Question

Oblicz wartość wyrażenia:
√5 (ctgα-tgα),jeśli sinα= 2/3

Sprawdź tożsamość:
1+ctgα=(sinα+cosα)/sinα

Answer 1

Zad. 1)

Dane:

sinα= 2/3

√5 (ctgα-tgα) = ?

tgα = sinα/cosα

ctgα = 1/tgα

sin²α + cos²α = 1

cos²α = 1 - sin²α

cos²α = 1 - (2/3)²

cos²α = 1 - 4/9

cos²α = 5/9

cos²α - 5/9 = 0

(cosα - √5/3)(cosα + √5/3) = 0

cosα - √5/3 = 0 ∨ cosα + √5/3 = 0

cosα = √5/3 ∨ cosα = - √5/3

tgα = sinα/cosα

tgα = (2/3): (√5/3) ∨ tgα = (2/3): (- √5/3)

tgα = 0,4√5 ∨ tgα = - 0,4√5

ctgα = 1/0,4√5 ∨ ctgα = - 1/0,4√5

ctgα = 0,5√5 ∨ ctgα = - 0,5√5

√5 (ctgα-tgα) = √5 (0,5√5 - 0,4√5) = √5*0,1√5 = 0,5

lub

5 (ctgα-tgα) = √5 (- 0,5√5 + 0,4√5) = - √5*0,1√5 = - 0,5

Zad. 2)

1 + ctgα = (sinα + cosα)/sinα

P = (sinα + cosα)/sinα = sinα/sinα + cosα/sinα =

= 1 + ctgα =

= P