Question

26.9 упростите выражение​

Answer 1

Ответ:

2.

$\sin(2 \beta ) - tg( \beta ) - \cos( 2\beta ) tg ( \beta ) = \\ = \sin( 2\beta ) - tg (\beta )(1 + \cos( 2\beta ) ) = \\ = \sin( 2\beta ) - tg (\beta ) (1 + \cos {}^{2} ( \beta ) - \sin {}^{2} ( \beta ) ) = \\ = \sin( 2\beta ) - tg (\beta ) \times 2 \cos {}^{2} ( \beta ) = \\ = \sin( 2\beta ) - \frac{ \sin( \beta ) }{ \cos( \beta ) } \times 2 \cos {}^{2} ( \beta ) = \\ = \sin( 2\beta ) - 2 \sin( \beta ) \cos( \beta ) = \\ = \sin( 2\beta ) \sin( 2\beta ) = 0$

4.

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

6.

$1 + \frac{1 - \cos(2x) + \sin(2x) }{1 + \cos(2x) + \sin(2x) } = \\ = \frac{1 + \cos(2x) + \sin(2x) + 1 - \cos(2x) + \sin(2x) }{ 1 + \cos(2x) + \sin(2x) } = \\ = \frac{2 + 2 \sin(2x) }{1 + \cos {}^{2} (x) - \sin {}^{2} (x) + \sin(2x) } = \\ = \frac{2(1 + 2 \sin(x) \cos(x)) }{2 \cos {}^{2} (x) + 2\sin(x) \cos(x) } = \\ = \frac{2( \sin {}^{2} (x) + 2 \cos(x) \sin(x) + \cos {}^{2} (x)) }{2 \cos(x)( \cos(x) + \sin(x)) } = \\ = \frac{ {( \sin(x) + \cos(x) )}^{2} }{ \cos(x)( \sin(x) + \cos(x)) } = \\ = \frac{ \sin(x) + \cos(x) }{ \cos(x) } = tgx + 1$