• Предмет: Алгебра
  • Автор: anbasova44
  • Вопрос задан 8 лет назад

прошу, хотя бы первые два, нужно доказать тождество

Приложения:

Ответы

Ответ дал: 7x8
0

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}

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