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Fakla
@Fakla
July 2022
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Нужно упросить дробь: (4cos^2(2a)-4cos^2(a)+3sin^2(a))/(4sin^2(a)-sin^2(2a))
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sedinalana
(4cos^2(2a)-4cos^2(a)+3sin^2(a))/(4sin^2(a)-sin^2(2a))=
=(4cos
²a-4sin²a-4cos²a+3sin²a)/(4sin²a-4sin²acos²a)=
=-sin²a/[4sin²a(1-cos²a)]=-1/4sin²a
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Answers & Comments
=(4cos²a-4sin²a-4cos²a+3sin²a)/(4sin²a-4sin²acos²a)=
=-sin²a/[4sin²a(1-cos²a)]=-1/4sin²a