Решить неравенство: sin4x + cos4x*ctg2x> 1

(sin(3п/2+a)+sin(2п+a)/(2cos(-a)sin(-a)+1)=

=-cos a + sin a / (2 cos a * (-sin a) +1) =

=-cos a + sin a / (-2 cos a * sin a + (cos a)^2 +2 cos a * sin a + (sin a)^2 )=

= -cos a + sin a /    ((cos a)^2  + (sin a)^2) =

= -cos a + sin a / 1 = sin a - cos a

5cos360*-7tg45*+8ctg90*-6sin0*=5*cos0*-7*tg45*+8ctg90*-6sin0*=5*1-7*1+8*0-6*0=5-7=-2

1) -2

2) 1/2 

tg (α+β) = sin (α + β) / cos (α + β) =

(sin α * cos β + cos α * sin β) / (cos α * cos β - sin α * sin β) =

(sin α * cos β + cos α * sin β)/(cos α * cos β) / (cos α * cos β - sin α * sin β)/(cos α * cos β) =  (tg α + tg β) / (1 - tg α * tg β)