Question

Нужна помощь срочно 26.15

Answer 1

Ответ:

1

$1 - 8 \sin(2 \beta ) \cos( 2\beta ) = 1 - 4 \sin(4 \beta )$

2

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

3

$\frac{2 \sin( \beta ) - \sin( 2\beta ) }{2 \sin( \beta ) + \sin( 2\beta ) } = \frac{2 \sin( \beta ) - 2 \sin( \beta ) \cos( \beta ) }{2 \sin( \beta ) + 2 \sin( \beta ) \cos( \beta ) } = \\ = \frac{2 \sin( \beta )(1 - \cos( \beta ) ) }{2 \sin( \beta ) (1 + \cos( \beta ) )} = \frac{1 - \cos( \beta ) }{1 + \cos( \beta ) }$

4

$\frac{ctg(45 - \beta ) }{1 - {ctg}^{2} (45 - \beta )} = - \frac{ctg(45 - \beta )}{ {ctg}^{2} (45 - \beta ) - 1} = \\ = - \frac{1}{2} \times \frac{2ctg(45 - \beta )}{ {ctg}^{2} (45 - \beta ) - 1} = - \frac{1}{2ctg(2(45 - \beta ))} = \\ = - \frac{1}{ctg(90 - 2\beta )} = - \frac{1}{tg2 \beta }$