Ответ:

1. Преобразование:

sin(2a)cos(2a)sin(a)cos(a) + 4sin(4a - 2ctg(2a - 31) - tg(2a + X))sin(z)sin(-cos(a))

= 2sin(a)cos(a)sin(2a)cos(2a) + 4sin(4a - 2ctg(2a - 31) - tg(2a + X))sin(z)sin(a)cos(a) (переносим sin(a)cos(a) в начало первого слагаемого)

= sin(2a)sin(2a) + 2sin(4a - 2ctg(2a - 31) - tg(2a + X))sin(z)sin(2a) (упрощение)

= sin(2a)[sin(2a) + 2sin(4a - 2ctg(2a - 31) - tg(2a + X))sin(z)]

2. Упрощение:

sin²x - sin²x cos²x + sin²x cos²x - cos²x = sin²x (1 - cos²x) + cos²x (sin²x - 1)

= sin²x sin²x + cos²x(-sin²x) = sin⁴x - sin²x cos²x

3. Доказательство:

a) (sin(a) + cos(a))² - 1 = sin²(a) + cos²(a) + 2sin(a)cos(a) - 1

= 2sin(a)cos(a) = ctg(a)

b) (1 - tg(x))(1 + tg(x)) = 1 - tg²(x) = 1 - (sin(x)/cos(x))²

= (cos²(x) - sin²(x))/cos²(x) = cos(2x)/cos²(x)

(1 + sina + cosa)² - 1 = 1 + 2sin(a) + 2cos(a) + sin²(a) + 2sin(a)cos(a) + cos²(a) - 1

= 2sin(a) + 2cos(a) + 2sin(a)cos(a) = 2(sin(a) + cos(a))(1 + sin(a)cos(a))

= 2(ctg(a))(1 + tg(a))

(ctg(a) - sin(a)cos(a))/(1 - tg(x)) = cos²(a)/(sin(x) - cos(x))

= cos²(a)/(-sin(a)/cos(a)) = -cos³(a)/sin(a)

2tg²(a) - 3 = 2(sin²(a)/cos²(a)) - 3

= (2sin²(a) - 3cos²(a))/cos²(a) = -cos(2a)/cos²(a)