posted by Nyke .
Find an equation of the tangent line to the curve at the given point.
y = sin(sin x), (pi, 0)
y = sin(sin x)
y' = cos(sin x)*cos x
y'(pi) = cos(0)*cos(pi) = 1(-1) = -1
so, you want the line through (pi,0) with slope = -1
(y-0)/(x-pi) = -1
y = pi - x