Question

510 номер СРОЧНОООО ПОМОГИТЕЕЕЕЕ

Answer 1

$(Sin\alpha+Cos\alpha)^{2}=(\frac{1}{5})^{2}\\\\Sin^{2}\alpha+2Sin\alpha Cos\alpha+Cos^{2} \alpha=\frac{1}{25}\\\\1+2Sin\alpha Cos\alpha=\frac{1}{25}\\\\2Sin\alpha Cos\alpha=-\frac{24}{25}\\\\Sin\alpha Cos\alpha=-\frac{12}{25}$

$1)Sin\alpha=-\frac{3}{5} ;Cos\alpha=\frac{4}{5}\\\\tg\frac{\alpha }{2}=\frac{Sin\alpha }{1+Cos\alpha }=\frac{-\frac{3}{5} }{1+\frac{4}{5} }=-\frac{3*5}{5*9} =-\frac{1}{3}\\\\2)Sin\alpha=\frac{4}{5};Cos\alpha =-\frac{3}{5}\\\\tg\frac{\alpha }{2}=\frac{Sin\alpha }{1+Cos\alpha }=\frac{\frac{4}{5} }{1-\frac{3}{5} }=\frac{4*5}{5*2}=2$

Answer 2

$sina+cosa=\frac{1}{5}\\\\sina+cosa=\frac{2\, tg\frac{a}{2}}{1+tg^2\frac{a}{2}}+\frac{1-tg^2\frac{a}{2}}{1+\frac{a}{2}}=\frac{-tg^2\frac{a}{2}+2tg\frac{a}{2}+1}{1+tg^2\frac{a}{2}}=\frac{1}{5}\; \; \to \\\\t=tg\frac{a}{2}\; ,\; \; \frac{-t^2+2t+1}{1+t^2}=\frac{1}{5}\; \; ,\; \; \frac{5(-t^2+2t+1)-(1+t^2)}{5(1+t^2)}=0\; \; ,\\\\\frac{-6t^2+10t+4}{1+t^2}=0\; \; ,\; \; \frac{-2\, (3t^2-5t-2)}{1+t^2}=0\; ,\\\\3t^2-5t-2=0\; ,\; \; D=49\; ,\; \; t_1=-\frac{1}{3}\; ,\; \; t_2=2\\\\Otvet:\; \; tg\frac{a}{2}=-\frac{1}{3}\; \; \; ili\; \; \; tg\frac{a}{2}=2\; .$