Cos(пи/33) * cos(2*пи/33) * cos(4*пи/33) * cos(8*пи/33) - вопрос №8579614332334338331633 от DANIL79NR 15.04.2022 00:33

Question

Алгебра: Cos(пи/33) * cos(2*пи/33) * cos(4*пи/33) * cos(8*пи/33) * cos(16*пи/33)

DANIL79NR · Answer

Sin 2x = 2 sin x  cos xsin (pi -x) = sin x cos(pi/33)*cos(2pi/33)*cos(4pi/33)*cos(8pi/33)*cos(16pi/33)  = sin(pi/33)*cos(pi/33)*cos(2pi/33)*cos(4pi/33)*cos(8pi/33)*cos(16pi/33) / sin(pi/33) = sin(2pi/33)*cos(2pi/33)*cos(4pi/33)*cos(8pi/33)*cos(16pi/33) / 2sin(pi/33) =  sin(4pi/33)*cos(4pi/33)*cos(8pi/33)*cos(16pi/33) / 4sin(pi/33) =  sin(8pi/33)*cos(8pi/33)*cos(16pi/33)/ 8sin(pi/33) = sin(16pi/33)*cos(16pi/33)/ 16sin(pi/33) = sin(32pi/33) / 32pi(33) = sin(pi-pi/33)/16sin(pi/33) = sin(pi/33) / 32sin(pi/33)...