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$$\frac{1+sin \alpha –cos2 \alpha - sin3 \alpha }{2sin^2 \alpha +sin \alpha -1}$$ задан 9 Янв '12 13:50 
Alekcandra |
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$$\frac{1+sin \alpha–cos2 \alpha - sin3 \alpha }{2sin^2 \alpha +sin \alpha -1}=\frac{1+sin \alpha –1+2sin^2 \alpha - 3sin \alpha +4sin^3\alpha}{2sin^2 \alpha +sin \alpha -1}=$$$$=\frac{2sin\alpha(2sin^2 \alpha +sin \alpha -1)}{2sin^2 \alpha +sin \alpha -1}=2sin\alpha.$$ отвечен 4 Янв '13 21:58 
Anatoliy |
