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seminartem98
@seminartem98
July 2022
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(4sinα+5cosα+3)/(12sinα−5cosα+21)=1/7.Найдите значение выражения: tgα. Здравствуйте, не могли помочь, как решать данные выражния.
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BOR48
(4sinα+5cosα+3)/(12sinα- 5cosα +21) =1/7⇔
7*(4sinα+5cosα+3) =1*(12sinα- 5cosα +21) ;
28sinα +35cosα +21 =12sinα -5cosα +21 ;
16sinα = -40cosα ;
sinα = (-5/2)*cosα ;
sinα/cosα = (-5/2)*cosα/cosα ;
tqα = -2,5.
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Answers & Comments
7*(4sinα+5cosα+3) =1*(12sinα- 5cosα +21) ;
28sinα +35cosα +21 =12sinα -5cosα +21 ;
16sinα = -40cosα ;
sinα = (-5/2)*cosα ;
sinα/cosα = (-5/2)*cosα/cosα ;
tqα = -2,5.