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МатематикаВопрос:
13) tg(-17π/4) ⋅ sin(-25π/6) =
Ответ:
Ответ: -1/2
Краткое пояснение: Используем периодичность и свойства синуса и тангенса.
tg(-17π/4) ⋅ sin(-25π/6) = -tg(17π/4) ⋅ (-sin(25π/6)) = tg(4π + π/4) ⋅ sin(4π + π/6) = tg(π/4) ⋅ sin(π/6) = 1 ⋅ (1/2) = 1/2
Ответ: 1/2
Похожие
- 1) sin(-π/4) = -sin π/4 = -√2/2
- 2) cos(-π/2) =
- 3) tg(-π/3) =
- 4) cos(-π/4) =
- 5) sin(-π/2) =
- 6) cos(-7π/3) =
- 7) tg(-13π/3) =
- 8) sin(-25π/4) =
- 9) sin(-25π/6) + tg(-π/4) =
- 10) ctg(-π/2) - cos(-7π/6) =
- 11) √2 sin(-π/4) ⋅ cos(-π/3) =
- 12) √6 cos(-9π/4) ⋅ ctg(-7π/3) =
- 14) 3 cos(-5π) ⋅ sin(-7π/2) =
- 15) sin²(π/3) + cos²(-π/4) - tg²(-19π/3) =
- 16) 4 cos²(-π/3) - 4 sin²(-π/3) + tg²(-π/4) =
- 1) sin(-α) + sin α =
- 2) tg α - tg (-α) =
- 3) sin(-α) / cos(-α) =
- 4) 2 cos(-α) / sin(-α) =
- 5) tg(-α) ⋅ ctg (-α) - sin²(-α) =
- 6) ctg(-α) + cos(-α) - sin(-α) / sin α =
- 7) tg(-α) ⋅ sin(-α) - 1/cos α =
- 8) ctg(-α) ⋅ sin(-α) - cos(-α) =
