alpha)\cos(\alpha)} = \\\\= \frac{\sin^2(\alpha) + \cos^2(\alpha) - 2\sin(\alpha)\cos(\alpha)}{\sin^2(\alpha) + \cos^2(\alpha) + 2\sin(\alpha)\cos(\alpha)} =$

$= \frac{(\cos(\alpha) - \sin(\alpha))^2}{(\cos(\alpha) + \sin(\alpha))^2}$