Question

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Answer 1

Ответ:

$\displaystyle \frac{sina+sin2a+sin3a}{cosa+cos2a+cos3a}=\frac{(sina+sin3a)+sin2a}{(cosa+cos3a)+cos2a}=\frac{2sin2a\cdot cosa+sin2a}{2\, cos2a\cdot cosa+cos2a}=\\\\\\=\frac{sin2a\cdot (2cosa+1)}{cos2a\cdot (2\, cosa+1)}=\frac{sin2a}{cos2a}=tg2a\\\\\\\\\boxed{\ sinx+siny=2\, sin\frac{x+y}{2}\cdot cos\frac{x-y}{2}\ \ ,\ \ cosx+cosy=2\, cos\frac{x+y}{2}\cdot cos\frac{x-y}{2}\ }$