Помогите, пожалуйста. Даю всего 99 баллов

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Universalka_zn · Answer 1 · 2018-05-18T05:07:57+0000

$Sin \frac{5 \pi }{6} = Sin( \pi - \frac{ \pi }{6} = Sin \frac{ \pi }{6} = \frac{1}{2}$
$Cos( \frac{5 \pi }{6}) = Cos( \pi - \frac{ \pi }{6}) = - Cos \frac{ \pi }{6} = - \frac{ \sqrt{3} }{2}$
$tg \frac{5 \pi }{6}= tg( \pi - \frac{ \pi }{6}) = - tg \frac{ \pi }{6} = - \frac{ \sqrt{3} }{3}$
$Sin \frac{7 \pi }{4} = Sin(2 \pi - \frac{ \pi }{4} ) = - Sin \frac{ \pi }{4} = - \frac{ \sqrt{2} }{2}$
$Cos \frac{7 \pi }{4} = Cos(2 \pi - \frac{ \pi }{4} ) = Cos \frac{ \pi }{4} = \frac{ \sqrt{2} }{2}$
$tg \frac{7 \pi }{4} = tg(2 \pi - \frac{ \pi }{4} = - tg \frac{ \pi }{4} = - 1$
$Sin(- \frac{7 \pi }{4} ) = - Sin \frac{7 \pi }{4}$ $= - Sin(2 \pi - \frac{ \pi }{4}) = Sin \frac{ \pi }{4} = \frac{ \sqrt{2} }{2}$
$Cos(- \frac{7 \pi }{4} ) = Cos \frac{7 \pi }{4} =$ $Cos (2 \pi - \frac{ \pi }{4} ) = Cos \frac{ \pi }{4} = \frac{ \sqrt{2} }{2}$
$tg(- \frac{7 \pi }{4} = - tg \frac{7 \pi }{4} = - tg(2 \pi - \frac{ \pi }{4} ) = tg \frac{ \pi }{4} = 1$
$Sin(- \frac{5 \pi }{6} ) = - Sin \frac{5 \pi }{6} = - Sin( \pi - \frac{ \pi }{6} ) = - Sin \frac{ \pi }{6} = - \frac{1}{2}$
$Cos(- \frac{5 \pi }{6} ) = Cos \frac{5 \pi }{6} = Cos( \pi - \frac{ \pi }{6})= - Cos \frac{ \pi }{6} = - \frac{ \sqrt{3} }{2}$
$tg(- \frac{5 \pi }{6} ) = - tg \frac{5 \pi }{6} = - tg( \pi - \frac{ \pi }{6}) = tg \frac{ \pi }{6} = \frac{ \sqrt{3} }{3}$
$Sin(- \frac{4 \pi }{3}) = - Sin \frac{4 \pi }{3} = - Sin( \pi + \frac{ \pi }{3} ) = Sin \frac{ \pi }{3} = \frac{ \sqrt{3} }{2}$
$Cos(- \frac{4 \pi }{3}) = Cos \frac{4 \pi }{3}= Cos( \pi + \frac{ \pi }{3} )= - Cos \frac{ \pi }{3} = - \frac{1}{2}$
$tg(- \frac{4 \pi }{3})= - tg( \frac{4 \pi }{3} = - tg( \pi + \frac{ \pi }{3} ) = tg \frac{ \pi }{3} = \sqrt{3}$
$Sin(- \frac{5 \pi }{3}) = - Sin(2 \pi - \frac{ \pi }{3} )= Sin \frac{ \pi }{3} =\frac{ \sqrt{3} }{2}$
$Cos(- \frac{5 \pi }{3} ) = Cos \frac{5 \pi }{3} = Cos(2 \pi - \frac{ \pi }{3} = Cos \frac{ \pi }{3} = \frac{1}{2}$
$tg(- \frac{5 \pi }{3}) = - tg \frac{5 \pi }{3} = - tg(2 \pi - \frac{ \pi }{3} ) = tg \frac{ \pi }{3} = \sqrt{3}$
$Sin \frac{13 \pi }{6} = Sin(2 \pi + \frac{ \pi }{6} ) = Sin \frac{ \pi }{6} = \frac{1}{2}$
$Cos \frac{13 \pi }{6}= Cos(2 \pi + \frac{ \pi }{6} ) = Cos \frac{ \pi }{6}= \frac{ \sqrt{3} }{2}$
$tg \frac{13 \pi }{6} = tg (2 \pi + \frac{ \pi }{6} ) = tg \frac{ \pi }{6} = \frac{ \sqrt{3} }{3}$
$Sin(- \frac{8 \pi }{3}) = - Sin( \frac{8 \pi }{3}) = - Sin \frac{2 \pi }{3} = - Sin( \pi - \frac{ \pi }{3}) =- Sin \frac{ \pi }{3}= - \frac{ \sqrt{3} }{2}$
<img src="https://tex.z-dn.net/?f=Cos%28-++%5Cfrac%7B8+%5Cpi+%7D%7B3%7D%29+%3D+Cos+%5Cfrac%7B8+%5Cpi+%7D%7B3%7D+%3D+Cos+%5Cfrac%7B2+%5Cpi+%7D%7B3%7D++%3D+Cos%28+%5Cpi+-++%5Cfrac%7B+%5Cpi+%7D%7B3%7D%29+%3D+-+Cos+%5Cfrac%7B+%5Cpi+%7D%7B3%7D+%3D+-++%5Cfrac%7B1%7D%7B2%7D++" id="TexFormula25" title="Cos(- \frac{8 \pi }{3}) = Cos \frac{8 \pi }{3} = Cos \frac{2 \pi }{3} = Cos( \pi - \frac{ \pi }{3}) = - Cos \frac{ \pi