Question

Wiedząc że dla pewnego kata ostrego α sinα=⅓ oblicz cos α, tg α, i ctg α.

Answer 1

sin α = 1/3
sin²α + cos²α = 1 ---> cos²α = 1 - sin²α
czyli cos²α = 1 - (1/3)² = 9/9 - 1/9 = 8/9
zatem cos α = √(8/9) = √8/√9 = (2√2)/3
tg α = sinα / cos α = (1/3) / [ (2√2)/3] = (1/3) * [ 3/ (2√2)] =
= 1/(2√2) = √2/4
ctg α = 1/ tg α = 4/√2

Answer 2

Sin2α + cos2α = 1
cos2α = 1 - sin2α
cos2α = 1 - 1/9
cos2α = 8/9
cosα = √8/3
cosα = (2√2)/3

tgα = sinα/cosα
tgα = (1/3)/(2√2/3)
tgα = 1/2√2
tgα = √2/4

ctgα = cosα/sinα
ctgα = 1/tgα
ctgα = 1/(√2/4)
ctgα = 4/√2
ctgα = 2√2

Odp: cosα = (2√2)/3 , tgα = √2/4 , ctgα = 2√2

(nie wszystkie nawiasy są potrzebne. Dodałem je dla lepszej widoczności)

Answer 3

Sinα=⅓
kąt α - ostry

Korzystam z jedynki trygonometrycznej

cos^2α = 1- sin^2α
cos^2α = 1- 1/9
cos^2α = 8/9

cosα = (2√2)/3

tgα = sinα/cosα
tgα = (1/3)/[(2√2)/3]
tgα = 1/(2√2)

tgα = (√2)/4

ctgα = 1/tgα
ctgα = 4/(√2)

ctgα = 2√2