Question

Задание во вложениях. Нужно только №2 подробное решение. Заранее благодарю)

Answer 1

$2sqrt{2}sinfrac{alpha}{2}sin(frac{alpha}{2}+frac{pi}{4})=2sqrt{2}sinfrac{alpha}{2}(sinfrac{alpha}{2}cosfrac{pi}{4}+cosfrac{alpha}{2}sinfrac{pi}{4})=\ \ =2sqrt{2}sinfrac{alpha}{2}(sinfrac{alpha}{2}frac{sqrt{2}}{2}+cosfrac{alpha}{2}frac{sqrt{2}}{2})=\ \ =2sqrt{2}*frac{sqrt{2}}{2}*sinfrac{alpha}{2}(sinfrac{alpha}{2}+cosfrac{alpha}{2})=2sin^2frac{alpha}{2}+2sinfrac{alpha}{2}cosfrac{alpha}{2}=\ \ =1-1+2sin^2frac{alpha}{2}+sinalpha=$

$1-1+2sin^2frac{alpha}{2}+sinalpha=1-(1-2sin^2frac{alpha}{2})+sinalpha=\ \ =1-cosalpha+sinalpha$

что и требовалось доказать

Answer 2

$1-cosalpha+sinalpha=2sqrt2sinfrac{alpha}{2}sin(frac{alpha}{2}+frac{pi}{4})\2sin^2(frac{alpha}{2})+2sin(frac{alpha}{2})*cos(frac{alpha}{2})=2sqrt2sinfrac{alpha}{2}sin(frac{alpha}{2}+frac{pi}{4})\2sinfrac{alpha}{2}((sin(frac{alpha}{2})+cos(frac{alpha}{2}))=2sqrt2sinfrac{alpha}{2}sin(frac{alpha}{2}+frac{pi}{4})\sqrt2sin(frac{alpha}{2}+frac{pi}{4})*2sinfrac{alpha}{2}=2sqrt2sinfrac{alpha}{2}sin(frac{alpha}{2}+frac{pi}{4})$
Тождество доказано.

Для справки:
$1-cosalpha=2sin^2frac{alpha}{2}\sinalpha=2sinfrac{alpha}{2}*cosfrac{alpha}{2}\(sin(frac{alpha}{2})+cos(frac{alpha}{2}))=sqrt2sin(frac{alpha}{2}+frac{pi}{4})$