Question

40 баловвв!!!! 3 примера
Упростить:
sin⁡2α+sin8α/cos2α-cos8α

Довести тождество:
(ctg(π/2-α)*(sin⁡(3π/2 -α)+sin⁡(π+α)) / (ctg(π+α)*(cos⁡(2π+α)-sin⁡(2π-α) ) = -tg^2 α

16cos π/9 cos 2π/9 cos 4π/9=2

Answer 1

$1); ; frac{sin2a+sin8a}{cos2a-cos8a}=frac{2cdot sin5acdot cos3a}{-2cdot sin5acdot sin(-3a)}=frac{cos3a}{sin3a}=ctg3a\\2); ; frac{ctg(frac{pi}{2}-a)cdot (sin(frac{3pi}{2}-a)+sin(pi +a))}{ctg(pi +a)cdot (cos(2pi +a)-sin(2pi -a))}=frac{tgacdot (-cosa-sina)}{ctgacdot (cosa+sina)}=\\= frac{-tgacdot (cosa+sina)}{frac{1}{tga}cdot (cosa+sina)} =-tg^2a$

$3); ; 16cdot cos frac{pi }{9}cdot cosfrac{2pi }{9}cdot cosfrac{4pi }{9} =\\= frac{8}{sinfrac{pi}{9}}cdot (underbrace {2, sin frac{pi }{9}cdot cosfrac{pi }{9}}_{sinfrac{2pi }{9}})cdot cosfrac{2pi }{9}cdot cosfrac{4pi }{9}=\\=frac{4}{sinfrac{pi}{9}}cdot (underbrace {2, sinfrac{2pi }{9}cdot cosfrac{2pi }{9}}_{sinfrac{4pi}{9}})cdot cosfrac{4pi }{9}=\\=frac{2}{sinfrac{pi}{9}}cdot (underbrace{2sin frac{4pi }{9}cdot cosfrac{4pi }{9}}_{sinfrac{8pi }{9}})=frac{2sinfrac{8pi}{9}}{sinfrac{pi}{9}} = frac{2sin(pi-frac{pi}{9})}{sinfrac{pi}{9}}=\\=frac{2sinfrac{pi}{9}}{sinfrac{pi}{9}} =2$

Answer 2

в самом первом ctg(3a)