Упростите, пожалуйста.. Created by NoizeR. algebra-ru

Упростите, пожалуйста...

gartenzie Ctg(3x/2 +5π/4)*(1 -sin(3x -π)) =ctg(3π/2-(π/4-3x/2 ))*(1+sin(π-3x)) =
(tg(π/4-3x/2)) *(1+sin3x) =(tqπ/4- tq3x/2)/(1+(tqπ/4)* (tq3x/2))*(1+sin3x)=
=(1- tq3x/2)/(1+tq3x/2)* (1+sin3x)=
(cos3x/2- sin3x/2)/(cos3x/2 + sin3x/2)* (cos3x/2+sin3x/2)²=
(cos3x/2- sin3x/2)*(cos3x/2+sin3x/2) =cos²3x/2- sin²3x/2=cos3x.

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NoizeR извините, а почему 2,5 пи?

oganesbagoyan тяжелый случай

NoizeR спасибо, Вы умница!

oganesbagoyan

Verified answer

$ctg( \frac{3x}{2} + \frac{ 5 \pi }{4} ) ( 1 - \sin{ ( 3x - \pi ) } ) = \frac{ 1 + \sin{3x} }{ tg( [ 3x + 2.5 \pi ] / 2 ) } = \\\\ = \frac{ 1 + \sin{3x} }{ ( 1 - \cos{ ( 3x + 2.5 \pi ) } ) / \sin{ ( 3x + 2.5 \pi ) } } = \frac{ 1 + \sin{3x} }{ 1 - \cos{ ( 3x + \frac{ \pi }{2} ) } } \sin{ ( 3x + \frac{ \pi }{2} ) } = \\\\ = -\sin{ ( - 3x - \frac{ \pi }{2} ) } \cdot \frac{ 1 + \sin{3x} }{ 1 - \cos{ ( - 3x - \frac{ \pi }{2} ) } } = \sin{ ( \frac{ \pi }{2} - 3x ) } \cdot \frac{ 1 + \sin{3x} }{ 1 + \cos{ ( \frac{ \pi }{2} - 3x ) } } = \\\\ = \cos{3x} \cdot \frac{ 1 + \sin{3x} }{ 1 + \sin{3x} } = \cos{3x} .$

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