Question

Сочч по алгебре ... прорешайте весь, только нормально а не абы как...
заранее ! Чем быстрее тем лучше ​

Answer 1

1.

а)

$1) - 120^{\circ} \times \frac{\pi}{180} = - \frac{2\pi}{ 3} \\$

3 четверть

$2)210 ^ {\circ} \times \frac{\pi}{180} = \frac{7\pi}{6} \\$

3 четверть

б)

$1) \frac{3\pi}{5} \times \frac{180 ^ {\circ}}{\pi} = 3 \times 36 ^ {\circ} = 108 ^ {\circ} \\$

2 четверть

$2) \frac{6\pi}{7} \times \frac{180 ^ {\circ}}{\pi} = \frac{1080 ^ {\circ}}{7} = (154 \frac{2}{7}) ^ {\circ} \\$

2 четверть

2.

$a) \sin(28 ^ {\circ}) \cos(17 ^ {\circ}) + \sin(17 ^ {\circ}) \cos(28 ^ {\circ}) = \\ = \sin(28 ^ {\circ} + 17 ^ {\circ}) = \sin(45 ^ {\circ}) = \frac{ \sqrt{2} }{2}$

$b) \sin( - \frac{\pi}{6} ) + {ctg}^{2} \frac{\pi}{6} = - \sin( \frac{\pi}{6} ) + (\frac{ \cos( \frac{\pi}{6} ) }{ \sin( \frac{\pi}{6} ) } ) {}^{2} = \\ = - \frac{1}{2} + {( \frac{ \sqrt{3} }{2} \times 2 )}^{2} = \frac{1}{2} + {( \sqrt{3} )}^{2} = 0.5 + 3 = 3.5$

3.

$a)ctg(\pi - \alpha ) \times \cos( \frac{\pi}{2} - \alpha ) - \sin( \frac{3\pi}{2} - \alpha ) = \\ = - ctg( \alpha ) \times \sin( \alpha ) + \cos( \alpha ) = \\ = - \frac{ \cos( \alpha ) }{ \sin( \alpha ) } \times \sin( \alpha ) + \cos( \alpha ) = \\ = - \cos( \alpha ) + \cos( \alpha ) = 0$

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

в)

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

4.

$\cos( \alpha ) = \frac{1}{ \sqrt{5} } \\ \sin( \alpha ) = \sqrt{1 - \cos {}^{2} ( \alpha ) } = \\ = \sqrt{1 - \frac{1}{5} } = \sqrt{ \frac{4}{5} } = \frac{2}{ \sqrt{5} } \\ tg( \alpha ) = \frac{ \sin(\alpha )}{ \cos( \alpha ) } = \frac{2}{ \sqrt{5} } \times \sqrt{5} = 2$