What is sin(inverse tangent(12/5)?

##sin(arctan(12/5)) = 12/13##
From the trig identity ##sin^2(theta) + cos^2(theta) = 1##, we divide both sides by ##sin^2(theta)##
##1 + cos^2(theta)/sin^2(theta) = 1/sin^2(theta)##
Since ##sin(theta)/cos(theta) = tan(theta)## we can rewrite the second term
##1 + 1/tan^2(theta) = 1/sin^2(theta)##
Taking the least common multiple,
##(tan^2(theta) + 1)/(tan^2(theta)) = 1/sin^2(theta)##
Inverting both sides
##tan^2(theta)/(tan^2(theta) + 1) = sin^2(theta)##
Subsituting ##theta = arctan(12/5)##
##(12/5)^2/((12/5)^2+1) = sin^2(arctan(12/5))##
##sin^2(arctan(12/5)) = 144/25 * 25/169 = 144/169##
Taking the root
##sin(arctan(12/5)) = +-12/13##
To pick the sign we look at the range of the arctangent, it only takes arguments on the first and fourth quadrants, during which the cosine is always positive. If, for ##12/5## the cosine is positive and the tangent is positive, then the sine must be positive too.
##sin(arctan(12/5)) = 12/13##
Also, protip, you can use either the function with a “^-1” or put an arc before it to notate the inverse trig functions, but usually there’s a lot less headache for everybody involved if you use the arc notation. (There’s no grounds for mistaking it for other functions).