5. Доказать тождество: (√2 cos α - 2 cos(π/4 - α)) / (2 sin(π/4 + α) - √3 sin α) = -√2 tg α.

Решение:

Рассмотрим левую часть тождества:

\( \frac{\sqrt{2} \cos \alpha - 2 \cos\(\frac{\pi}{4} - \alpha\)}{2 \(\sin\)\(\frac{\pi}{4} + \alpha\) - \(\sqrt{3}\) \(\sin\) \(\alpha\)} = \(\frac\){\(\sqrt{2}\) \(\cos\) \(\alpha\) - 2 \(\cos \frac{\pi}{4} \cos \alpha + \sin \frac{\pi}{4} \sin \alpha\)}{2 \(\sin \frac{\pi}{4} \cos \alpha + \cos \frac{\pi}{4} \sin \alpha\) - \(\sqrt{3}\) \(\sin\) \(\alpha\)} = \(\frac\){\(\sqrt{2}\) \(\cos\) \(\alpha\) - 2 \(\frac{\sqrt{2}}{2} \cos \alpha + \frac{\sqrt{2}}{2} \sin \alpha\)}{2 \(\frac{\sqrt{2}}{2} \cos \alpha + \frac{\sqrt{2}}{2} \sin \alpha\) - \(\sqrt{3}\) \(\sin\) \(\alpha\)} = \(\frac{\sqrt{2} \cos \alpha - \sqrt{2} \cos \alpha - \sqrt{2} \sin \alpha}{\sqrt{2} \cos \alpha + \sqrt{2} \sin \alpha - \sqrt{3} \sin \alpha}\) = \(\frac{-\sqrt{2} \sin \alpha}\){\(\sqrt{2}\) \(\cos\) \(\alpha\) + \(\sqrt{2} - \sqrt{3}\) \(\sin\) \(\alpha\)}

Далее, для правой части:

\( -\(\sqrt{2}\) \(\operatorname{tg}\) \(\alpha\) = -\(\sqrt{2}\) \(\frac{\sin \alpha}{\cos \alpha}\) = \(\frac{-\sqrt{2} \sin \alpha}{\cos \alpha}\)

Из-за несовпадения знаменателей в левой и правой частях, тождество не доказано.

Ответ: Тождество не доказано.