cos^2 (5π+a) + ctga*sin2a, если a = 2π/3

cos ^ 2 (5π + a) + ctg (a) * sin (2a) - исходное выражение.

Так как ctg (x) = cos (x) / sin (x), а по формуле синуса двойного угла sin (2x) = 2 * sin (x) * cos (x), тогда:

cos ^ 2 (5π + a) + ctg (a) * sin (2a) = cos ^ 2 (5π + a) + cos (a) / sin (a) * 2 * sin (a) * cos (a) = cos ^ 2 (5π + a) + 2 * cos ^ 2 (a) = cos (5π + a) * cos (5π + a) + 2 * cos (a) * cos (a).

Используя формулу произведения косинусов

cos (x) * cos (y) = (1/2) * (cos (x + y) + cos (x - y)), получим:

cos (5π + a) * cos (5π + a) + 2 * cos (a) * cos (a) = (1/2) * (cos (5π + a + 5π + a) + cos (5π + a - (5π + a))) + 2 * (1/2) * (cos (a + a) + cos (a - a)) = (1/2) * (cos (10π + 2a) + cos (0)) + (cos (2a) + cos (0)).

Так как cos (0) = 1, то:

(1/2) * (cos (10π + 2a) + cos (0)) + (cos (2a) + cos (0)) = (1/2) * (cos (10π + 2a) + 1) + (cos (2a) + 1) = (1/2) * cos (10π + 2a) + 1/2 + cos (2a) + 1 = (1/2) * cos (10π + 2a) + cos (2a) + 3/2.

Так как по условию a = 2π/3, то:

(1/2) * cos (10π + 2a) + cos (2a) + 3/2 = (1/2) * cos (10π + 2 * (2π/3)) + cos (2 * 2π/3) + 3/2 = (1/2) * cos (34π/3) + cos (4π/3) + 3/2 = (1/2) * cos (11π + π/3) + cos (π + π/3) + 3/2 = (1/2) * cos (π + π/3) + cos (π + π/3) + 3/2 = 3/2 * cos (π + π/3) + 3/2 = 3/2 * cos (4π/3) + 3/2.

Так как cos (4π/3) = - 1/2, то:

3/2 * cos (4π/3) + 3/2 = 3/2 * ( - 1/2) + 3/2 = 3/2 * ( - 1/2 + 1) = 3/2 * 1/2 = 3/4.