Witaj :)
Dane:
[tex]\cos(90^\circ-\alpha)=\frac{\sqrt{6}}{4}\\\alpha\in (0;90^\circ)[/tex]
Ze wzorów redukcyjnych zauważamy, że:
[tex]\cos(90^\circ-\alpha)=\sin \alpha[/tex]
Więc:
[tex]\sin\alpha=\frac{\sqrt{6}}{4}[/tex]
- Korzystając z jedynki trygonometrycznej obliczam wartość cosα:
[tex]\sin^2\alpha+\cos^2\alpha=1\\\\(\frac{\sqrt{6}}{4})^2+\cos^2\alpha =1\\\\\frac{6}{16}+\cos^2\alpha=1\\\\\frac{3}{8}+\cos^2\alpha=1\\\\\cos^2\alpha=1-\frac{3}{8}\\\\\cos^2\alpha=\frac{5}{8}\implies \cos\alpha=\sqrt{\frac{5}{8} }=\frac{\sqrt{10}}{4}\ \ \vee \ \ \ \cos\alpha=-\sqrt{\frac{5}{8}}=-\frac{\sqrt{10}}{4}\\\\Poniewaz\ \alpha\in (0;90^\circ)\implies \cos\alpha >0 \\\\\boxed{\cos\alpha=\frac{\sqrt{10}}{4} }[/tex]
[tex]tg\alpha= \frac{\sin\alpha}{\cos\alpha}\\\\tg\alpha=\frac{\frac{\sqrt{6}}{4} }{\frac{\sqrt{10}}{4} } \\\\tg\alpha=\frac{\sqrt{6}}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}}=\frac{\sqrt{60}}{10}=\frac{\sqrt{4\cdot15}}{10} =\frac{2\sqrt{15}}{10}=\frac{\sqrt{15}}{10}\\\\\boxed{tg\alpha=\frac{\sqrt{15}}{10} }[/tex]
[tex]ctg\alpha=\frac{1}{tg\alpha}\\\\ctg\alpha=\frac{1}{\frac{\sqrt{15}}{5} } \\\\ctg\alpha=\frac{5}{\sqrt{15}} \cdot\frac{\sqrt{15}}{\sqrt{15}}=\frac{5\sqrt{15}}{15}=\frac{\sqrt{15}}{3}\\\\\boxed{ctg\alpha=\frac{\sqrt{15}}{3}}[/tex]
ODP.:
[tex]\sin\alpha=\frac{\sqrt{6} }{4}\\\\\cos\alpha=\frac{\sqrt{10}}{4}\\\\tg\alpha=\frac{\sqrt{15}}{10}\\\\ctg\alpha=\frac{\sqrt{15}}{3}[/tex]
Wyjaśnienie do obliczenia cosα:
[tex]\sqrt{\frac{5}{8}}=\frac{\sqrt{5}}{\sqrt{8}}=\frac{\sqrt{5}}{\sqrt{4\cdot2}} =\frac{\sqrt{5}}{\sqrt{4}\cdot \sqrt{2}} =\frac{\sqrt{5}}{2\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} =\frac{\sqrt{5\cdot2}}{2\sqrt{2}\cdot\sqrt{2}} =\frac{\sqrt{10}}{2\cdot(\sqrt{2})^2} =\frac{\sqrt{10}}{2\cdot2} =\frac{\sqrt{10}}{4}[/tex]
