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WenZ07er
@WenZ07er
November 2021
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5
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Здравствуйте! Помогите, пожалуйста, упростить два выражения. Не понимаю,какую именно формулу нужно применить .
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Answers & Comments
m11m
1)
cosα
=
cos(2*(α/2))
=
cos(α/2) + sin(α/2) cos(α/2) + sin(α/2)
=
cos²(α/2) - sin²(α/2)
=
(cos(α/2) - sin(α/2)) (cos(α/2) + sin(α/2))
=
cos(α/2) + sin(α/2) cos(α/2) + sin(α/2)
=cos(α/2) - sin(α/2)
2) 2cos²(135° -2.5α) - 1 = -sin5α
По действиям:
1) cos(135° -2.5α)=cos(90° +45° -2.5α) = -sin(45° - 2.5α) =
= -(sin45° cos2.5α - cos45° sin2.5α) =-
√2
cos2.5α +
√2
sin2.5α =
2 2
=
√2
(sin2.5α - cos2.5α)
2
2) 2*(
√2
)² (sin2.5α -cos2.5α)² -1 =
2*2
(sin2.5α-cos2.5α)² -1 =
( 2 ) 4
=(sin2.5α-cos2.5α)² -1 = sin² 2.5α - 2cos2.5α sin2.5α +cos² 2.5α -1 =
=(cos² 2.5α + sin² 2.5α) - 1- 2sin2.5α cos2.5α =
=1-1 - sin(2*2.5α)=
=-sin5α
4 votes
Thanks 4
WenZ07er
Огромное спасибо! Очень подробное решение! Теперь все понятно.
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Answers & Comments
cos(α/2) + sin(α/2) cos(α/2) + sin(α/2)
= cos²(α/2) - sin²(α/2) = (cos(α/2) - sin(α/2)) (cos(α/2) + sin(α/2)) =
cos(α/2) + sin(α/2) cos(α/2) + sin(α/2)
=cos(α/2) - sin(α/2)
2) 2cos²(135° -2.5α) - 1 = -sin5α
По действиям:
1) cos(135° -2.5α)=cos(90° +45° -2.5α) = -sin(45° - 2.5α) =
= -(sin45° cos2.5α - cos45° sin2.5α) =- √2 cos2.5α + √2 sin2.5α =
2 2
=√2 (sin2.5α - cos2.5α)
2
2) 2*(√2)² (sin2.5α -cos2.5α)² -1 = 2*2 (sin2.5α-cos2.5α)² -1 =
( 2 ) 4
=(sin2.5α-cos2.5α)² -1 = sin² 2.5α - 2cos2.5α sin2.5α +cos² 2.5α -1 =
=(cos² 2.5α + sin² 2.5α) - 1- 2sin2.5α cos2.5α =
=1-1 - sin(2*2.5α)=
=-sin5α