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Marisabel10061999
@Marisabel10061999
August 2021
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3,4)упростите)задание во вложении
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oganesbagoyan
Verified answer
3)
= (sin9α+sin7α -sin8α)/(cos9α+cos7α -cos8α) =
(2sin8αcosα -sin8α)/ (2cos8αcosα-cos8α) =sin8α(2cosα-1)/(cos8α(2cosα -1) =
tq8α.
4)
(cos6αcos3α+sin8αsinα)/cos5α =1/2(cos3α +cos9α +cos7α - cos9α)/cos5α
=1/2(cos7α+cos3α)/cos5α = (1/2*2cos(7α+3α)/2*cos(7α -3α)/2) / cos5α
=
cos2α .
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Answers & Comments
Verified answer
3) = (sin9α+sin7α -sin8α)/(cos9α+cos7α -cos8α) =(2sin8αcosα -sin8α)/ (2cos8αcosα-cos8α) =sin8α(2cosα-1)/(cos8α(2cosα -1) =
tq8α.
4) (cos6αcos3α+sin8αsinα)/cos5α =1/2(cos3α +cos9α +cos7α - cos9α)/cos5α
=1/2(cos7α+cos3α)/cos5α = (1/2*2cos(7α+3α)/2*cos(7α -3α)/2) / cos5α
= cos2α .