zad.6
[tex]\cos\alpha\cdot tg\alpha=\cos\alpha\cdot \frac{\sin\alpha}{\cos\alpha} =\sin\alpha[/tex]
Odp.C
zad.7
[tex]\alpha - rozwarty[/tex] , [tex]tg\alpha=-\frac{1}{2}[/tex]
[tex]\frac{\sin\alpha}{\cos\alpha} =-\frac{1}{2} \ |\cdot \cos\alpha\\\sin\alpha=-\frac{1}{2}\cos\alpha \ = > \ \sin^2\alpha=(-\frac{1}{2}\cos\alpha)^2=\frac{1}{4} \cos^2\alpha[/tex] i wstawiam do jedynki trygonometrycznej
[tex]\sin^2\alpha+\cos^2\alpha=1\ = > \frac{1}{4} \cos^2\alpha+\cos^2\alpha=1\\\frac{5}{4} \cos^2\alpha=1 \ | :\frac{5}{4} \\\\\cos^2\alpha=\frac{4}{5} \ = > \ \cos\alpha=\frac{2}{\sqrt{5} } , \ \cos\alpha=-\frac{2}{\sqrt{5}}[/tex], ale [tex]\alpha-rozwarty[/tex], więc :
[tex]\cos\alpha=-\frac{2}{\sqrt{5}} =-\frac{2\sqrt{5} }{5}\\\sin\alpha=-\frac{1}{2} \cos\alpha=\frac{\sqrt{5} }{5}[/tex]
