Question

Zadanie2
Podaj wartości pozostałych funkcji trygonometrycznych
Zadanie3
Oblicz wartość wyrażenia​

Answer 1

Verified answer

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

[tex]sin^{2}\alpha =1-\frac{1}{16} \\sin^{2} \alpha =\frac{15}{16} /\sqrt{}[/tex]

[tex]sin\alpha =-\frac{\sqrt{15} }{4}[/tex]     (- ponieważ to 4 ćwiartka ,tam sinus jest ujemny)

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

[tex]tg\alpha =\frac{\frac{\sqrt{15} }{-4} }{\frac{1}{4} } =-\frac{\sqrt{15} }{4} *\frac{4}{1} =-\sqrt{15}[/tex]

[tex]ctg\alpha =\frac{1}{tg\alpha } \\ctg\alpha =\frac{1}{-\sqrt{15} } =\frac{-\sqrt{15} }{15}[/tex]

b)

[tex]tg\alpha =-3[/tex]   α∈(90,180)

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

[tex]9cos^{2} \alpha +cos^{2} \alpha =1\\10cos^{2}\alpha =1/:(10)\\ cos^{2}\alpha =\frac{1}{10} /\sqrt{}[/tex]

[tex]cos\alpha =-\frac{1}{\sqrt{10} } =-\frac{\sqrt{10} }{10}[/tex]

[tex]sin\alpha =-3cos\alpha \\sin\alpha =\frac{3\sqrt{10} }{10}[/tex]

[tex]ctg\alpha =\frac{1}{tg\alpha } \\ctg\alpha =\frac{1}{-3} =-\frac{1}{3}[/tex]

Zad3

[tex]\frac{-3cos\alpha +6sin\alpha /:(sin\alpha) }{3cos\alpha +8sin\alpha /:(sin\alpha )} =\frac{-3ctg\alpha +6}{3ctg\alpha +8} =\frac{-3*(-2)+6}{3*(-2)+8} =\frac{6+6}{-6+8} =\frac{12}{2} =6[/tex]

[tex]ctg\alpha =-2[/tex]