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gaga5115
@gaga5115
September 2021
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пожалуйста,'помогите 1 задание по производной
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LFP
Verified answer
Не очень уверена --- нужно ли было cos(pi/8) подставлять)))
как-то так...
1 votes
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oganesbagoyan
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F '(x) = ( tq(x/4 -x) =tq(-3x/4) = -tq3x/4 ;
f '(x)
= ( -tq3x/4 )' = -(tq3x/4 )' = -(1/cos²3x/4) *(3x/4)' = -(1/cos²3x/4) *(3/4)*(x) '
= -(1/cos²3x/4) *(3/4)*(1) =
-3/4cos²3x/4 * ** = - 3/2(1+cos3x/2)
.
f '(3π/4)
= -3/4cos²(3*3π/4)/4 =-3/4cos²(9π/16)= -3/2(1+cos2*(9π/16) )=.
-3/2(1+cos9π/8) = -3/2(1+cos(π+π/8) ) =
-3/2(1- cosπ/8) ,
но cosπ/8 =√((1+cosπ/4)/2) = √(1+(√2)/2)/2 = √(2+√2) /2
,
окончательно :
f '(3π/4)
= -3/
(
2-√(2+√2)
) .
* * * cosα =
±
√
(
(1+cos2α)/2
) * * *
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Answers & Comments
Verified answer
Не очень уверена --- нужно ли было cos(pi/8) подставлять)))как-то так...
Verified answer
F '(x) = ( tq(x/4 -x) =tq(-3x/4) = -tq3x/4 ;f '(x) = ( -tq3x/4 )' = -(tq3x/4 )' = -(1/cos²3x/4) *(3x/4)' = -(1/cos²3x/4) *(3/4)*(x) '
= -(1/cos²3x/4) *(3/4)*(1) = -3/4cos²3x/4 * ** = - 3/2(1+cos3x/2).
f '(3π/4) = -3/4cos²(3*3π/4)/4 =-3/4cos²(9π/16)= -3/2(1+cos2*(9π/16) )=.
-3/2(1+cos9π/8) = -3/2(1+cos(π+π/8) ) = -3/2(1- cosπ/8) ,
но cosπ/8 =√((1+cosπ/4)/2) = √(1+(√2)/2)/2 = √(2+√2) /2 ,
окончательно : f '(3π/4) = -3/( 2-√(2+√2) ) .
* * * cosα =±√( (1+cos2α)/2 ) * * *