Ответы на вопрос:
Tgx = 1-√2 1-tgx=√2 tgπ/4 - tgx=√2 ; tgα-tgβ = sin(α-β)/cosαcosβ sin(45-x)/(cos45·cosx =√2 sin(45-x)[√2/2·cosx] = √2 ⇔ sin(45-x)/cosx = √2·√2/2 =1 sin(45 - x)= cosx cos(90-(45-x)) -cosx=0 cos(45+x) - cosx =0 -2sin[(45+2x)/2] ·sin(45/2) =0 ⇒ sin(π/8 +x) =0 π/8 +x = πk ; k ∈ z x = -π/8 +πk ; k ∈ z
Популярно: Алгебра
-
hjhthy18.11.2021 02:53
-
lizasyper615.04.2022 18:59
-
zeriss23ozvp1825.03.2023 04:32
-
и5п5п28.03.2023 13:24
-
НезнайкаЛёша28.01.2021 19:31
-
BlankDawn24.03.2023 16:12
-
Назмина13413.07.2021 21:06
-
Snihdsfg10.09.2020 11:12
-
емдетей14.06.2022 18:41
-
silverside03.09.2021 15:02