Question

4. Oblicz: sin² a, sin a i tga, jeśli a jest kątem rozwartym oraz.
a) cos a = - 1/4
b) cos a = - 2/3
c) cos a = - √5/5​

Answer 1

Odpowiedź:

Szczegółowe wyjaśnienie:

α∈(90°,180°)  ⇒ II ćw.

a.

[tex]cos\alpha =-\frac{1}{4}[/tex]

     [tex]sin^2\alpha +cos^2=1[/tex]

[tex]sin^2\alpha=1-cos^2\alpha =1-(cos\alpha )^2=1-(-\frac{1}{4} )^2=1-\frac{1}{16} =\frac{15}{16}[/tex]

     [tex]sin^2\alpha =\frac{15}{16}[/tex]

[tex]sin_1\alpha =\frac{\sqrt{15} }{4}[/tex] ∈ II ćw.         lub    [tex]sin_2\alpha =-\frac{\sqrt{15} }{4}[/tex] ∉ II ćw.

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

[tex]tg\alpha =\frac{\frac{\sqrt{15} }{4} }{-\frac{1}{4} } =\frac{\sqrt{15} }{4}*(-4)=-\sqrt{15}[/tex] ∈ II ćw.

b.

[tex]cos\alpha =-\frac{2}{3}[/tex]

    [tex]sin^2\alpha +cos^2=1[/tex]

[tex]sin^2\alpha=1-cos^2\alpha =1-(cos\alpha )^2=1-(-\frac{2}{3} )^2=1-\frac{4}{9} =\frac{5}{9}[/tex]

     [tex]sin^2\alpha =\frac{5}{9}[/tex]

[tex]sin_1\alpha =\frac{\sqrt{5} }{3}[/tex] ∈ II ćw.         lub    [tex]sin_2\alpha =-\frac{\sqrt{5} }{3}[/tex] ∉ II ćw.

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

[tex]tg\alpha =\frac{\frac{\sqrt{5} }{3} }{-\frac{2}{3} } =\frac{\sqrt{5} }{3}*(-\frac{3}{2} )=-\frac{1}{2} \sqrt{5}[/tex] ∈ II ćw.

c.

[tex]cos\alpha =-\frac{\sqrt{5} }{5}[/tex]

    [tex]sin^2\alpha +cos^2=1[/tex]

[tex]sin^2\alpha=1-cos^2\alpha =1-(cos\alpha )^2=1-(-\frac{\sqrt{5} }{5} )^2=1-\frac{5}{25} =1-\frac{1}{5}= \frac{4}{5}[/tex]

     [tex]sin^2\alpha =\frac{20}{25}[/tex]

[tex]sin_1\alpha =\frac{\sqrt{20} }{5}=\frac{2\sqrt{5} }{5}[/tex] ∈ II ćw.         lub    [tex]sin_2\alpha =-\frac{\sqrt{20} }{5}=-\frac{2\sqrt{5} }{5}[/tex] ∉ II ćw.

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

[tex]tg\alpha =\frac{\frac{2\sqrt{5} }{5} }{-\frac{\sqrt{5} }{5} } =\frac{2\sqrt{5} }{5}*(-\frac{5}{\sqrt{5} } )=-2[/tex] ∈ II ćw.