Question

№2,4,5 в каждом варианте. Пожалуйста!
40 Б​

Answer 1

1 вариант

2.

$2 - 4 \sin {}^{2} ( \alpha ) = 2 - 4(1 - \cos {}^{2} ( \alpha ) ) = \\ = 2 - 4 + 4 \cos {}^{2} ( \alpha ) = - 2 + 4 \cos {}^{2} ( \alpha ) = \\ \\ = - 2 + 4 \times \frac{3}{4} = - 2 + 3 = 1$

Ответ: 1

4.

$\frac{1 - {tg}^{2} \alpha }{1 + {tg}^{2} \alpha } = \frac{1 - \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } }{1 + \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } } = \\ = \frac{ \cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } \times \frac{ \cos {}^{2} ( \alpha ) }{ \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) } = \\ = \frac{ \cos(2 \alpha ) }{1} = \cos(2 \alpha )$

5.

$\sin( - \frac{49\pi}{6} ) = - \sin( \frac{48 + 1}{6} \pi) = \\ = - \sin(8\pi + \frac{\pi}{6} ) = - \sin( \frac{\pi}{6} ) = - 0.5$

2 вариант

2.

$6 \sin {}^{2} ( \alpha ) - 4 = 6(1 - \cos {}^{2} ( \alpha )) - 4 = \\ = 6 - 6 \cos {}^{2} ( \alpha ) - 4 = 2 - 6 \cos {}^{2} ( \alpha ) = \\ \\ = 2 - 6 \times \frac{3}{4} = 2 - 4.5 = - 2.5$

4.

$\frac{2tg \alpha }{1 + tg {}^{2} \alpha } = \frac{2tg \alpha }{ \frac{1}{ \cos {}^{2} ( \alpha ) } } = 2 \times \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \times \cos {}^{2} ( \alpha ) = \\ = 2 \sin( \alpha ) \cos( \alpha ) = \sin(2 \alpha )$

5.

$\cos( \frac{17\pi}{4} ) = \cos( \frac{16+ 1}{4}\pi ) = \\ = \cos(4\pi + \frac{\pi}{4} ) = \cos( \frac{\pi}{4} ) = \frac{ \sqrt{2} }{2}$