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BezymFox
@BezymFox
July 2023
1
10
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Упростите: 1) 4cos3asin3acos6a 2)
cos4a
---------
cos2a-sin2a
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alohadens
1)4cos(3a)sin(3a)cos(6a) = 2) 2sin(6a)cos(6a)sin(3a)
cos(4a)/(cos(2a)-sin(2a)) = (cos²(2a)+sin²(2a))cos(4a)/(cos²(2a)-sin²(2a)) = cos²(2a)cos(4a)/(cos²(2a)-sin²(2a)) = cos(2a)cos(4a)/(cos(2a)+sin(2a))(cos(2a)-sin(2a)) = cos(2a)cos(4a)/(cos²(2a)-sin²(2a)) = cos(2a)cos(4a)/cos(4a) = cos(2a)
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BezymFox
Можеш по подробней обьяснить 1 Пожалуйста
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Answers & Comments
cos(4a)/(cos(2a)-sin(2a)) = (cos²(2a)+sin²(2a))cos(4a)/(cos²(2a)-sin²(2a)) = cos²(2a)cos(4a)/(cos²(2a)-sin²(2a)) = cos(2a)cos(4a)/(cos(2a)+sin(2a))(cos(2a)-sin(2a)) = cos(2a)cos(4a)/(cos²(2a)-sin²(2a)) = cos(2a)cos(4a)/cos(4a) = cos(2a)