Question

1)5*sin98/(sin49*sin41)

2)8/(sin(-27/пи/4)*cos(31*пи/4))

Answer 1

$5frac{sin 98}{sin 49*sin 41} = 5frac{2sin 49 cos 49}{sin 49*sin 41} =\ = 10frac{cos 49}{sin 41} = 10frac{sin (90-49)}{sin 41} =10$

 

$frac{8}{sin(-27pi/4)*cos(31pi/4)} = frac{8}{sin(pi/4)*cos(-pi/4)} = \ =frac{8}{sqrt{2}/2*sqrt{2}/2} =8*2=16$