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Lidanext
@Lidanext
August 2021
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срочно,докажите тождество
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oganesbagoyan
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Доказать тождество :
(1 - 2sin²α)/sin2α -(cos3α - cosα)/(sin3α+sinα) =1/sin2α.
--------------------------------------------------------------------------------
(1 - 2sin²α)/sin2α -(cos3α - cosα)/(sin3α+sinα) =
(1 - 2sin²α)/2sinα*cosα - (-2sin2α*sinα)/2sin2α*cosα =
(1 - 2sin²α)/2sinα*cosα +sinα/cosα =(1-2sin²α +2sinα*sinα)/2sinα*cosα =1/sin2α.
************************* использованы формулы ********************************
sin2α =2sinα*cosα ;
cosα - cosβ = -2 sin(α-β)/2 * sin(α+β)/2 ;
sinα + sinβ = 2 sin(α+β)/2 * cos(α-β)/2
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Answers & Comments
Verified answer
Доказать тождество :(1 - 2sin²α)/sin2α -(cos3α - cosα)/(sin3α+sinα) =1/sin2α.
--------------------------------------------------------------------------------
(1 - 2sin²α)/sin2α -(cos3α - cosα)/(sin3α+sinα) =
(1 - 2sin²α)/2sinα*cosα - (-2sin2α*sinα)/2sin2α*cosα =
(1 - 2sin²α)/2sinα*cosα +sinα/cosα =(1-2sin²α +2sinα*sinα)/2sinα*cosα =1/sin2α.
************************* использованы формулы ********************************
sin2α =2sinα*cosα ;
cosα - cosβ = -2 sin(α-β)/2 * sin(α+β)/2 ;
sinα + sinβ = 2 sin(α+β)/2 * cos(α-β)/2