SOLUTION: sin[arctan(-31/2)] =
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Question 821451: sin[arctan(-31/2)] =
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
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=√(31^2+2^2)=√(961+4)=√965
sin[arctan(-31/2)]=sin x=-31/√985=(-31√965)/965
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