Question

Помогите упростить выражения

Answer 1

Ответ:

1.

а)

$1 - { \sin}^{2} \alpha = { \cos}^{2} \alpha$

б)

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

в)

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

г)

${ \sin }^{2} \alpha - 3 { \cos}^{2} \alpha - 1 = \\ = - (1 - { \sin }^{2} \alpha ) - 3 { \cos}^{2} \alpha = \\ = - { \cos }^{2} \alpha - 3 { \cos}^{2} \alpha = - 4 { \cos}^{2} \alpha$

д)

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

е)

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

2.

а)

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

б)

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

в)

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

г)

${( \cos(x) tg(x))}^{2} + {( \sin(x) ctg(x))}^{2} = \\ = { \sin }^{2} x + { \cos }^{2} x = 1$