438 Найти значение выражения: 1) sin COS - sin cos cos; 4 4 3 6 2) 2 tg2-ctg²- sin cos; 6 6 3 3) tg- ctg ctg+ tg; 4 3 4 6 4) 2 cos2 π 6 - sin²+tg tgctg ctg T 3 6 3

1) sin $$ \frac{\pi}{4} $$ cos $$ \frac{\pi}{4} $$ - sin $$ \frac{\pi}{3} $$ cos $$ \frac{\pi}{6} $$ = $$ \frac{\sqrt{2}}{2} $$ × $$ \frac{\sqrt{2}}{2} $$ - $$ \frac{\sqrt{3}}{2} $$ × $$ \frac{\sqrt{3}}{2} $$ = $$ \frac{2}{4} $$ - $$ \frac{3}{4} $$ = -$$ \frac{1}{4} $$

Ответ: -$$ \frac{1}{4} $$

2) 2 tg² $$ \frac{\pi}{3} $$ - ctg² $$ \frac{\pi}{6} $$ - sin $$ \frac{\pi}{6} $$ cos $$ \frac{\pi}{3} $$ = 2 × (√3)² - (√3)² - $$ \frac{1}{2} $$ × $$ \frac{1}{2} $$ = 2 × 3 - 3 - $$ \frac{1}{4} $$ = 6 - 3 - $$ \frac{1}{4} $$ = 3 - $$ \frac{1}{4} $$ = $$ \frac{11}{4} $$

Ответ: $$ \frac{11}{4} $$

3) (tg $$ \frac{\pi}{4} $$ - ctg $$ \frac{\pi}{3} $$) (ctg $$ \frac{\pi}{4} $$ + tg $$ \frac{\pi}{6} $$) = (1 - $$ \frac{\sqrt{3}}{3} $$) (1 + $$ \frac{\sqrt{3}}{3} $$) = 1 - ( $$ \frac{\sqrt{3}}{3} $$)² = 1 - $$ \frac{3}{9} $$ = 1 - $$ \frac{1}{3} $$ = $$ \frac{2}{3} $$

Ответ: $$ \frac{2}{3} $$

4) 2 cos² $$ \frac{\pi}{6} $$ - sin² $$ \frac{\pi}{3} $$ + tg $$ \frac{\pi}{6} $$ ctg $$ \frac{\pi}{3} $$ = 2 × ( $$ \frac{\sqrt{3}}{2} $$)² - ( $$ \frac{\sqrt{3}}{2} $$)² + $$ \frac{\sqrt{3}}{3} $$ × $$ \frac{\sqrt{3}}{3} $$ = 2 × $$ \frac{3}{4} $$ - $$ \frac{3}{4} $$ + $$ \frac{3}{9} $$ = $$ \frac{6}{4} $$ - $$ \frac{3}{4} $$ + $$ \frac{1}{3} $$ = $$ \frac{3}{4} $$ + $$ \frac{1}{3} $$ = $$ \frac{9 + 4}{12} $$ = $$ \frac{13}{12} $$

Ответ: $$ \frac{13}{12} $$