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ЭмилияСмит
@ЭмилияСмит
August 2022
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Как вычисляется arccos(-cos(-x))?
Всем привет!) Нужна помощь. Если нужен конкретный случай, то вот: arccos(-cos(-43π/6)).
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oganesbagoyan
Verified answer
Task/24970505
----------------------
arccos(-cos(-43π/6) ) - ?
cos(- 43π/6)
=cos(43π/6)=cos(6π +π+π/6) =cos(π+π/6) = -cos(π/6) =
-(√3)/2.
arccos(-cos(-43π/6) )
=arccos(√3)/2 )
=
π/6
.
ответ :
π/6 .
------------------
формулу arccosa +arccos(-a) =π ,
|a| ≤ 1 не использовал
sin(16π/3)
= sin(4π+π+π/3) = sin(π+π/3) = -sin(π/3) = -(√3)/2 ;
arcsin(sin(16π/3)
= arcsin(-(√3)/2) =
- π/3 ;
arccos(cos(-15π/8) )
=arccos(cos(15π/8) )=arccos(cos(2π -π/8) )=
arccos(cos(2π -π/8) ) =arccos(cos(π/8) )
= π
/8
35*arccos(-cos(-43π/6) ) /(arcsin(sin(16π/3)+arccos(cos(-15
π/8) )
=
35*(π/6) / ( - π/3+π/8 ) =(35π/6) / (-5π/24)
= -28 .
3 votes
Thanks 4
ЭмилияСмит
Хорошо, узнаю тогда решение в пт. Это B8 из демонстрационного ЦТ 2017, если вдруг интересно) Но спасибо за помощь! Хорошего вечера)))
oganesbagoyan
arcsin(sin(16π/3) )=arcsin(sin(-π/3)= - π/3
oganesbagoyan
arccos(cos(-15π/8) )=arccos(cos(15π/8) ) =arccos(cos(π/8) )=π/8
oganesbagoyan
arcsin(sinα)=α , если -π/2 ≤ α ≤ π/2
oganesbagoyan
arccos(cosα)=α , ,если 0 ≤ α ≤ π
ЭмилияСмит
Огромное вам спасибо! Давно не была на сайте, не видела вашего решения. Безумно благодарна вам!)))
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Verified answer
Task/24970505----------------------
arccos(-cos(-43π/6) ) - ?
cos(- 43π/6)=cos(43π/6)=cos(6π +π+π/6) =cos(π+π/6) = -cos(π/6) = -(√3)/2.
arccos(-cos(-43π/6) )=arccos(√3)/2 ) = π/6 .
ответ : π/6 .
------------------
формулу arccosa +arccos(-a) =π , |a| ≤ 1 не использовал
sin(16π/3) = sin(4π+π+π/3) = sin(π+π/3) = -sin(π/3) = -(√3)/2 ;
arcsin(sin(16π/3) = arcsin(-(√3)/2) = - π/3 ;
arccos(cos(-15π/8) )=arccos(cos(15π/8) )=arccos(cos(2π -π/8) )=
arccos(cos(2π -π/8) ) =arccos(cos(π/8) ) = π/8
35*arccos(-cos(-43π/6) ) /(arcsin(sin(16π/3)+arccos(cos(-15π/8) )=
35*(π/6) / ( - π/3+π/8 ) =(35π/6) / (-5π/24) = -28 .