Question

Практическая работа «Применение формул тригонометрии»

Answer 1

Ответ:

3.1

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

3.2

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

3.3

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

3.4

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

3.5

$\frac{ { \sin }^{2} ( \frac{\pi}{2} + \alpha) - { \cos}^{2} ( \alpha - \frac{3\pi}{2} ) }{ {tg}^{2}( \frac{3\pi}{2} + \alpha ) - {ctg}^{2} ( \alpha - \frac{\pi}{2}) } = \\ = \frac{ { \cos}^{2} (\alpha ) - { \sin }^{2} (\alpha ) }{ {ctg}^{2}( \alpha ) - {tg}^{2}( \alpha ) } = \\ = \frac{ { \cos}^{2} ( \alpha ) - { \sin }^{2} (\alpha ) }{ \frac{ { \cos }^{2} (\alpha )}{ { \sin }^{2} (\alpha) } - \frac{ { \sin}^{2} ( \alpha )}{ { \cos }^{2} (\alpha )} } = \\ = ( { \cos }^{2} (\alpha ) - { \sin}^{2} (\alpha )) \times \frac{ { \sin }^{2} (\alpha ) { \cos }^{2} ( \alpha )}{ { { \cos }^{4}( \alpha )} - { \sin }^{4} (\alpha ) } = \\ = ( { \cos }^{2} (\alpha) - { \sin }^{2} (\alpha )) \times \frac{ { \sin }^{2} (\alpha) { \cos }^{2} (\alpha ) }{( { \cos }^{2} ( \alpha) - { \sin }^{2} (\alpha ))( { \cos }^{2} (\alpha) + { \sin}^{2}( \alpha ) } = \\ = { \sin}^{2} (\alpha ) { \cos }^{2} ( \alpha )$