Все ** фото. Буду благодарен! Баллов накину))

Question 1

Все на фото. Буду благодарен! Баллов накину))

Question 2

$\sin(x - \frac{\pi}{4} ) = \\ sinxcos \frac{\pi}{4} - \cos(x) \sin( \frac{\pi}{4} ) = \\ \frac{ \sqrt{2} }{2} \sin(x) - \frac{ \sqrt{2} }{2} \cos(x) = \\ \frac{ \sqrt{2} }{2} ( \sin(x) - \cos(x) )$

$tg( \frac{\pi}{6} + x) = \frac{tg \frac{\pi}{6} + tgx}{1 - tg \frac{\pi}{6} tgx} = \\ \frac{ \frac{ \sqrt{3} }{3} + tgx }{1 - \frac{ \sqrt{3} }{3}tgx }$

$sin150 = \sin(180 - 30) = \frac{1}{2}$

$\cos(75) = \cos(45 + 30) = \\ \cos(45) \cos(30) - \sin(45) \sin(30) = \\ \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} - \frac{ \sqrt{2} }{2} \times \frac{1}{2 } = \\ \frac{ \sqrt{6} }{4} - \frac{ \sqrt{2} }{4} = \frac{ \sqrt{6} - \sqrt{2} }{4}$

$tg105 = tg(60 + 45) = \\ \frac{tg60 + tg45}{1 - tg60tg45} = \frac{ \sqrt{3} + 1 }{1 - \sqrt{3} } = \\ \frac{( \sqrt{3 } + 1 )(1 + \sqrt{3}) }{(1 - \sqrt{3} )(1 + \sqrt{3}) } = \frac{ \sqrt{3} + 3 + 1 + \sqrt{3} }{1 - 3} = \\ - \frac{4 + 2 \sqrt{3} }{2} = - 2 - \sqrt{3}$

$\sin(x) = - \frac{2}{5} \\ \cos(x) = \sqrt{1 - { (\frac{2}{5}) }^{2} } = \\ \sqrt{1 - \frac{4}{25} } = \sqrt{ \frac{21}{25} }$

так как угол третьей четверти то cosx <0</p>

$\sin(2x) = 2 \sin(x) \cos(x) = \\ 2( - \frac{2}{5} )( \frac{ \sqrt{21} }{5} ) = - \frac{4 \sqrt{21} }{5}$

$cosx = - 0.3 \\ sinx = \sqrt{1 - 0.3 ^{2} } = \sqrt{1 - 0.09 } = \sqrt{0.91} \\ cos2x = cos ^{2} x - {sin}^{2} x = \\ ( { - 0.3}^{2} ) - (0.91) = 0.09 - 0.91 = 0.82$

$tgx = 2 \\ tg2x = \frac{2tgx}{1 - {tg}^{2} x} = \frac{4}{1 - 4} = - \frac{3}{4}$

$2 \sin( \frac{\pi}{6} ) \cos( \frac{\pi}{6} ) = \sin( \frac{\pi}{3} ) = \frac{ \sqrt{3} }{2}$

$tg120 = \frac{2tg60}{ 1 - {tg}^{2} 60} = \frac{2 \sqrt{3} }{1 - 3} = \frac{ 2\sqrt{3} }{ - 2} = - \sqrt{3}$