# Maclaurin expansion of $\sqrt{\cos 2x}$ and $\tan^2 x$ up to degree 4

Find the Maclaurin expansion $\sqrt{\cos(2x)}$ and $\tan^2x$ up to degree $4$.

I tried differentiation but it gives me something really horrible.

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Are you sure they want you to find the Maclaurin expansion of $\sqrt{\cos(2x)}$? It really is just horrible differentiation with no visible pattern, but it doesn't seem like a very useful exercise if that is the case. The one for $\tan^2(x)$ can be found more easily by simply squaring the Maclaurin expansion for $\tan(x)$. –  august Feb 5 '13 at 20:05
@august I have to find both up to degree 4 so squaring would be a bit of a pain, wouldn't it? –  bbr4in Feb 5 '13 at 20:14
@user52187: if all you need is the expression use wolframalpha.com, it saves your time –  Alex Feb 5 '13 at 21:17
@Alex working is needed –  bbr4in Feb 5 '13 at 22:44
Not really, if you only need up to degree 4 then squaring will give you the 4 terms with relative ease. –  august Feb 6 '13 at 4:57

Working up to degree $4$ is the key here; otherwise the whole affair would be hopeless. You are given the license to ignore all powers of $x$ above $4$. So, $$\cos 2x = 1-\frac{(2x)^2}{2}+\frac{(2x)^4}{24} +\dots$$ Write this as $1+u$ because $1+u$ is the kind of thing you want to have under the square root. There's no need to worry about $u^3$ and higher powers, they will contain only powers of $x$ above $4$. So, $$\sqrt{1+u}= 1+\frac12 u - \frac{1}{8}u^2 + \dots$$ And the conclusion is $$\sqrt{\cos 2x} = 1+ \frac12\left(-\frac{(2x)^2}{2}+\frac{(2x)^4}{24}\right) -\frac18 \left(-\frac{(2x)^2}{2}\right)^2 +\dots$$
With $\tan^2x$ it's easier: $$\tan^2x=(x-x^3/6+\cdots)^2 (1-x^2/2+\cdots)^{-2} = (x^2- 2x^4/6 +\cdots) (1+x^2+\cdots)$$
To "40 votes". Your expansion of $\cos 2x$ is clearly wrong. What you have written is expansion for $\cos x$. Similar is case with $\tan^{2}x$. Anyway the OP wants to get a maclaurin series upto $x^{4}$ and in this case it is best to do four times differentiation rather than using series manipulation techniques. –  Paramanand Singh Jul 28 '13 at 4:47
@ParamanandSingh Thanks, I put in the missing $2$. –  40 votes Jul 28 '13 at 4:50
check the expansion for $\tan^{2}x$ too. It seems this has been typed in hurry. Ideally it should be $(x - x^{3}/6 + \cdots)^{2}(1 - x^{2}/2 + \cdots)^{-2}$. –  Paramanand Singh Jul 28 '13 at 4:53