Question

Решите номер 22 под цифрой 3 и 5 пожалуйста.

Answer 1

$-cos720^0+mathrm{tg}30^0-mathrm{ctg}210^0+sin120^0= \ =-cos(2cdot360^0)+mathrm{tg}30^0-mathrm{ctg}(180^0+30^0)+sin(180^0-60^0)= \ =-cos0^0+mathrm{tg}30^0-mathrm{ctg}30^0+sin60^0= -1+ frac{ sqrt{3} }{3}- sqrt{3} + frac{ sqrt{3} }{2} = \ =- frac{6}{6} + frac{2 sqrt{3} }{6}- frac{6sqrt{3} }{6} + frac{3 sqrt{3} }{6} = frac{-6+2 sqrt{3}-6sqrt{3}+ 3 sqrt{3}}{6}=frac{-6-sqrt{3}}{6}$

$-9cos frac{ pi }{2} +3sin frac{ pi }{2} -3mathrm{ctg} frac{5 pi }{4} -5mathrm{tg}0^0= \ =-9cdot0 +3cdot1 -3cdot1 -5cdot0=0+3-3-0=0$