Ответы на вопрос:
1(sinπ/4cosa+cosπ/4sina-cosπ/4cosa+sinπ/4sina)/(sinπ/4cosa+cosπ/4sina+cosπ/4cosa-sinπ/4sina)=(√2/2cosa+√2/2sina-√2/2cosa+√2/2sina)/(2/2cosa+√2/2sina+√2/2cosa-√2/2sina)=(√2sina)/(√2cosa)=tga 2 (sinacosb-sinbcosa): (sina/cosb-sinb/cosb)= =(sinacosb-sinbcosa)*cosacosb/(sinacosb-sinbcosa)=cosacosb
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