Odpowiedź:
Jedynka trygonometryczna:
[tex]sin^2\alpha +cos^2\alpha =1[/tex]
Inne tożsamości, z których należy skorzystać aby rozwiązać zadanie:
[tex]tg\alpha =\frac{sin\alpha}{cos\alpha}[/tex]
[tex]ctg\alpha=\frac{cos\alpha }{sin\alpha }[/tex]
a)
[tex]sin\alpha =0,7=\frac{7}{10}\\cos\alpha=0,3=\frac{3}{10}[/tex]
[tex](\frac{7}{10})^2+(\frac{3}{10})^2=1\\\frac{49}{100}+\frac{9}{100}=1\\\frac{58}{100}\neq 1[/tex]
Nie istnieje taki kąt.
b)
[tex]sin\alpha =0,25=\frac{1}{4}\\tg\alpha =2[/tex]
[tex]2=\frac{\frac{1}{4}}{cos\alpha}\\2cos\alpha =\frac{1}{4}\\cos\alpha=\frac{1}{8}[/tex]
[tex](\frac{1}{4})^2+(\frac{1}{8})^2=1\\\frac{1}{16}+\frac{1}{64}=1\\\frac{4}{64}+\frac{1}{64}=1\\\frac{5}{64}\neq 1[/tex]
c)
[tex]cos\alpha =0,3=\frac{3}{10}\\ctg\alpha =1,5=\frac{3}{2}[/tex]
[tex]\frac{3}{2}=\frac{\frac{3}{10}}{sin\alpha}\\\frac{3}{2}sin\alpha =\frac{3}{10}\\sin\alpha =\frac{3}{10}*\frac{2}{3}\\sin\alpha =\frac{1}{5}[/tex]
[tex](\frac{1}{5})^2+(\frac{3}{10})^2=1\\\frac{1}{25}+\frac{9}{100}=1\\\frac{4}{100}+\frac{9}{100}=1\\\frac{13}{100}\neq 1[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
Odpowiedź:
Jedynka trygonometryczna:
[tex]sin^2\alpha +cos^2\alpha =1[/tex]
Inne tożsamości, z których należy skorzystać aby rozwiązać zadanie:
[tex]tg\alpha =\frac{sin\alpha}{cos\alpha}[/tex]
[tex]ctg\alpha=\frac{cos\alpha }{sin\alpha }[/tex]
a)
[tex]sin\alpha =0,7=\frac{7}{10}\\cos\alpha=0,3=\frac{3}{10}[/tex]
[tex]sin^2\alpha +cos^2\alpha =1[/tex]
[tex](\frac{7}{10})^2+(\frac{3}{10})^2=1\\\frac{49}{100}+\frac{9}{100}=1\\\frac{58}{100}\neq 1[/tex]
Nie istnieje taki kąt.
b)
[tex]sin\alpha =0,25=\frac{1}{4}\\tg\alpha =2[/tex]
[tex]tg\alpha =\frac{sin\alpha}{cos\alpha}[/tex]
[tex]2=\frac{\frac{1}{4}}{cos\alpha}\\2cos\alpha =\frac{1}{4}\\cos\alpha=\frac{1}{8}[/tex]
[tex]sin^2\alpha +cos^2\alpha =1[/tex]
[tex](\frac{1}{4})^2+(\frac{1}{8})^2=1\\\frac{1}{16}+\frac{1}{64}=1\\\frac{4}{64}+\frac{1}{64}=1\\\frac{5}{64}\neq 1[/tex]
Nie istnieje taki kąt.
c)
[tex]cos\alpha =0,3=\frac{3}{10}\\ctg\alpha =1,5=\frac{3}{2}[/tex]
[tex]ctg\alpha=\frac{cos\alpha }{sin\alpha }[/tex]
[tex]\frac{3}{2}=\frac{\frac{3}{10}}{sin\alpha}\\\frac{3}{2}sin\alpha =\frac{3}{10}\\sin\alpha =\frac{3}{10}*\frac{2}{3}\\sin\alpha =\frac{1}{5}[/tex]
[tex]sin^2\alpha +cos^2\alpha =1[/tex]
[tex](\frac{1}{5})^2+(\frac{3}{10})^2=1\\\frac{1}{25}+\frac{9}{100}=1\\\frac{4}{100}+\frac{9}{100}=1\\\frac{13}{100}\neq 1[/tex]
Nie istnieje taki kąt.