ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 781

\[1)\ \frac{\sqrt{2} - \cos x - \sin x}{\sin x - \cos x} =\]

\[= \frac{\sqrt{2} - \left( \sin\left( \frac{\pi}{2} - x \right) + \sin x \right)}{\sin x - \sin\left( \frac{\pi}{2} - x \right)} =\]

\[= \frac{\sqrt{2} - 2 \bullet \sin\frac{\pi}{4} \bullet \cos\left( \frac{\pi}{4} - x \right)}{2 \bullet \sin\left( x - \frac{\pi}{4} \right) \bullet \cos\frac{\pi}{4}} =\]

\[= \frac{\sqrt{2} - 2 \bullet \frac{\sqrt{2}}{2} \bullet \cos\left( \frac{\pi}{4} - x \right)}{2 \bullet \frac{\sqrt{2}}{2} \bullet \sin\left( x - \frac{\pi}{4} \right)} =\]

\[= \frac{\sqrt{2} - \sqrt{2}\cos\left( \frac{\pi}{4} - x \right)}{\sqrt{2}\sin\left( x - \frac{\pi}{4} \right)} =\]

\[= \frac{1 - \cos\left( x - \frac{\pi}{4} \right)}{\sin\left( x - \frac{\pi}{4} \right)} =\]

\[= \frac{2\sin^{2}\left( \frac{x}{2} - \frac{\pi}{8} \right)}{2\sin\left( \frac{x}{2} - \frac{\pi}{8} \right) \bullet \cos\left( \frac{x}{2} - \frac{\pi}{8} \right)} =\]

\[= \frac{\sin\left( \frac{x}{2} - \frac{\pi}{8} \right)}{\cos\left( \frac{x}{2} - \frac{\pi}{8} \right)} = tg\left( \frac{x}{2} - \frac{\pi}{8} \right).\]

\[2)\ \frac{1 + \cos x + \sin x + tg\ x}{\sin x + \cos x} =\]

\[= \frac{1 + \cos x}{\cos x} = 1 + \frac{1}{\cos x}.\]