Question

Помогите решить. Срочно ​

Answer 1

Ответ:

1.

а)

$\frac{ \cos( - 240°) ctg(1000°)}{ \sin(550°) } = \\ = \frac{ ( - \cos( 240°)) \times( - ctg(1000° ))}{ - \sin(550°) }$

Ответ: знак минус

б)

$tg( - 1.6) \times \cos(2.3) \times \sin(5) = - tg( 1.6) \times \cos(2.3) \times \sin(5)$

Ответ: знак минус

2.

а)

$ctg( - 750°) = ctg( - 720° - 30°) = - ctg(30°) = - \sqrt{3}$

б)

$\cos( - \frac{22\pi}{3} ) = \cos( - 7\pi - \frac{\pi}{3} ) = \\ = - \cos( \frac{\pi}{3} ) = - 0.5$

3.

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

б)

$(1 + \sin(t)) (tgt + ctgt)(1 - \sin(t)) = \\ = (1 - { \sin }^{2} t) \times (tgt + \frac{1}{tgt} ) = \\ = { \cos}^{2} t \times \frac{ {tg}^{2} t + 1}{tgt} = \\ =\frac{ { \cos }^{2} t( 1 + \frac{ { \sin}^{2}t }{ { \cos}^{2}t } )}{ tgt} = \\ = ( { \cos}^{2} t + { \sin }^{2}t ) \times \frac{1}{tgt} = \\ = \frac{1}{tgt}$

4.

$\sin( \frac{\pi}{2} + \alpha ) \times \cos(2\pi + \alpha ) - \cos( \frac{3\pi}{2} - \alpha ) \times \sin(3\pi - \alpha ) = \\ = \cos( \alpha ) \times \cos( \alpha ) - ( - \sin( \alpha )) \times \sin( \alpha ) = \\ = { \cos}^{2} \alpha + { \sin }^{2} \alpha = 1$

5.

$tg( \alpha ) = 2 \sqrt{3} \\ \\ \sin(( \frac{\pi}{6} - 2 \alpha )) = \\ = \sin( \frac{\pi}{6} ) \cos( 2\alpha ) - \cos( \frac{\pi}{6} ) \sin( 2\alpha ) = \\ = \frac{1}{2} \cos(2 \alpha ) - \frac{ \sqrt{3} }{2} \sin( 2\alpha )$

необходимо найти sin2a и cos2a.

Используем формулу:

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

$\cos( \alpha ) = \sqrt{ \frac{1}{1 + 4 \times 3} } = \sqrt{ \frac{1}{1 + 12} } = \\ = \sqrt{ \frac{1}{13} } = \frac{1}{ \sqrt{13} }$

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

$\sin( \alpha ) = \sqrt{1 - \frac{1}{13} } = \frac{ \sqrt{12} }{ \sqrt{13} } = \\ = \frac{2 \sqrt{3} }{ \sqrt{13} }$

$\sin( 2\alpha ) = 2 \sin( \alpha ) \cos( \alpha ) = \\ = 2 \times \frac{1}{ \sqrt{13} } \times \frac{2 \sqrt{3} }{ \sqrt{13} } = \frac{4 \sqrt{3} }{13}$

$\cos(2 \alpha ) = { \cos }^{2} \alpha - { \sin}^{2} \alpha = \\ = \frac{1}{13} - \frac{12}{13} = - \frac{11}{13}$

$\sin( \frac{\pi}{6} - 2 \alpha ) = \\ = \frac{1}{2} \times ( - \frac{11}{13}) - \frac{ \sqrt{3} }{2} \times \frac{4 \sqrt{3} }{13} = \\ = \frac{ - 11 - 13}{26} = - \frac{24}{26} = - \frac{12}{13}$

Ответ: -12/13