Question

Proszę o szybką pomoc

Answer 1

Odpowiedź:

sinα=3/5 ; α ∈ ( 90° ,180°)

W tym przedziale mamy:

sinα = sinα

cosα =  - cosα

tgα = - tgα

ctgα = - ctgα

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sinα = 3/5

sin²α = (3/5)²= 9/25

1- cos²α = 9/25

cos²α= 1- 9/25 = 25/25- 9/25 = 16/25

cosα = √(16/25) = 4/5

cos(180°- α) = - cosα = - 4/5

tgα = sinα/cosα = 3/5 :4/5 = 3/5 * 5/4 = 3/4

tg(180 - α) = - tgα = - 3/4

ctgα = 1/tgα= 4/3

ctg(180° - α) = - ctgα = - 4/3= - 1 1/3