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oxjuckova2015
@oxjuckova2015
August 2021
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Помогите по алгебре
1)2 cos^2 15градусов * tg15градусов
2)1-sin^2*22°30'
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nKrynka
Решение
1)2 cos^2 15градусов * tg15градусов = (cos²15 * sin15)/ cos15 =
= (1/2)*sin30 = 1/4
2)1-sin^2*22°30' = cos²22°30' = (1 + cos45)/2 = 1/2 + (1/2)*√2/2 = 1/2 + √2/4
1 votes
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oxjuckova2015
Спасибо большое!
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Answers & Comments
1)2 cos^2 15градусов * tg15градусов = (cos²15 * sin15)/ cos15 =
= (1/2)*sin30 = 1/4
2)1-sin^2*22°30' = cos²22°30' = (1 + cos45)/2 = 1/2 + (1/2)*√2/2 = 1/2 + √2/4