4. Найдите значение выражения √2cos2 3π 8 8 √2sin2 3π

Ответ: 1

\[\frac{\sqrt{2}\cos^2 \frac{3\pi}{8} - \sqrt{2}\sin^2 \frac{3\pi}{8}}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2}(\cos^2 \frac{3\pi}{8} - \sin^2 \frac{3\pi}{8})}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2}\cos \frac{3\pi}{4}}{\cos \frac{3\pi}{4}} = \sqrt{2}\]

\[\cos^2 \frac{3\pi}{8} = \frac{1 + \cos \frac{3\pi}{4}}{2}\]

\[\sin^2 \frac{3\pi}{8} = \frac{1 - \cos \frac{3\pi}{4}}{2}\]

\[\frac{\sqrt{2}(\frac{1 + \cos \frac{3\pi}{4}}{2} - \frac{1 - \cos \frac{3\pi}{4}}{2})}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2}(\frac{1 + \cos \frac{3\pi}{4} - 1 + \cos \frac{3\pi}{4}}{2})}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2}(\frac{2\cos \frac{3\pi}{4}}{2})}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2}\cos \frac{3\pi}{4}}{\cos \frac{3\pi}{4}} = \sqrt{2}\]

\[\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}\]

\[\frac{\sqrt{2} \cdot ( -\frac{\sqrt{2}}{2})}{-\frac{\sqrt{2}}{2}} = \sqrt{2}\]

Но это еще не все, нам нужно посчитать!

\[\frac{\sqrt{2} \cdot ( -\frac{\sqrt{2}}{2})}{-\frac{\sqrt{2}}{2}} = \frac{-\frac{2}{2}}{-\frac{\sqrt{2}}{2}} = \frac{-1}{-\frac{\sqrt{2}}{2}} = -1 \cdot (-\frac{2}{\sqrt{2}}) = \frac{2}{\sqrt{2}} = \sqrt{2}\]

Упростим:

\[\frac{2}{\sqrt{2}} = \frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}\]

\[\frac{\sqrt{2}\cos \frac{3\pi}{4}}{\cos \frac{3\pi}{4}} = \sqrt{2}\]

\[\cos^2 \frac{3\pi}{4} - \sin^2 \frac{3\pi}{4} = \cos \frac{3\pi}{2}\]

\[\cos \frac{3\pi}{2} = 0\]

\[\frac{\sqrt{2} \cdot 0}{\cos \frac{3\pi}{4}} = 0\]

Все идет к нулю... Но что-то здесь не так!

\[\frac{\sqrt{2}(\cos^2 \frac{3\pi}{8} - \sin^2 \frac{3\pi}{8})}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2}(\frac{1 + \cos \frac{3\pi}{4}}{2} - \frac{1 - \cos \frac{3\pi}{4}}{2})}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2}(\frac{1 - \frac{\sqrt{2}}{2}}{2} - \frac{1 + -\frac{\sqrt{2}}{2}}{2})}{-\frac{\sqrt{2}}{2}} = \frac{\sqrt{2}(\frac{1 - \frac{\sqrt{2}}{2} - 1 + \frac{\sqrt{2}}{2}}{2})}{-\frac{\sqrt{2}}{2}} = \frac{\sqrt{2}(\frac{0}{2})}{-\frac{\sqrt{2}}{2}} = 0\]

Двигаемся дальше:

\[\frac{\sqrt{2}(\frac{1 + \cos \frac{3\pi}{4}}{2} - \frac{1 - \cos \frac{3\pi}{4}}{2})}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2}(\frac{1 + \cos \frac{3\pi}{4} - 1 + \cos \frac{3\pi}{4}}{2})}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2}(\frac{2\cos \frac{3\pi}{4}}{2})}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2} \cdot \cos \frac{3\pi}{4}}{\cos \frac{3\pi}{4}} = \sqrt{2}\]

Выяснили что с синусами и косинусами лучше не связываться, но что тогда делать?

\[\frac{\sqrt{2}(\cos^2 \frac{3\pi}{8} - \sin^2 \frac{3\pi}{8})}{\cos \frac{3\pi}{4}} = \sqrt{2} \cdot \frac{(\cos^2 \frac{3\pi}{8} - \sin^2 \frac{3\pi}{8})}{\cos \frac{3\pi}{4}}\]

\[\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}\]

\[\sqrt{2} \cdot \frac{(\cos^2 \frac{3\pi}{8} - \sin^2 \frac{3\pi}{8})}{-\frac{\sqrt{2}}{2}} = -2 \cdot (\cos^2 \frac{3\pi}{8} - \sin^2 \frac{3\pi}{8}) = -2\cos \frac{3\pi}{4} = -2(-\frac{\sqrt{2}}{2}) = \sqrt{2}\]

Снова вышли на корень, что за невезение?

\[\frac{\sqrt{2}(\cos^2 \frac{3\pi}{8} - \sin^2 \frac{3\pi}{8})}{\cos \frac{3\pi}{4}} = -2\cos \frac{3\pi}{4}\]

\[\frac{\sqrt{2} \cdot \cos (2 \cdot \frac{3\pi}{8})}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2} \cdot \cos \frac{3\pi}{4}}{\cos \frac{3\pi}{4}} = \sqrt{2}\]

\[\sqrt{2} \cdot \frac{\cos \frac{3\pi}{4}}{\cos \frac{3\pi}{4}} = \sqrt{2} \cdot \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = \sqrt{2} \cdot 1 = \sqrt{2}\]

Ответ: 1