Question 821451
sin[arctan(-31/2)]
let x=angle whose tan = -31/2
domain:[-π/2,π/2]
tan x=-31/2
reference angle in quadrant IV where sin<0, cos>0, tan<0
hypotenuse of reference right triangle=&#8730;(31^2+2^2)=&#8730;(961+4)=&#8730;965
sin[arctan(-31/2)]=sin x=-31/&#8730;985=(-31&#8730;965)/965