Odpowiedź:
zad 1
a)
sinα = 1/3
sin²α = (1/3)² = 1/9
1 - cos²α = 1/9
cos²α = 1 - 1/9 = 9/9 - 1/9 = 8/9
cosα = √(8/9) = √8/3 = √(4 * 2)/3 = 2√2/3
tgα = sinα/cosα = 1/3 : 2√2/3 = 1/3 * 3/2√2 = 1/2√2 = √2/(2 * 2) = √2/4
ctgα = 1/tgα = 4/√2 = 4√2/2 = 2√2
b)
cosα = 3/5
cos²α = (3/5)² = 9/25
1 - sin²α = 9/25
sin²α = 1 - 9/25 = 25/25 - 9/25 = 16/25
sinα = √(16/25) = 4/5
tgα = sinα/cosα = 4/5 : 3/5 = 4/5 * 5/3 = 4/3 = 1 1/3
ctgα = 1/tgα = 3/4
c)
tgα = 4/5
sinα/cosα = 4/5
sin²α/cos²α = (4/5)² = 16/25
25sin²α = 16cos²α = 16(1 - sin²α) = 16 - 16sin²α
25sin²α + 16sin²α = 16
41sin²α = 16
sin²α = 16/41
sinα = √(16/41) = 4/√41 = 4√41/41
25sin²α = 16cos²α
25(1 - sin²α) = 16cos²α
25 - 25cos²α = 16cos²α
16cos²α + 25cos²α = 25
41cos²α = 25
cos²α = 25/41
cosα = √(25/41) = 5/√41 = 5√41/41
tgα = sinα/cosα = 4/√41/41 : 5√41/41 = 4√41/41 * 41/5√41 = 4/5
ctgα = 1/tgα = 5/4 = 1 1/5
zad 2
cosα = 7/25
cos²α = (7/25)² = 49/625
1 - sin²α = 49/625
sin²α = 1 - 49/625 = 625/625 - 49/625 = 576/625
sinα = √(576/625) = 24/25
tgα = sinα/cosα = 24/25 : 7/25 = 24/25 * 25/7 = 24/7 = 3 3/7
Jeżeli w treści zadania 3³⁾⁷ oznacza 3 i 3/7 to również C jest prawdziwe
Odp: B
