Sprawdz czy istnieje taki kat ostry alfa ze cos alfa = 1/pierwiastek 6 - pierwiastek 5.



Odpowiedź :

[tex]cos\alpha=\frac{1}{\sqrt{6}-\sqrt{5}}\\\\sin^2\alpha+cos^2\alpha=1\\sin^2\alpha+(\frac{1}{\sqrt6-\sqrt5})^2=1\\sin^2\alpha+\frac{1}{6-2\sqrt{30}+5}=1\\sin^2\alpha+\frac{1}{11-2\sqrt{30}}=1\\sin^2\alpha=1-\frac{1}{11-2\sqrt{30}}\\sin^2\alpha=\frac{11-2\sqrt{30}-1}{11-2\sqrt{30}}\\sin^2\alpha=\frac{10-2\sqrt{30}}{11-2\sqrt{30}}\\sin^2\alpha=\frac{(10-2\sqrt{30})(11+2\sqrt{30})}{121-120}\\[/tex]

[tex]sin^2\alpha=110+20\sqrt{30}-22\sqrt{30}-120\\sin^2\alpha=-10-2\sqrt{30} < 0\\\text{Nie istnieje taki kat ostry } \alpha[/tex]