Question

√2cosх-2 cos(pi/4-x)/ 2sin(pi/6+x)-√3 sinx

Answer 1

$\frac{\sqrt{2}cosx-2cos(\frac{\pi}{4}-x)}{2sin(\frac{\pi}{6}+x)-\sqrt{3}sinx}=\frac{\sqrt{2}cosx-2cos\frac{\pi}{4}cosx+2sin\frac{\pi}{4}sinx}{2sin\frac{\pi}{6}cosx+2sinxcos\frac{\pi}{6}-\sqrt{3}sinx}=\frac{\sqrt{2}cosx-\sqrt{2}cosx+\sqrt{2}sinx}{cosx+\sqrt{3}sinx-\sqrt{3}sinx}=\\\\=\frac{\sqrt{2}sinx}{cosx}=\sqrt{2}tgx$

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$cos(\alpha-\beta)=cos\alpha \cdot cos\beta+sin\alpha \cdot sin\beta\\ sin(\alpha+\beta)=sin\alpha \cdot cos\beta+sin\beta \cdot cos\alpha \\ \\ sin\frac{\pi}{4}=cos\frac{\pi}{4}=\frac{\sqrt{2}}{2} \\ 2sin\frac{\pi}{4}=2cos\frac{\pi}{4}=\sqrt{2} \\ \\ sin\frac{\pi}{6}=\frac{1}{2} \\ 2sin\frac{\pi}{6}=1 \\ \\ cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} \\ 2cos\frac{\pi}{6}=\sqrt{3}$

Answer 2

на фото.....................