Sin (-23pi/4) tg19 pi/6

Question

Алгебра

Ангеля

3 мая, 19:06

Sin (-23pi/4) tg19 pi/6

+4

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Лора · Answer 1

Sin (-23π/4) · tg (19π/6) = sin (-24π/4 - π/4) · tg (18π/6 + π/6) =

= sin (-6π - π/4) · tg (3π + π/6) = sin ( - π/4) · tg (3π + π/6) =

=-0.5√2 · tg (π + π/6) = - 0.5√2 · tg (π + π/6) = - 0.5√2 · tg (π/6) =

= - 0.5√2 ·√3 / 3 = - √6 / 6 = - 1/√6