Прошу, хотя бы первые два, нужно доказать тождество

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Проверено экспертом

Автор - 7x8

12.

$frac{sinalpha+sinfrac{alpha}{2}}{1+cosalpha+cosfrac{alpha}{2}}=frac{sin(2cdotfrac{alpha}{2})+sinfrac{alpha}{2}}{1+cos(2cdotfrac{alpha}{2)}+cosfrac{alpha}{2}}=$

$frac{2sinfrac{alpha}{2} cosfrac{alpha}{2}+sinfrac{alpha}{2}}{1+cos^2frac{alpha}{2}-sin^2frac{alpha}{2}+cosfrac{alpha}{2}}=frac{sinfrac{alpha}{2}(2 cosfrac{alpha}{2}+1)}{1-sin^2frac{alpha}{2}+cos^2frac{alpha}{2}+cosfrac{alpha}{2}}=$

$frac{sinfrac{alpha}{2}(2 cosfrac{alpha}{2}+1)}{cos^2frac{alpha}{2}+cos^2frac{alpha}{2}+cosfrac{alpha}{2}}=frac{sinfrac{alpha}{2}(2 cosfrac{alpha}{2}+1)}{2cos^2frac{alpha}{2}+cosfrac{alpha}{2}}=$

$frac{sinfrac{alpha}{2}(2 cosfrac{alpha}{2}+1)}{cosfrac{alpha}{2}(2cosfrac{alpha}{2}+1)}=frac{sinfrac{alpha}{2}}{cosfrac{alpha}{2}}=tgfrac{alpha}{2}$

13.

$frac{1+cosfrac{alpha}{2}-sinfrac{alpha}{2}}{1-cosfrac{alpha}{2}-sinfrac{alpha}{2}}=frac{1+cos(2cdotfrac{alpha}{4})-sin(2cdotfrac{alpha}{4})}{1-cos(2cdotfrac{alpha}{4})-sin(2cdotfrac{alpha}{4})}=$

$frac{1+cos^2frac{alpha}{4}-sin^2frac{alpha}{4}-2sinfrac{alpha}{4}cosfrac{pi}{4}}{1-cos^2frac{alpha}{4}+sin^2frac{pi}{4}-2sinfrac{alpha}{4} cosfrac{pi}{4}}=frac{1-sin^2frac{alpha}{4}+cos^2frac{alpha}{4}-2sinfrac{alpha}{4}cosfrac{pi}{4}}{sin^2frac{alpha}{4}+sin^2frac{pi}{4}-2sinfrac{alpha}{4} cosfrac{pi}{4}}=$

$frac{cos^2frac{alpha}{4}+cos^2frac{alpha}{4}-2sinfrac{alpha}{4}cosfrac{pi}{4}}{2sin^2frac{pi}{4}-2sinfrac{alpha}{4} cosfrac{pi}{4}}=frac{2cos^2frac{alpha}{4}-2sinfrac{alpha}{4}cosfrac{pi}{4}}{2sin^2frac{pi}{4}-2sinfrac{alpha}{4} cosfrac{pi}{4}}=$

$frac{-2cosfrac{alpha}{4}(-cosfrac{alpha}{4}+sinfrac{alpha}{4})}{2sinfrac{pi}{4}(sinfrac{pi}{4}-cosfrac{pi}{4})}=frac{-cosfrac{alpha}{4}}{sinfrac{pi}{4}}=-ctgfrac{pi}{4}$

14.

$frac{4sin^4frac{alpha}{4}}{1-cos^2frac{alpha}{2}}=frac{4sin^4frac{alpha}{4}}{sin^2frac{alpha}{2}}=$

$frac{4sin^4frac{alpha}{4}}{sin^2(2cdotfrac{alpha}{4})}=frac{4sin^4frac{alpha}{4}}{(sin(2cdotfrac{alpha}{4}))^2}=$

$frac{4sin^4frac{alpha}{4}}{(2sinfrac{alpha}{4}cosfrac{alpha}{4})^2}=frac{4sin^4frac{alpha}{4}}{4sin^2frac{alpha}{4}cos^2frac{alpha}{4}}=$

$frac{sin^2frac{alpha}{4}}{cos^2frac{alpha}{4}}=tg^2frac{alpha}{4}$

15.

$frac{2sinalpha-sin2alpha}{2sinalpha+sin2alpha}=frac{2sinalpha-2sinalpha cosalpha}{2sinalpha+2sinalpha cosalpha}=$

$frac{2sinalpha(1-cosalpha)}{2sinalpha(1+cosalpha)}=frac{1-cosalpha}{1+cosalpha}=$

$frac{1-cos(2cdotfrac{alpha}{2})}{1+cos(2cdotfrac{alpha}{2})}=frac{1-cos^2frac{alpha}{2}+sin^2frac{alpha}{2}}{1+cos^2frac{alpha}{2}-sin^2frac{alpha}{2}}=$

$frac{sin^2frac{alpha}{2}+sin^2frac{alpha}{2}}{1-sin^2frac{alpha}{2}+cos^2frac{alpha}{2}}=frac{2sin^2frac{alpha}{2}}{cos^2frac{alpha}{2}+cos^2frac{alpha}{2}}=$

$frac{2sin^2frac{alpha}{2}}{2cos^2frac{alpha}{2}}=tg^2frac{alpha}{2}$

прошу, хотя бы первые два, нужно доказать тождество

Ответ

Проверено экспертом

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