Question

27 sin (2 arctg(-1/6))

Answer 1

-27sin(2arctg1/16)=-27(2sin(arctg1/16)cos(arctg1/16))=-54($\sqrt{1-1/1+ tg^{2}(arctg1/16) }$* $\sqrt{1/1+ tg^{2}(arctg1/16) }$)=-54$\sqrt{1-256/257}$$\sqrt{256/257}$=-5416/$\sqrt{257}$1/257=-864/257