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olesik300899
@olesik300899
August 2022
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Тригонометрия. 10 класс.
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Verified answer
( sin²2,5α - sin²1,5α)/ (sin4α*sinα +cos3α*cos2α) =
(
(1-cos2*2,5α)/2 -(1-cos2*1,5α)
)
/ (sin4α*sinα +cos3α*cos2α)=
(cos3α -cos5α) / (2*sin4α*sinα +2cos3α*cos2α) =
(cos3α - cos5α) / (cos3α - cos5α +cos5α+ cosα)=
(cos3α -cos5α) / (cos3α +cosα) =2sinα*sin4α / 2cos2α*cosα=
sinα*sin2*2α / cos2α*cosα=
sinα*2sin2α*cos2α / cos2α*cosα =
2tqα*sin2α
.
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Answers & Comments
Verified answer
( sin²2,5α - sin²1,5α)/ (sin4α*sinα +cos3α*cos2α) =( (1-cos2*2,5α)/2 -(1-cos2*1,5α)) / (sin4α*sinα +cos3α*cos2α)=
(cos3α -cos5α) / (2*sin4α*sinα +2cos3α*cos2α) =
(cos3α - cos5α) / (cos3α - cos5α +cos5α+ cosα)=
(cos3α -cos5α) / (cos3α +cosα) =2sinα*sin4α / 2cos2α*cosα=
sinα*sin2*2α / cos2α*cosα=
sinα*2sin2α*cos2α / cos2α*cosα = 2tqα*sin2α.