N№7 Найдите значение выражения 366tg-sin π π 6 4

$$36\sqrt{6}tg\frac{\pi}{6} - sin\frac{\pi}{4} = 36\sqrt{6} \cdot \frac{\sqrt{3}}{\sqrt{3}\sqrt{3}} - \frac{\sqrt{2}}{2} = 36\sqrt{6} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{2}}{2} = 36\sqrt{3} - \frac{\sqrt{2}}{2}$$.

$$36\sqrt{6} \cdot \frac{\sqrt{3}}{3} = 12\sqrt{18} = 12 \cdot 3 \sqrt{2} = 36\sqrt{2}$$

$$36\sqrt{2} - \frac{\sqrt{2}}{2} = \frac{72\sqrt{2} - \sqrt{2}}{2} = \frac{71\sqrt{2}}{2}$$

$$36\sqrt{6} \cdot \frac{\sqrt{3}}{3} - \frac{\sqrt{2}}{2} = 12\sqrt{2} - \frac{\sqrt{2}}{2} = \frac{71\sqrt{2}}{2}$$

$$tg\frac{\pi}{6} = \frac{\sqrt{3}}{3}$$.

$$sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$.

$$36\sqrt{6} \cdot \frac{\sqrt{3}}{3} - \frac{\sqrt{2}}{2} = 36 \cdot \frac{\sqrt{18}}{3} - \frac{\sqrt{2}}{2} = 12 \sqrt{9 \cdot 2} - \frac{\sqrt{2}}{2} = 12 \cdot 3 \sqrt{2} - \frac{\sqrt{2}}{2} = 36\sqrt{2} - \frac{\sqrt{2}}{2} = \frac{72\sqrt{2} - \sqrt{2}}{2} = \frac{71\sqrt{2}}{2}$$

Ответ: 3