Question

50 ! выражения (a-альфа) 1. 2=tg

Answer 1

$1)\frac{(Sin\frac{\alpha}{2}+Cos\frac{\alpha}{2})^{2}}{1+Sin\alpha }=\frac{Sin^{2}\frac{\alpha}{2}+2Sin\frac{\alpha}{2}Cos\frac{\alpha}{2}+Cos^{2}\frac{\alpha}{2}}{1+Sin\alpha}=\frac{1+Sin\alpha}{1+Sin\alpha} =1$

$2)\frac{2Sin\alpha-Sin2\alpha}{2Sin\alpha+Sin2\alpha} =\frac{2Sin\alpha-2Sin\alpha Cos\alpha}{2Sin\alpha+2Sin\alpha Cos\alpha}=\frac{2Sin\alpha(1-Cos\alpha)}{2Sin\alpha(1+Cos\alpha)}=\frac{1-Cos\alpha}{1+Cos\alpha} =\frac{2Sin^{2}\frac{\alpha}{2}}{2Cos^{2}\frac{\alpha}{2}}=tg^{2}\frac{\alpha}{2}\\\\tg^{2}\frac{\alpha}{2} =tg^{2}\frac{\alpha}{2}$

Тождество доказано

Answer 2

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