Question

довести тотожність
а) tg8a-ctg8a=-2ctg16a
б) (cos(π/2-5a)-sin(π+3a))(sin(π/2+3a)-cos(π+5a))/1+cos(2π-2a)=sin8a

Answer 1

$1)tg8\alpha-Ctg8\alpha=\frac{Sin8\alpha}{Cos8\alpha}-\frac{Cos8\alpha}{Sin8\alpha}=\frac{Sin^{2}8\alpha-Cos^{2}8\alpha}{Sin8\alpha Cos8\alpha}=\frac{-Cos16\alpha}{Sin8\alpha Cos8\alpha}=\frac{-2Cos16\alpha}{2Sin8\alpha Cos8\alpha}=\frac{-2Cos16\alpha}{Sin16\alpha}=-2Ctg16\alpha\\\\-2Ctg16\alpha=-2Ctg16\alpha$

$2)\frac{(Cos(\frac{\pi}{2}-5\alpha)-Sin(\pi+3\alpha))(Sin(\frac{\pi }{2}+3\alpha)-Cos(\pi+5\alpha))}{1+Cos(2\pi-2\alpha)} =\frac{(Sin5\alpha+Sin3\alpha)(Cos3\alpha+Cos5\alpha)}{1+Cos2\alpha}=\frac{2Sin4\alpha Cos\alpha*2Cos4\alpha Cos\alpha}{1+Cos2\alpha}=\frac{2Sin4\alpha*Cos4\alpha*2Cos^{2}\alpha }{2Cos^{2}\alpha}=Sin8\alpha\\\\Sin8\alpha=Sin8\alpha$