Tg y+ctg y+tg 3y+ctg 3y=8cos^2 2y/sin 6y - вопрос №8857710338226 от машина34 19.02.2020 15:01

Question

Алгебра: Tg y+ctg y+tg 3y+ctg 3y=8cos^2 2y/sin 6y

машина34 · Answer

tg y+ctg y+tg 3y+ctg 3y=8cos^2 2y/sin 6y
tg y+ctg y+tg 3y+ctg 3y=tgy + 1/tgy + tg3y + 1/tg3y =
= (tg²y +1)/tgy + (tg²3y +1)/tg3y=
=1/(Cos²ytgy) + 1/(Cos²3ytg3y) = 1/(CosySiny) + 1/Cos3ySin3y=
= 2/Sin2y + 2/Sin6y=2(1/Sin2y + 1/Sin6y) = 2*(Sin6y + Sin2y)/Sin2ySin6y=
=2*2Sin4yCos2y/Sin2ySin6y= 4*Sin4yCos2y/Sin2ySin6y=
=4*2Sin2yCos2yCos2y/Sin2ySin6y = 8Cos²2y/Sin6y