Question

Помогите! 50 баллов!
Упростить выражения (a-альфа)
1.$frac{(sinfrac{a}{2}+cosfrac{a}{2} )^{2} }{1+sina}$
2$frac{2sina-sin2a}{2sina+sin2a}$=tg$^{2} frac{a}{2}$

Answer 1

1) (sinα/2 + cosα/2)²/(1 + sinα) = (sin²α/2 + 2sinα/2·cosα/2 + cos²α/2)/(1 + sinα) = (1 + sinα)/(1 + sinα) = 1.

2) (2sinα - sin2α)/(2sinα + sin2α) = (2sinα - 2sinαcosα)/(2sinα + 2sinαcosα) = 2sinα(1 - cosα)/2sinα(1 + cosα) = (1 - cosα)/(1 + cosα) = (2sin²α/2)/(2cos²α/2) = tg²α/2

Answer 2

$1)frac{(Sinfrac{alpha}{2}+Cosfrac{alpha}{2})^{2}}{1+Sinalpha }=frac{Sin^{2}frac{alpha}{2}+2Sinfrac{alpha}{2}Cosfrac{alpha}{2}+Cos^{2}frac{alpha}{2}}{1+Sinalpha}=frac{1+Sinalpha}{1+Sinalpha} =1$

$2)frac{2Sinalpha-Sin2alpha}{2Sinalpha+Sin2alpha} =frac{2Sinalpha-2Sinalpha Cosalpha}{2Sinalpha+2Sinalpha Cosalpha}=frac{2Sinalpha(1-Cosalpha)}{2Sinalpha(1+Cosalpha)}=frac{1-Cosalpha}{1+Cosalpha} =frac{2Sin^{2}frac{alpha}{2}}{2Cos^{2}frac{alpha}{2}}=tg^{2}frac{alpha}{2}\\tg^{2}frac{alpha}{2} =tg^{2}frac{alpha}{2}$

Тождество доказано