13106 (no MarepHaJIaM EI 3) BbrpaxeHHX 36sqrt (6)tg(pi )/(6)sin(pi )/(4) 4sqrt (2)cos(pi )/(4)cos(7pi )/(3) -4sqrt (3)cos(-750^circ ) 2sqrt (3)tg(-300^circ ) (14sin409^circ )/(sin49^circ ) sqrt (3)cdot sin(pi )/(3)cdot cos2pi +sqrt (2)cos(pi )/(4)cdot sin(3pi )/(2)

Let's go through each expression step by step:<br /><br />1. $36\sqrt {6}tg\frac {\pi }{6}sin\frac {\pi }{4}$<br /> - $tg\frac {\pi }{6} = \frac{1}{\sqrt{3}}$<br /> - $sin\frac {\pi }{4} = \frac{\sqrt{2}}{2}$<br /> - $36\sqrt {6} \cdot \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{2}}{2} = 36\sqrt{2} \cdot \frac{1}{\sqrt{3}} = 36\sqrt{\frac{2}{3}} = 36 \cdot \frac{\sqrt{6}}{3} = 12\sqrt{6}$<br /><br />2. $4\sqrt {2}cos\frac {\pi }{4}cos\frac {7\pi }{3}$<br /> - $cos\frac {\pi }{4} = \frac{\sqrt{2}}{2}$<br /> - $cos\frac {7\pi }{3} = cos(2\pi + \frac{\pi}{3}) = cos\frac{\pi}{3} = \frac{1}{2}$<br /> - $4\sqrt{2} \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = 4 \cdot \frac{1}{2} = 2$<br /><br />3. $-4\sqrt {3}cos(-750^{\circ })$<br /> - $cos(-750^{\circ}) = cos(-2\pi \cdot 375) = cos(0) = 1$<br /> - $-4\sqrt{3} \cdot 1 = -4\sqrt{3}$<br /><br />4. $2\sqrt {3}tg(-300^{\circ })$<br /> - $tg(-300^{\circ}) = tg(-360^{\circ} + 60^{\circ}) = tg60^{\circ} = \sqrt{3}$<br /> - $2\sqrt{3} \cdot \sqrt{3} = 6$<br /><br />5. $\frac {14sin409^{\circ }}{sin49^{\circ }}$<br /> - $sin409^{\circ} = sin(360^{\circ} + 49^{\circ}) = sin49^{\circ}$<br /> - $\frac{14 \cdot sin49^{\circ}}{sin49^{\circ}} = 14$<br /><br />6. $\sqrt {3}\cdot sin\frac {\pi }{3}\cdot cos2\pi +\sqrt {2}cos\frac {\pi }{4}\cdot sin\frac {3\pi }{2}$<br /> - $sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$<br /> - $cos2\pi = 1$<br /> - $sin\frac{3\pi}{2} = -1$<br /> - $\sqrt{3} \cdot \frac{\sqrt{3}}{2} \cdot 1 + \sqrt{2} \cdot \frac{\sqrt{2}}{2} \cdot (-1) = \frac{3}{2} - \frac{1}{2} = 1$<br /><br />So, the final answers are:<br />1. $12\sqrt{6}$<br />2. $2$<br />3. $-4\sqrt{3}$<br />4. $6$<br />5. $14$<br />6. $1$