(с+1)(с^2+3c+2) =c^3+3c^2+2c
<em>(х-7)(х+7)-(х-4)²=(х²-49)-(х²-8х+16)=х²-49-х²+8х-16=<u>8х-65</u>.</em>
Решение
1)cos5βcos2β + sin5βsin2β = cos(5β - 2β) = cos3β
2)
Левая сторонаα)
(1 + sin2α)/cos2α = 1/cos2α + tg2α =
= (1 + tg²α)/(1 - tg²α) + 2tgα / (1 - tg²α) =
= (1 + tgα)² / (1 - tg²α) = (1 + tgα)² / (1 - tgα)(1 + tgα) = (1 + tgα)/(1 - tgα)
Правая сторона
tg(π/4 + α) = (tgπ/4 + tgα)/(1 - tgπ/4 * tgα) = (1 + tgα)/(1 - tgα)
Левая часть равна правой
(1 + tgα)/(1 - tgα) = (1 + tgα)/(1 - tgα)
доказано
3)
cos120° = - 1/2
sin(- 13/6) = - sin(2π + π/6) = - sin(π/6) = - 1/2
Вот как-то так.
Попробуй понять