Question

cos n/6-√2 sin n/4+√3 tg n/3

Answer 1

Cos n/6-√2 sin n/4+√3 tg n/3 = 0,5√3 - √2 · 0,5√2 + √3 ·√3 =
= 0,5√3 - 1 + 3 = 2 + 0,5√3 ≈ 2,866

Answer 2

$cos \frac{\pi}{6} - \sqrt{2} sin \frac{\pi}{4} + \sqrt{3} tg \frac{\pi}{3} = \\ = \frac{ \sqrt{3} }{2} - \sqrt{2} \frac{ \sqrt{2} }{2} + \sqrt{3} \times \sqrt{3} = \\ = \frac{ \sqrt{3} }{2} - 1 + 3 = \frac{ \sqrt{3} }{2} + 2 = \\ = \frac{ \sqrt{3} + 4 }{2}$