Question

Помогите решить .............................

Answer 1

$6)Sin(\frac{35\pi }{2}-\alpha)+Cos(68\pi -\alpha)= Sin(16\pi+(\frac{3\pi }{2}-\alpha))+Cos\alpha=Sin(\frac{3\pi }{2}-\alpha)+Cos\alpha=-Cos\alpha+Cos\alpha=0\\\\7)tg(9\pi-\alpha)+Ctg(\frac{57\pi }{2}+\alpha)=tg(-\alpha)-Ctg(28\pi+(\frac{\pi }{2}+\alpha))=-tg\alpha+Ctg(\frac{\pi }{2}+\alpha)=-tg\alpha-tg\alpha=-2tg\alpha$

$8)\frac{tg(\frac{11\pi }{2}+\alpha)*Cos(\frac{7\pi }{2}-\alpha)*Cos(\alpha-4\pi)}{Ctg(5\pi-\alpha)*Sin(\frac{11\pi}{2}+\alpha)}= \frac{Cos(3\pi+(\frac{\pi }{2}-\alpha)*Cos\alpha }{Ctg(-\alpha )*Cos(6\pi-(\frac{\pi }{2}-\alpha))}=\frac{Cos(\frac{\pi }{2}-\alpha)*Cos\alpha}{-Ctg\alpha*Cos(\frac{\pi }{2}-\alpha)} =-\frac{Cos\alpha*Sin\alpha}{Cos\alpha}=-Sin\alpha$

$9)Sin(7\pi -\alpha )*Cos(\frac{15\pi }{2}+\beta)-Sin(\frac{19\pi }{2}-\alpha)*Cos(6\pi-\beta)=Sin(6\pi+(\pi-\alpha))*Cos(8\pi-(\frac{\pi }{2}-\beta))-Sin(10\pi-(\frac{\pi }{2}+\alpha))*Cos\beta=Sin(\pi-\alpha)*Cos(\frac{\pi }{2}-\beta)+Sin(\frac{\pi }{2}+\alpha)*Cos\beta=Sin\alpha *Sin\beta+Cos\alpha Cos\beta=Cos(\alpha-\beta)$