Originally Posted by
hazecraze Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x))
y'=cos[(sinx)(cosx)]
You wrote that wierd, so I'm not sure if you got it wrong, or just used bad notation.
y'3π=cos[(sinx)(-1)]
cos [(0)(-1)]
cos(0)=1
y-0=1(x-3π)
y= x-3πWhat's wrong with my answer?