Odpowiedź:
Szczegółowe wyjaśnienie:
1.
[tex]cos2\alpha =\frac{1}{4} \\2cos\alpha^2\alpha -1=\frac{1}{4} \\\\2cos^2\alpha -1\frac{1}{4}=0 /:2\\cos^2\alpha -\frac{5}{8} \\(cos\alpha -\sqrt{\frac{5}{8}})(cos\alpha +\sqrt{\frac{5}{8} } ) } =0\\cos\alpha_1 =\sqrt{ \frac{5}{8} }\\cos\alpha _2=-\sqrt{\frac{5}{8} }[/tex]
α∈(0°, 90°) więc cosα=[tex]\sqrt{\frac{5}{8} } =\frac{1}{4} \sqrt{10} \\[/tex]
[tex]\\sin^2\alpha +cos^2\alpha =1\\sin^2\alpha +(\frac{1}{4}\sqrt{10})^2} =1\\sin^2\alpha +\frac{10}{16} -1=0\\sin^2\alpha -\frac{6}{16} =0\\(sin\alpha -\frac{\sqrt{6} }{4} )(sin\alpha +\frac{\sqrt{6}}{4})=0\\sin\alpha _1=\frac{\sqrt{6}}{4}\\sin\alpha _2=-\frac{\sqrt{6}}{4}\\[/tex]
α∈(0°, 90°) więc sinα=[tex]\frac{\sqrt{6} }{4}[/tex]
[tex]tg\alpha =\frac{sin\alpha }{cos\alpha } \\tg\alpha =\frac{\frac{\sqrt{6} }{4} }{\frac{\sqrt{10} }{4} } =\frac{\sqrt{6} }{4} *\frac{4}{\sqrt{10} } =\sqrt{\frac{3}{5} } =\frac{\sqrt{15} }{5}\\[/tex]
[tex]\\ctg\alpha =\frac{1}{tg\alpha } \\ctg\alpha =\frac{1}{\frac{\sqrt{15} }{5} } =\frac{\sqrt{15} }{3} \\[/tex]
Zad.2
sin15°=sin(60°-15°)=sin60°cos45°-cos60°sin45°=[tex]\\\frac{\sqrt{3} }{2} *\frac{\sqrt{2} }{2} -\frac{1}{2} *\frac{\sqrt{2} }{2}=\frac{\sqrt{6} }{4} -\frac{\sqrt{2} }{4} =\frac{1}{4} (\sqrt{6} -\sqrt{2} )[/tex]
√6≈2,45
√2≈1,41
sin15°≈0,26
cos105°=cos(60°+45°)=cos60°cos45°-sin60°sin45°=[tex]\frac{1}{2} *\frac{\sqrt{2} }{2}-\frac{\sqrt{3} }{2}*\frac{\sqrt{2} }{2}=\frac{\sqrt{2} }{4}-\frac{\sqrt{6} }{4}=\frac{1}{4} (\sqrt{2} -\sqrt{6} )\\[/tex]
cos105≈-0,26