Question

Помогите с 5 вариантом пожалуйста

Answer 1

Ответ:

$1) 2sin^2(a) + cos(2a) - 1 = 2sin^2(a) + (1 - 2sin^2(a)) - 1 = 0\\2) sin(-\frac{11\pi }{6} ) + cos^2(\frac{5\pi }{2} ) - ctg(\frac{5\pi }{4} ) = -sin(\frac{11\pi }{6} ) + cos^2(\frac{\pi }{2} + 2\pi) - ctg(\frac{\pi }{4} + \pi) =\\= -sin(-\frac{\pi }{6} + 2\pi ) + cos^2(\frac{\pi }{2}) - ctg(\frac{\pi }{4}) =\\= -sin(-\frac{\pi }{6} ) + 0-1 = sin(\frac{\pi }{6}) -1 = \frac{1}{2} - 1 = -\frac{1}{2}$