Odpowiedzi

2010-02-13T12:23:13+01:00
Tgα + (1/tgα) = 3/cos α , dla 0⁰ < α < 90⁰

sinα/cosα + cosα / sinα = 3 /cosα
[sin²α + cos²α]/[sinα cosα] = 3 /cosα
1/[sinα cos α] = 3/ cosα
cosα = 3 sinα cosα
3 sin α = 1
sin α = 1/3

cos²α = 1 - sin²α = 1 -(1/3)² = 9/9 - 1/9 = 8/9
cos α = √8/√9 = 2√2 /3

tg α = sinα / cosα = (1/3) /(2√2 /3) = (1/3)*(3/2√2) = 1/2√2 =√2/4

ctg α = 1 /tgα = 1/(√2/4) = 4/√2 = 2√2
1 5 1