; α∈ (
)
sin α, tg α, ctg α - ?
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Т.к. α∈(
) ⇒ все функции кроме sin α будут отрицательными
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Из основного тригонометрического тождества sin²α + cos²α = 1 выразим sin α
sin α = √1-cos²α

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

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

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Ответ:
,
, 