Question

СРОЧНО помогите пожалуйста по 2 с каждого задания

Answer 1

1 задание

1.

$\sin( \alpha + \beta ) - \sin( \beta ) \cos( \alpha ) = \\ = \sin( \alpha ) \cos( \beta ) + \sin( \beta ) \cos( \alpha ) - \sin( \beta ) \cos( \alpha ) = \\ = \sin( \alpha ) \cos( \beta )$

2.

$\cos(63^{\circ} ) \cos(18^{\circ} ) + \sin(63^{\circ} ) \sin(18^{\circ} ) = \\ = \cos(63^{\circ} - 18^{\circ} ) = \cos(45^{\circ} ) = \frac{ \sqrt{2} }{2}$

2 задание

1.

$\frac{ \sin( 2\beta ) }{ \sin {}^{2} ( \beta ) } = \frac{2 \sin( \beta ) \cos( \beta ) }{ \sin {}^{2} ( \beta ) } = 2 \times \frac{ \cos( \beta ) }{ \sin( \beta ) } = 2ctg( \beta )$

2.

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

3 задание

1.

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

2.

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

4 задание

1.

$\cos( \frac{\pi}{2} + \alpha ) = - \sin( \alpha )$

2.

$\cos(360^{\circ} + \alpha ) = \cos( \alpha )$

5 задание

1.

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

2.

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