Odpowiedź:
[tex]a)\\sin\alpha=\frac{12}{13}\\cos^2\alpha=1-(\frac{12}{13})^2\\cos^2\alpha=1-\frac{144}{169}\\cos^2\alpha=\frac{25}{169}\\cos\alpha=\frac{5}{13}\\\\tg\alpha=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{13}*\frac{13}5=\frac{12}5=2\frac25\\\\ctg\alpha=\frac{1}{\frac{12}5}=1*\frac5{12}=\frac5{12}[/tex]
[tex]b)\\\\cos\alpha=\frac8{17}\\sin^2\alpha=1-(\frac8{17})^2\\sin^2\alpha=1-\frac{64}{289}\\sin^2\alpha=\frac{225}{289}\\sin\alpha=\frac{15}{17}\\\\tg\alpha=\frac{\frac{15}{17}}{\frac{8}{17}}=\frac{15}{17}*\frac{17}8=\frac{15}8=1\frac78\\\\ctg\alpha=\frac{1}{\frac{15}8}=1*\frac8{15}=\frac8{15}[/tex]
[tex]c)\\\\tg\alpha=4\\\frac{sin\alpha}{cos\alpha}=4\\\frac{sin^2\alpha}{cos^2\alpha}=16\\\frac{sin^2\alpha}{1-sin^2\alpha}=16\\sin^2\alpha=16(1-sin^2\alpha)\\sin^2\alpha=16-16sin^2\alpha /+16sin^2\alpha\\17sin^2\alpha=16 /:17\\sin^2\alpha=\frac{16}{17}\\sin\alpha=\frac{4}{\sqrt{17}}=\frac{4\sqrt{17}}{17}\\\\cos^2\alpha=1-\frac{16}{17}\\cos^2\alpha=\frac1{17}\\cos\alpha=\frac{\sqrt{17}}{17}\\\\ctg\alpha=\frac{1}4[/tex]
[tex]d)\\\\ctg\alpha=\frac{11}{60}\\tg\alpha*ctg\alpha=1\\tg\alpha*\frac{11}{60}=1 /*\frac{60}{11}\\tg\alpha=\frac{60}{11}\\\\\frac{sin\alpha}{cos\alpha}=\frac{60}{11}\\\frac{sin^2\alpha}{cos^2\alpha}=\frac{3600}{121}\\\frac{1-cos^2\alpha}{cos^2\alpha}=\frac{3600}{121}\\3600cos^2\alpha=121(1-cos^2\alpha)\\3600cos^2\alpha=121-121cos^2\alpha /+121cos^2\alpha\\3721cos^2\alpha=121 /:3721\\cos^2\alpha=\frac{121}{3721}\\cos\alpha=\frac{11}{61}\\[/tex]
[tex]sin^2\alpha=1-\frac{121}{3721}\\sin^2\alpha=\frac{3600}{3721}\\sin\alpha=\frac{60}{61}[/tex]
Szczegółowe wyjaśnienie:
[tex]sin^2\alpha+cos^2\alpha=1\\tg\alpha=\frac{sin\alpha}{cos\alpha}\\ctg\alpha=\frac{1}{tg\alpha}[/tex]
