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kosi69
@kosi69
July 2022
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формулы двойного и половинного угла.вариант В1,задание 3
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Cosα cosα cosα ·sin(α/2)
------------------- = ---------------------------------------- = ----------------------------=
ctg(α/2)-sinα cos(α/2) cos(α/2)(1-2sin(α/2))
----------- -2sin(α/2)cos(α/2)
sin(α/2)
cosα ·sin(α/2) sin(α/2)
= ----------------------- = -------------- = tg (α/2)
cos(α/2) cosα cos(α/2)
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Answers & Comments
Verified answer
Cosα cosα cosα ·sin(α/2)------------------- = ---------------------------------------- = ----------------------------=
ctg(α/2)-sinα cos(α/2) cos(α/2)(1-2sin(α/2))
----------- -2sin(α/2)cos(α/2)
sin(α/2)
cosα ·sin(α/2) sin(α/2)
= ----------------------- = -------------- = tg (α/2)
cos(α/2) cosα cos(α/2)