Strona 1 z 1
Trygonometria
: 08 gru 2016, 14:13
autor: BladyWZ
zad 1. Oblicz
a). \(\frac{1}{\sin ^2 \alpha } + \frac{1}{\cos ^2 \alpha } - \frac{1}{ \sin ^2 \alpha \cos ^2 \alpha }\)
b)/ \(\tg \alpha + \frac{1}{tg \alpha } - \frac{1}{\sin \alpha \cos \alpha}\)
zad 2. WIedząc że \(\sin 22.5^ \circ = \frac{ \sqrt{2- \sqrt{2} } }{2}\) oblicz \(\cos22.5^ \circ i \tg 22.5^ \circ\)
: 08 gru 2016, 14:54
autor: Galen
a)
\(\frac{1}{sin^2\alpha}+ \frac{1}{cos^2\alpha}- \frac{1}{sin^2\alpha cos^2\alpha}= \frac{cos^2\alpha+sin^2\alpha}{sin^2\alpha cos^2\alpha}- \frac{1}{sin^2\alpha cos^2\alpha}= \frac{1-1}{sin^2\alpha cos^2\alpha}=0\)
b)
\(tg\alpha+ctg\alpha- \frac{1}{sin\alpha cos\alpha}= \frac{sin\alpha}{cos\alpha}+ \frac{cos\alpha}{sin\alpha}- \frac{1}{sin\alpha cos\alpha}= \frac{sin^2\alpha+cos^2\alpha}{sin\alpha cos\alpha}- \frac{1}{sin\alpha cos\alpha}= \frac{1-1}{sin\alpha cos\alpha}=0\)
: 08 gru 2016, 15:20
autor: Galen
Zad.2
\(sin22,5^o= \frac{ \sqrt{2- \sqrt{2} } }{2}\)
\(cos22,5^o= \sqrt{1-sin^222,5^o}= \sqrt{1- \frac{2- \sqrt{2} }{4} }= \sqrt{ \frac{4-2+ \sqrt{2} }{4} }= \sqrt{ \frac{2+ \sqrt{2} }{4} }= \frac{ \sqrt{2+ \sqrt{2} } }{2}\)
\(tg\alpha= \frac{sin\alpha}{cos\alpha}\)
\(tg22,5^o= \frac{ \frac{ \sqrt{2- \sqrt{2} } }{2} }{ \frac{ \sqrt{2+ \sqrt{2} } }{2} } = \frac{ \sqrt{2- \sqrt{2} } }{\sqrt{2+ \sqrt{2}} }= \sqrt{ \frac{2- \sqrt{2} }{2+ \sqrt{2} } }= \sqrt{ \frac{(2- \sqrt{2})^2 }{(2+ \sqrt{2})(2- \sqrt{2}) } }= \sqrt{ \frac{(2- \sqrt{2})^2 }{4-2} }= \frac{2- \sqrt{2} }{ \sqrt{2} }= \frac{2 \sqrt{2}-2 }{2}= \sqrt{2}-1\)