докажите тождество: (2x-3y)^3-(2x+3y)^3=-18y(4x^2+3y^2)^3, - вопрос №2135969023323318423232 от w11Maxim11w 06.03.2020 06:53

Question

Алгебра: докажите тождество: (2x-3y)^3-(2x+3y)^3=-18y(4x^2+3y^2)^3, ^2 - это степень

w11Maxim11w · Answer

Sin 2000 = sin (1800+200) = sin (360*5+180+20) = sin (180+20) = -sin 20
sin (-1000) = sin (1080-1000) = sin (360*3-1000) = sin 80 = sin (90-10) = cos 10
cos 1509 = cos (1440+69) = cos(360*4+69) = cos 69 = cos (90-21) = sin 21
cos (-2900) = cos (2880-2900) = cos (360*8-2900) = cos (-20) = cos 20
tg 606 = tg (540+66) = tg (180*3+66) = tg 66 = tg (90-24) = ctg 24
tg (-400) = tg (360-400) = tg (-40) = -tg 40
ctg 1315 = ctg (1260+55) = ctg (7*180+55) = ctg 55 = ctg (90-35) = tg 35
ctg (-320) = ctg (360-320) = ctg 40
sin (-7,3pi) = -sin (7,3pi) = -sin (7pi+0,3pi) = sin (0,3pi)
cos (34pi/9) = cos (36pi/9-2pi/9) = cos (4pi-2pi/9) = cos (-2pi/9) = cos (2pi/9)
tg (-17pi/18) = tg (pi-17pi/18) = tg (pi/18)
ctg (-41pi/45) = ctg (pi-41pi/45) = ctg (4pi/45)