Question

Упростить выражение умоляю от ​

Answer 1

9

$\frac{ \cos(5x) - \cos(x) }{ \sin(5x) + \sin(x) } = \frac{ - 2 \sin( \frac{5x - x}{2} ) \sin( \frac{5x + x}{2} )} {2 \sin( \frac{5x + x}{2} ) \cos( \frac{5x - x}{2} ) } = \\ = - \frac{ \sin(2x) \sin(3x) }{ \sin(3x) \cos(2x) } = - tg(2x)$

10

$\frac{ \sin(2x) - \sin(x) }{ \cos(2x) + \cos(x) } = \frac{2 \sin( \frac{2x - x}{2} ) \cos( \frac{2x + x}{2} ) }{2 \cos( \frac{2x + x}{2} ) \cos( \frac{2x - x}{2} ) } = \\ = \frac{ \sin( \frac{x}{2} ) }{ \cos( \frac{x}{2} ) } = tg( \frac{x}{2} )$

11

$\frac{ \sin(2x) + \sin(6x) }{ \cos(2x) - \cos(6x) } = \frac{2 \sin( \frac{2x + 6x}{2} ) \cos( \frac{2x - 6x}{2} ) }{ - 2 \sin( \frac{2x - 6x}{2} ) \sin( \frac{2x + 6x}{2} ) } = \\ = - \frac{ \sin(4x) \cos(2x) }{ ( - \sin(2x)) \sin(4x) } = ctg(2x)$

12

$\frac{ \cos(2x) - \cos(3x) }{ \sin(2x) - \sin(3x) } = \frac{ - 2 \sin( \frac{2x - 3x}{2} ) \sin( \frac{2x + 3x}{2} ) }{2 \sin( \frac{2x - 3x}{2} ) \cos( \frac{2x + 3x}{2} ) } = \\ = - \frac{ \sin( \frac{5x}{2} ) }{ \cos( \frac{5x}{2} ) } = - tg( \frac{5x}{2})$

13

$\frac{ \sin( 4\alpha ) - \sin(6x) }{ \cos(3 \alpha ) + \cos( 7\alpha ) } = \frac{2 \sin( \frac{4 \alpha - 6x}{2} ) \cos( \frac{4 \alpha + 6x}{2} ) }{2 \cos( \frac{3 \alpha + 7 \alpha }{2} ) \cos( \frac{3 \alpha - 7 \alpha }{2} ) } = \\ = \frac{ \sin(2 \alpha - 3x) \cos(2 \alpha + 3x) }{ \cos(5 \alpha ) \cos( 2\alpha ) }$

14

$\frac{ \sin( 7\beta ) + \sin( 11\beta ) }{ \cos(10 \beta ) - \cos( 8\beta ) } = \frac{2 \sin( \frac{7 \beta + 11 \beta }{2} ) \cos( \frac{7 \beta - 11 \beta }{2} ) }{ - 2 \sin( \frac{10 \beta - 8\beta }{2} ) \sin( \frac{10 \beta + 8 \beta }{2} ) } = \\ = - \frac{ \sin(9 \beta ) \cos( 2\beta ) }{ \sin( \beta ) \sin(9 \beta ) } = - \frac{ \cos( 2\beta ) }{ \sin( \beta ) }$

15

$\frac{ \cos(45^{\circ} - \alpha ) + \cos(45^{\circ} + \alpha ) }{ \sin(45^{\circ} - \alpha ) + \sin(45 ^{\circ} + \alpha ) } = \\ = \frac{2 \cos( \frac{45 ^{\circ} - \alpha + 45^{\circ} + \alpha }{2} ) \cos( \frac{45 ^{\circ} - \alpha - 45 ^{\circ} - \alpha }{2} ) }{2 \sin( \frac{45^{\circ} - \alpha + 45^{\circ} + \alpha }{2} ) \cos( \frac{45^{\circ} - \alpha - 45^{\circ} - \alpha }{2} ) } = \\ = \frac{ \cos(45^{\circ} ) \cos( \alpha ) }{ \sin(45^{\circ} ) \cos( \alpha ) } = 1$

16

$\frac{ \sin( 45 ^{\circ} + \alpha ) + \sin( 45 ^{\circ} - \alpha ) }{ \sin(45^{\circ} + \alpha ) - \sin(45 ^{\circ} - \alpha ) } = \\ = \frac{2 \sin( \frac{45^{\circ} + \alpha + 45^{\circ} - \alpha }{2} ) \cos( \frac{45^{\circ} + \alpha - 45^{\circ} + \alpha }{2} ) }{2 \sin( \frac{45 ^{\circ} + \alpha - 45^{\circ} + \alpha }{2} ) \cos( \frac{45^{\circ} + \alpha + 45^{\circ} - \alpha }{2} ) } = \\ = \frac{ \sin(45^{\circ} ) \cos( \alpha ) }{ \sin( \alpha ) \cos(45^{\circ} ) } = ctg( \alpha )$