Question

упростите выражения пожалуйста ​

Answer 1

Ответ:

1.

${ \cos }^{2} \alpha - { \sin }^{2} \alpha - \cos( 2\alpha ) = \\ \cos(2 \alpha ) - \cos(2 \alpha ) = 0$

2.

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

3.

$ctg \gamma - \sin( 2\gamma ) - ctg \gamma \cos(2 \gamma ) = \\ = ctg \gamma (1 - \cos(2 \gamma ) ) - \sin(2 \gamma ) = \\ = ctg \gamma \times 2 { \sin }^{2} \gamma - \sin( 2\gamma ) = \\ = \frac{ \cos( \gamma ) }{ \sin( \gamma ) } \times 2 { \sin }^{2} \gamma - \sin(2 \gamma ) = \\ = \sin(2 \gamma ) - \sin( 2\gamma ) = 0$

4.

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

5.

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

6.

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

7.

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

8.

$1 + \frac{ \cos( \alpha ) }{ \sin( \alpha ) } - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } - 2ctg2 \alpha = \\ = \frac{ \sin( \alpha ) \cos( \alpha ) + { \cos}^{2} \alpha - { \sin }^{2} \alpha - 2ctg 2\alpha \times \sin( \alpha ) \cos( \alpha ) }{ \sin( \alpha ) \cos( \alpha ) } = \\ = \frac{ \sin( \alpha ) \cos( \alpha ) + \cos( 2\alpha ) - \sin( 2\alpha ) \times \frac{ \cos(2 \alpha ) }{ \sin( 2\alpha ) } }{ \sin( \alpha ) \cos( \alpha ) } = \\ = \frac{ \sin( \alpha ) \cos( \alpha ) + \cos(2 \alpha ) - \cos(2 \alpha ) }{ \sin( \alpha ) \cos( \alpha ) } = 1$

9.

${ \cos}^{2} \alpha - 4 { \sin }^{2} ( \frac{ \alpha }{2} ) { \cos}^{2} ( \frac{ \alpha }{2} ) - \cos( 2\alpha ) = \\ = { \cos}^{2} \alpha - { \sin}^{2}2 \alpha - \cos( 2\alpha ) = \cos(2 \alpha ) - \cos(2 \alpha ) = 0$

10.

$ctg2 \alpha - ctg \alpha = \frac{ \sin( - 2 \alpha + \alpha ) }{ \sin( 2\alpha ) \sin( \alpha ) } = \\ = - \frac{ \sin( \alpha ) }{ \sin(2 \alpha ) \sin( \alpha ) } = - \frac{1}{ \sin( 2\alpha ) }$