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Trygonometria
: 06 lut 2015, 15:20
autor: sara1996
Oblicz \(\frac{2 \sin \alpha -3 \cos \alpha }{2 \sin \alpha +3 \cos \alpha }\) jeśli wiadomo, że tg alfa =2
: 06 lut 2015, 15:59
autor: eresh
\(\frac{2\sin\alpha-3\cos\alpha}{2\sin\alpha+3\cos\alpha}=\frac{\cos\alpha(2\tg\alpha-3))}{\cos\alpha(2\tg\alpha+3)}=\frac{2\cdot 2-3}{2\cdot 2+3}=\frac{1}{7}\)
: 06 lut 2015, 21:53
autor: josselyn
\(\frac{2sinx-3cosx}{2sinx+3cosx}= \frac{ \frac{2sinx}{cosx}-3 \frac{cosx}{cosx} }{2 \frac{sinx}{cosx}+3 \frac{cosx}{cosx} }= \frac{2tgx-3}{2tgx+3}= \frac{4-3}{4+3}= \frac{1}{7}\)