Question

прошу помогите решить 161 и 162​

Answer 1

$161)Sin\alpha =2Sin\frac{\alpha}{2}Cos\frac{\alpha }{2}\\\\Cos\alpha=Cos^{2}\frac{\alpha }{2}-Sin^{2}\frac{\alpha }{2}\\\\Cos\alpha =1 -2Sin^{2}\frac{\alpha }{2}\\\\Cos\alpha =2Cos^{2}\frac{\alpha }{2}-1\\\\tg\alpha=\frac{2tg\frac{\alpha }{2}}{1-tg^{2}\frac{\alpha }{2}}$

$Sin4\alpha=2Sin2\alpha Cos2\alpha\\\\Cos4\alpha=Cos^{2}2\alpha -Sin^{2}2\alpha\\\\Cos4\alpha=1-2Sin^{2}2\alpha \\\\Cos4\alpha=2Cos^{2}2\alpha-1\\\\tg4\alpha=\frac{2tg2\alpha }{1-tg^{2}2\alpha}$

$162)\frac{Sin\alpha }{2Cos^{2}\frac{\alpha }{2}}=\frac{2Sin\frac{\alpha }{2} Cos\frac{\alpha }{2}}{2Cos^{2}\frac{\alpha }{2}}=\frac{Sin\frac{\alpha }{2}}{Cos\frac{\alpha }{2}}=tg\frac{\alpha }{2}\\\\\frac{Sin4\beta }{Cos2\beta}=\frac{2Sin2\beta Cos2\beta}{Cos2\beta}=2Sin2\beta$

$\frac{Cos\beta}{Cos\frac{\beta}{2}+Sin\frac{\beta }{2}}=\frac{Cos^{2}\frac{\beta}{2}-Sin^{2}\frac{\beta}{2}}{Cos\frac{\beta}{2}+Sin\frac{\beta}{2}}=\frac{(Cos\frac{\beta}{2}-Sin\frac{\beta}{2})(Cos\frac{\beta}{2}+Sin\frac{\beta}{2})}{Cos\frac{\beta}{2}+Sin\frac{\beta}{2}} =Cos\frac{\beta}{2}-Sin\frac{\beta}{2}$

$\frac{Cos2\alpha-Sin2\alpha}{Cos4\alpha}=\frac{Cos2\alpha-Sin2\alpha}{Cos^{2}2\alpha-Sin^{2}2\alpha}=\frac{Cos2\alpha-Sin2\alpha}{(Cos2\alpha-Sin2\alpha)(Cos2\alpha+Sin2\alpha)}=\frac{1}{Cos2\alpha+Sin2\alpha}$