# How do you find the Maclaurin Series for cos (x)^2?

Mar 25, 2017

cos^2(x) = 1+sum_(k=1)^(oo) ((-1)^(k)(2)^(2k-1))/((2k)!) * x^(2k)

$\textcolor{red}{\mathmr{and}}$

cos^2(x) = 1 - (2x^2)/(2!) + (8x^4)/(4!) - (32x^6)/(6!)+...

#### Explanation:

I am assuming that $\cos {\left(x\right)}^{2}$ refers to ${\cos}^{2} \left(x\right)$.

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First, find the first 4 or 5 derivatives of $f \left(x\right) = {\cos}^{2} \left(x\right)$

This is much easier if we use the identity ${\cos}^{2} \left(\theta\right) = \frac{1 + \cos \left(2 \theta\right)}{2}$.

Therefore, the first few derivatives of $f \left(x\right)$ are:

$f \left(x\right) = \frac{1 + \cos \left(2 x\right)}{2}$

$f ' \left(x\right) = - \sin \left(2 x\right)$

$f ' ' \left(x\right) = - 2 \cos \left(2 x\right)$

$f ' ' ' \left(x\right) = 4 \sin \left(2 x\right)$

${f}^{\left(4\right)} \left(x\right) = 8 \cos \left(2 x\right)$

Since we are finding a Maclaurin series, plug in $0$ to each derivative.

$f \left(0\right) = \frac{1 + \cos \left(0\right)}{2} = 1$

$f ' \left(0\right) = - \sin \left(0\right) = 0$

$f ' ' \left(0\right) = - 2 \cos \left(0\right) = - 2$

$f ' ' ' \left(0\right) = 4 \sin \left(0\right) = 0$

${f}^{\left(4\right)} \left(0\right) = 8 \cos \left(0\right) = 8$

It is easy to see that the coefficients will continue on in this way, with odd derivatives always being 0, and even derivatives being 2 to the power of one less than the order of the derivative, with alternating positive and negative signs. Therefore, the Maclaurin series can be written as:

cos^2(x) = 1 - (2x^2)/(2!) + (8x^4)/(4!) - (32x^6)/(6!)+...

Or, using summation notation and a little bit of intuition,

cos^2(x) = 1+sum_(k=1)^(oo) ((-1)^(k)(2)^(2k-1))/((2k)!) * x^(2k)