Question

Wyznacz pozostałe funkcje trygonometryczne kąta ostrego ∝, gdy:

a) cos∝=60/61
b) tg∝=2 2/5

Answer 1

Odpowiedź:

Szczegółowe wyjaśnienie:

a) boki: a = ?

b = 60

c = 61

[tex]a^{2} = 61^{2} - 60^{2}\\ a^{2} = 3721 - 3600\\ a^{2} = 121 / \sqrt{}\\ a = 11[/tex]

sin = 11/61

tg = 11/60

ctg = 60/11

b) boki:

12/5 = 2 2/5

a = 5

b = 12

c = [tex]\sqrt{144 + 25}[/tex] = [tex]\sqrt{169}[/tex]

c = 13

sin = 12/13

cos = 5/13

ctg = 5/12

Answer 2

Odpowiedź:

Szczegółowe wyjaśnienie:

a.

[tex]cos\alpha =\frac{60}{61}[/tex]       α∈(0°,90°)  ⇒ I ćw.

        [tex]sin^2\alpha +cos^2\alpha =1[/tex]

[tex]sin^2\alpha +(\frac{60}{61} )^2=1\\sin^2\alpha +\frac{3600}{3721} -1=0\\sin^2\alpha -\frac{121}{3721} =0\\(sin\alpha -\frac{11}{61} )(sin\alpha +\frac{11}{61} )=0[/tex]

[tex]sin\alpha =\frac{11}{61}[/tex]  ∈  I ćw.     v    [tex]sin\alpha =-\frac{11}{61}[/tex]  ∉ I ćw.

         [tex]tg\alpha =\frac{sin\alpha }{cos\alpha }[/tex]

[tex]tg\alpha =\frac{\frac{11}{61} }{\frac{60}{61} } =\frac{11}{61} *\frac{61}{60} =\frac{11}{60}[/tex]  ∈ I ćw.

        [tex]ctg\alpha =\frac{1}{tg\alpha }[/tex]

[tex]ctg\alpha =\frac{1}{\frac{11}{60} } =1*\frac{60}{11} =5\frac{5}{11}[/tex]  ∈ I ćw.

b.

[tex]tg\alpha =2\frac{2}{5} =\frac{12}{5}[/tex]      α∈(0°,90°)  ⇒ I ćw.

       [tex]ctg\alpha =\frac{1}{tg\alpha }[/tex]

[tex]ctg\alpha =\frac{1}{\frac{12}{5} } =\frac{5}{12}[/tex]   ∈ I ćw.

       [tex]tg\alpha =\frac{sin\alpha }{cos\alpha }[/tex]

[tex]\frac{12}{5} =\frac{sin\alpha }{cos\alpha }[/tex]    /*cosα

[tex]sin\alpha =\frac{12}{5} cos\alpha[/tex]    [***]

          [tex]sin^2\alpha +cos^2\alpha =1[/tex]

[tex](\frac{12}{5} cos\alpha )^2+cos^2\alpha =1\\\frac{144}{25} cos^2\alpha +cos^2\alpha =1\\\frac{169}{25} cos^2\alpha -1=0\\cos^2\alpha-\frac{25}{169} =0\\(cos\alpha -\frac{5}{13} )(cos\alpha +\frac{5}{13} )=0[/tex]

[tex]cos\alpha =\frac{5}{13}[/tex]  ∈ I ćw.      v           [tex]cos\alpha =-\frac{5}{13}[/tex]  ∉ I ćw.

  [***]

[tex]sin\alpha =\frac{12}{5} *\frac{5}{13}=\frac{12}{13}[/tex]  ∈ I ćw.