Question

Związki między funkcjami trygonometrycznymi.

zad.1
Wyznacz wartość funkcji trygonometrycznych kąta ostrego α,jeśli:
a) sin(90°-α)=3/10
b) cos(90°-α)√3/2
c) ctg(90°-α)=2/5

Answer 1

A)
sin (90⁰ - α ) = 3/10 < => cos α = 3/10
sin²α = 1 - cos²α = 1 - (3/10)² = 1 - 9/100 = 91/100
sin α = √91 : √100 = √91/10
tg α = sin α : cos α = √91/10 : (3/10) = √91/10 *(10/3) = √91/3
ctg α = 1/ tg α = 3/√91
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b)
cos ( 90⁰ - α) = √3/2 < => sin α = √3/2
cos²α = 1 - sin²α = 1 - (√3/2)² = 1 - 3/4 = 1/4
cos α = 1/2
tg α = (√3/2) : (1/2) = √3
ctg α = 1 /√3
α = 60⁰
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c)
ctg ( 90⁰ - α) = 2/5 <=> tg α = 2/5
tg = y/x --> y = 2 ; x = 5
r² = x² + y² = 5² + 2² =25 + 4 = 29
r = √29
sin α = y/r = 2/√29
cos α = x/r = 5/√29
ctg α = 1/ (2/5) = 5/2 = 2,5
===================================

Answer 2

Zad. 1)

a)
sin(90°- α) = 3/10

sin(90°- α) = cosα

cosα = 0,3

sin²α + cos²α = 1

sin²α + 0,3² = 1

sin²α = 1 - 0,3²

sin²α = 1 - 0,09

sin²α = 0,91

sin²α - 0,91 = 0

(sinα - √0,91)(sinα + √0,91) = 0

sinα - √91/10 = 0 lub sinα + √91/10 = 0

sinα = √91/10 lub sinα = - √91/10

tgα = sinα/cosα

tgα = (√91/10)/0,3 lub tgα = - (√91/10)/0,3

tgα = √91/3 lub tgα = - √91/3

ctgα = 1/tgα

ctgα = 3√91/91 lub ctgα = - 3√91/91


b)
cos(90°- α) = √3/2

cos(90°- α) = sinα

sinα = √3/2

sin²α + cos²α = 1

(3/4)² + cos²α = 1

cos²α = 1 - (3/4)²

cos²α = 1 - 9/16

cos²α = 7/16

cos²α - (√7/4)² = 0

(cosα - √7/4)(cosα + √7/4) = 0

cosα - √7/4 = 0 lub cosα + √7/4 = 0

cosα = √7/4 lub cosα = - √7/4

tgα = sinα/cosα

tgα = (√3/2)/(√7/4) lub tgα = - (√3/2)/(√7/4)

tgα = 2√21/7 lub tgα = - 2√21/7

ctgα = 1/tgα

ctgα = √21/6 lub ctgα = - √21/6


c)
ctg(90°- α) = 2/5

ctg(90°- α) = tgα

tgα = 0,4

ctgα = 2,5

tgα = sinα/cosα

sinα = 0,4cosα

sin²α + cos²α = 1

(0,4cosα)² + cos²α = 1

1,16cos²α = 1 /:(116/100)

cos²α = 100/116

cosα = 5√29/29 lub cosα = - 5√29/29

sinα = 2√29/29 lub sinα = - 2√29/29