The problem is as the title says. This is an example we went through during the lecture and therefore I have the solution. However there is a particular step in the solution which I do not understand. Using the Taylor series we will write sin(x) as: sin(x) = x - (x^3)/6 + (x^5)B(x) and arctan(x) = x - (x^3)/3 + (x^5)C(x) B and C I believe are functions restricted near 0. Anyway since our function is sin(arctan (x)) it is appropriate to insert arctan in every x in the sin(x) function. Therefore we get sin (arctan(x) ) = ( x - (x^3)/3 + (x^5)C(x) ) - ( x - (x^3)/3 + (x^5)C(x) )^3 / 6 + (( x - (x^3)/3 + (x^5)C(x) )^5) B(x) As I see it, it will take an enourmous amount of work to simplfy this polynom because we have long expressions with the power both 3 and 5: However in the lecture the lecturer says that the above is equal to... x - x^3)/3 - (x^3)/6 + (x^4)D(x), where D has the same def. as B and C have above. Anyway he does this in one single step. Could anyone please explain to me how he came to this conclusion? What am I not seeing? Is there a simple way to get to this conclusion? If so please demonstrate. Thank you!