Question

Вычислите; а)arcsin1/2+3arctg1; б)аrccos√3/2-2arcctg√3 пож

Answer 1

2 arcsin(-1/2) + arctg(-1) + arccos(sgrt2/2)= - pi/6 +(-pi/4) + pi/4= - pi/6;arctg(-sgrt3/3) + arccos(-1/2) + arctg1 = - pi/6 + 2pi/3 + pi/4 = 7pi/12
вроде так.

Answer 2

A)
$arcsin(1/2)+3arctg(1)= \frac{\pi} {6}+3* \frac{\pi } {4}= \frac{ \pi } {6}+ \frac{3 \pi }{4} = \frac{2 \pi +9 \pi }{12}= \frac{11 \pi }{12}$

B)
$arcsin( \sqrt{3}/2)-2arccot( \sqrt{3})= \frac{} \pi {3}-2* \frac{ \pi } {6}= \frac{ \pi } {3} - \frac{ \pi }{3}=0$