Maclaurin of tan x

Problem

Write the Maclaurin expansion of $\tan x$ up to the term in $x^3$ (hint: $\tan x = \sin x/\cos x$ and long division of the series).

Solution

I use the expansions $\sin x = x - \dfrac{x^3}{6} + o(x^4)$ and $\cos x = 1 - \dfrac{x^2}{2} + o(x^3)$. Then $\tan x = \frac{x - x^3/6 + o(x^4)}{1 - x^2/2 + o(x^3)} = \Bigl(x - \frac{x^3}{6}\Bigr)\Bigl(1 + \frac{x^2}{2} + o(x^2)\Bigr) + o(x^3),$ having used $\dfrac{1}{1-u} = 1+u+o(u)$ with $u=x^2/2$. Expanding the product and keeping the terms up to $x^3$: $\tan x = x + \frac{x^3}{2} - \frac{x^3}{6} + o(x^3) = \boxed{x + \frac{x^3}{3} + o(x^3)}.$

Links

Topics: Derivatives
Concepts: Maclaurin polynomial · Standard expansion
Functions: Tangent
Methods: Taylor polynomial
Skills: Use formulae
Exercise type: Derivative computation