Question

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

Answer 1

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

____________________________

$cos(alpha-beta)=cosalpha cdot cosbeta+sinalpha cdot sinbeta\ sin(alpha+beta)=sinalpha cdot cosbeta+sinbeta cdot cosalpha \ \ sinfrac{pi}{4}=cosfrac{pi}{4}=frac{sqrt{2}}{2} \ 2sinfrac{pi}{4}=2cosfrac{pi}{4}=sqrt{2} \ \ sinfrac{pi}{6}=frac{1}{2} \ 2sinfrac{pi}{6}=1 \ \ cosfrac{pi}{6}=frac{sqrt{3}}{2} \ 2cosfrac{pi}{6}=sqrt{3}$

Answer 2

на фото.....................