Предмет: Алгебра, автор: rom4ikilin

Сравните
arcsin (-1/2) и arctg(-1)
arccos(-√3/2) и arctg√3

Ответы

Автор ответа: xERISx

0

$arcsinbigg(-dfrac 12bigg)=-dfrac {pi}6~~ >~~-dfrac {pi}4=arctg(-1)\\boxed{boldsymbol{arcsinbigg(-dfrac 12bigg) >arctg(-1)}}\\\arccosbigg(-dfrac {sqrt3}2bigg)=dfrac {5pi}6~~>~~dfrac {pi}3=arctg(sqrt3)\\boxed{boldsymbol{arccosbigg(-dfrac {sqrt3}2bigg)>arctg(sqrt3)}}$

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