Задача 23409 ...
Условие
и a_{n+1}=3a_n-4a_n^3a Найдите (a_1a_{2018})^2
Решение
sin^2(Pi/5)=1-cos^2(Pi/5)=1-(1+2sqrt(5)+5)/16=
=(10-2sqrt(5))/16=(5-sqrt(5)/8
sin(Pi/5)=sqrt((5-sqrt(5))/8)
формула
sin^3 альфа =(3sin альфа -sin3 альфа )/4
a1=sin(Pi/5)
a2=3sin(Pi/5)-4(sin(Pi/5))^3=sin(3Pi/5)
a_(3)=3sin(3Pi/5)-4(sin(3Pi/5))^3=sin(9Pi/5)
a_(4)=3sin(9Pi/5)-4(sin(9Pi/5))^3=sin(27Pi/5)
a_(5)=sin(81Pi/5)=sin(Pi/5)
цикл повторится
значит
a_(2018)=sin(3^(2017)*Pi/5)=sin((3^(2016)*3)Pi/5)=
=sin(3Pi/5)=sin(2Pi/5) - синус двойного угла=
a_(1)*a_(2018)=
=sqrt((5-sqrt(5))/8)*2sqrt((5-sqrt(5))/8)*((1+sqrt(5))/4)=
=2*((5-sqrt(5))/8)*((1+sqrt(5))/4)=
=sqrt(5) *((1-sqrt(5))/4)*((1+sqrt(5))/4)=
=sqrt(5)*(1-5)/16=-sqrt(5)/4
(a_(1)*a_(2018))^2=5/16