Ответ: -2,5

Объяснение:

[tex]1) ~ \sin (-x )= -\sin x \\\\ 2) ~ \cos (-x)=\cos x[/tex]

[tex]\rm \displaystyle 4 \cos (-\tfrac{\pi }{6} ) \cdot \sin (-\tfrac{\pi }{3} ) + tg^2 (-\tfrac{\pi }{4} ) -\cos ^2 (-\tfrac{\pi }{4} ) = \\\\\\ \displaystyle 4 \cos (\tfrac{\pi }{6} ) \cdot (-\sin (\tfrac{\pi }{3} ) )+ tg^2 (-\tfrac{\pi }{4} ) -\cos ^2 (\tfrac{\pi }{4} ) = \\\\\\ 4\cdot \frac{\sqrt{3} }{2} \cdot \bigg (- \frac{\sqrt{3} }{2} \bigg ) + (-1)^2 - \bigg (\frac{\sqrt{2} }{2} \bigg )^2 = \\\\\\ -3+1-\frac{1}{2} = -2,5[/tex]