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SielandSatan
@SielandSatan
August 2022
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Найти 4sin^2b, если cos2b=0.12
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sedinalana
Verified answer
Cos2b=1-2sin²b
2sin²b=1-cos2b
4sin²b=2*(1-cos2b)=2*(1-0,12)=2*0,88=1,76
3 votes
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Dимасuk
Verified answer
Cos2b = 1 - 2sin²b, отсюда
2sin²b = 1 - cos2b
Тогда 4sin²b = 2(1 - cos2b) = 2(1 - 0,12) = 2·0,88 = 1,76
2 votes
Thanks 2
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Answers & Comments
Verified answer
Cos2b=1-2sin²b2sin²b=1-cos2b
4sin²b=2*(1-cos2b)=2*(1-0,12)=2*0,88=1,76
Verified answer
Cos2b = 1 - 2sin²b, отсюда2sin²b = 1 - cos2b
Тогда 4sin²b = 2(1 - cos2b) = 2(1 - 0,12) = 2·0,88 = 1,76