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N1keee
@N1keee
August 2021
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Докажите, пожалуйста, тождество и упростите выражение
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COS(π-α)= - COS(α), SIN(π/2-α)=COS(α), SIN(2α)=2SIN(α)COS(α)
[SIN(2α)+COS(π-α)SIN(α)]/SIN(π/2-α)=[2SIN(α)COS(α)-COS(α)SIN(α)]/COS(α) = SIN(α)COS(α)/COS(α) = SIN(α)
2) COS(2α) / (SIN(α)-COS(α))↑2-SIN(2α)
COS(2α) = COS²(α) - SIN²(α) = (COS(α) - SIN(α) )·(COS(α) + SIN(α) )
[COS(2α)/(
SIN(α) -
COS(α))]²-SIN(2α) =
=[ (COS(α) - SIN(α) )·(COS(α) + SIN(α) )/(SIN(α) -COS(α)) ]²-
-2COS(α) SIN(α)=
= COS²(α) + 2COS(α)SIN(α) +SIN²(α)-2COS(α) SIN(α)=
=COS²
(α) +SIN²(α)=1
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Answers & Comments
Verified answer
COS(π-α)= - COS(α), SIN(π/2-α)=COS(α), SIN(2α)=2SIN(α)COS(α)[SIN(2α)+COS(π-α)SIN(α)]/SIN(π/2-α)=[2SIN(α)COS(α)-COS(α)SIN(α)]/COS(α) = SIN(α)COS(α)/COS(α) = SIN(α)
2) COS(2α) / (SIN(α)-COS(α))↑2-SIN(2α)
COS(2α) = COS²(α) - SIN²(α) = (COS(α) - SIN(α) )·(COS(α) + SIN(α) )
[COS(2α)/( SIN(α) - COS(α))]²-SIN(2α) =
=[ (COS(α) - SIN(α) )·(COS(α) + SIN(α) )/(SIN(α) -COS(α)) ]²-
-2COS(α) SIN(α)=
= COS²(α) + 2COS(α)SIN(α) +SIN²(α)-2COS(α) SIN(α)=
=COS²(α) +SIN²(α)=1