Question

Sin( a+B) - sin B cos a

Cos 63° cos 19° +sin 63° sin 18°

Cos² a - cos2 a

1+cos 2 a: sin2 a

Ctg² a (1-cos2 a )²
Пожалуйста, очень срочно

Answer 1

Ответ:

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.

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

3

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

4

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

5

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