task/29633771 11. Упростите выражение :
a) [ 2sinαcosβ - sin(α-β) ] / [ cos(α-β) - 2sinαsinβ ] =
[ sin(α+β) +sin(α-β) -sin(α-β) ] / [cos(α-β)-(cos(α-β)-cos(α+β) )]=sin(α+β) /cos(α+β) = tg(α+β) .
б) (1 - cosα +cos2α) / (sin2α -sinα) = (2cos²α - cosα ) / (2sinα*cosα -sinα) =
cosα(2cosα - 1 ) / sinα(2cosα -1) = ctgα .
в) (√2 cosα - 2cos(π/4 +α) ) / ( 2sin(π/4 +α) -√2sinα ) =
(√2 cosα - 2( cos(π/4)*cosα - sin(π/4)* sinα ) ) /
/ ( 2(sin(π/4)*cosα) + cos(π/4)*sinα) - √2sinα ) =
(2cosα- √2cosα +√2sinα )/(√2cosα +√2sinα -√2sinα )=√2sinα√2cosα=tgα .
г) ctg²α(1 - cos2α) + cos²α = ctg²α*2sin²α + cos²α = 2cos²α + cos²α =3cos²α.
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task/29633771 11. Упростите выражение :
a) [ 2sinαcosβ - sin(α-β) ] / [ cos(α-β) - 2sinαsinβ ] =
[ sin(α+β) +sin(α-β) -sin(α-β) ] / [cos(α-β)-(cos(α-β)-cos(α+β) )]=sin(α+β) /cos(α+β) = tg(α+β) .
б) (1 - cosα +cos2α) / (sin2α -sinα) = (2cos²α - cosα ) / (2sinα*cosα -sinα) =
cosα(2cosα - 1 ) / sinα(2cosα -1) = ctgα .
в) (√2 cosα - 2cos(π/4 +α) ) / ( 2sin(π/4 +α) -√2sinα ) =
(√2 cosα - 2( cos(π/4)*cosα - sin(π/4)* sinα ) ) /
/ ( 2(sin(π/4)*cosα) + cos(π/4)*sinα) - √2sinα ) =
(2cosα- √2cosα +√2sinα )/(√2cosα +√2sinα -√2sinα )=√2sinα√2cosα=tgα .
г) ctg²α(1 - cos2α) + cos²α = ctg²α*2sin²α + cos²α = 2cos²α + cos²α =3cos²α.
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Всё решаем по формулам.