Odpowiedź:
zad 5
a)
cosα = 1/5
cos²α=(1/5)² = 1/25
1 - sin²α = 1/25
sin²α = 1 - 1/25 = 25/25 - 1/25 = 24/25
sinα = √(24/25) = √24/5 = √(4 * 6)/5 = 2√6/5
tgα = sinα : cosα = 2√6/5 : 1/5 = 2√6/5 * 5 = 2√6
ctgα = 1/tgα = 1/2√6 = √6/(2 * 6) = √6/12
b)
tgα = 3
tg²α = 3² = 9
sin²α/cos²α = 9
sin²α = 9cos²α = 9(1 - sin²α) = 9 - 9sin²α
sin²α + 9sin²α = 9
10sin²α = 9
sin²α = 9/10
sinα = √(9/10) = 3/√10 = 3√10/10
sin²α = 1 - cos²α
cos²α = 1 - sin²α = 1 - 9/10 = 1/10
cosα = √(1/10) = 1/√10 = √10/10
tgα = sinα/cosα = 3√10/10 : √10/10 = 3√10/10 * 10/√10 = 3
ctgα = 1/tgα = 1/3
c)
sinα = 8/17
sin²α =(8/17)² = 64/289
1 - cos²α = 64/289
cos²α = 1 - 64/289 = 225/289
cosα = √(225/289) = 15/17
tgα = sinα/cosα = 8/17 : 15/17 = 8/17 * 17/15 = 8/15
ctgα = 1/tgα = 15/8 = 1 7/8
