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

Jun 16, 2015

#### Answer:

$1 - {x}^{2} + {x}^{4} / 3 - \frac{2}{45} {x}^{6} + {x}^{8} / 315 + \setminus \cdots$

#### Explanation:

There are two methods.

1) Let $f \left(x\right) = {\cos}^{2} \left(x\right)$ and use f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+\cdots

We have, by the Chain Rule and/or Product Rule,

$f ' \left(x\right) = - 2 \cos \left(x\right) \sin \left(x\right)$, $f ' ' \left(x\right) = 2 {\sin}^{2} \left(x\right) - 2 {\cos}^{2} \left(x\right)$

$f ' ' ' \left(x\right) = 4 \sin \left(x\right) \cos \left(x\right) + 4 \cos \left(x\right) \sin \left(x\right) = 8 \cos \left(x\right) \sin \left(x\right)$

$f ' ' ' ' \left(x\right) = - 8 {\sin}^{2} \left(x\right) + 8 {\cos}^{2} \left(x\right)$, $f ' ' ' ' ' \left(x\right) = \setminus \cdots = - 32 \cos \left(x\right) \sin \left(x\right)$,

$f ' ' ' ' ' ' \left(x\right) = 32 {\sin}^{2} \left(x\right) - 32 {\cos}^{2} \left(x\right)$, etc...

Hence, $f \left(0\right) = 1$, $f ' \left(0\right) = 0$, $f ' ' \left(0\right) = - 2$, $f ' ' ' \left(0\right) = 0$, $f ' ' ' ' \left(0\right) = 8$, $f ' ' ' ' ' \left(0\right) = 0$, $f ' ' ' ' ' ' \left(0\right) = - 32$, etc...

Since 2! =2, 8/(4!)=8/24=1/3, and (-32)/(6!)=(-32)/720=-2/45, this much calculation leads to an answer of

$1 - {x}^{2} + {x}^{4} / 3 - \frac{2}{45} {x}^{6} + \setminus \cdots$

This does happen to converge for all $x$ and it does happen to equal ${\cos}^{2} \left(x\right)$ for all $x$. You can also check on your own that the next non-zero term is $+ {x}^{8} / 315$

2) Use the well-known Maclaurin series cos(x)=1-x^2/(2!)+x^4/(4!)-x^6/(6!)+\cdots

$= 1 - {x}^{2} / 2 + {x}^{4} / 24 - {x}^{6} / 720 + \setminus \cdots$ and multiply it by itself (square it).

To do this, first multiply the first term $1$ by everything in the series to get

${\cos}^{2} \left(x\right) = \left(1 - {x}^{2} / 2 + {x}^{4} / 24 - {x}^{6} / 720 + \setminus \cdots\right) + \setminus \cdots$

Next, multiply $- {x}^{2} / 2$ by everything in the series to get

${\cos}^{2} \left(x\right) = \left(1 - {x}^{2} / 2 + {x}^{4} / 24 - {x}^{6} / 720 + \setminus \cdots\right) + \left(- {x}^{2} / 2 + {x}^{4} / 4 - {x}^{6} / 48 + {x}^{8} / 1440 + \setminus \cdots\right) + \setminus \cdots$

Then multiply ${x}^{4} / 24$ by everything in the series to get

${\cos}^{2} \left(x\right) = \left(1 - {x}^{2} / 2 + {x}^{4} / 24 - {x}^{6} / 720 + \setminus \cdots\right) + \left(- {x}^{2} / 2 + {x}^{4} / 4 - {x}^{6} / 48 + {x}^{8} / 1440 + \setminus \cdots\right) + \left({x}^{4} / 24 - {x}^{6} / 48 + {x}^{8} / 576 - {x}^{10} / 17280 + \setminus \cdots\right) + \setminus \cdots$

etc...

If you go out far enough and combine "like-terms", using the facts, for instance, that $- \frac{1}{2} - \frac{1}{2} = - 1$ and $\frac{1}{24} + \frac{1}{4} + \frac{1}{24} = \frac{8}{24} = \frac{1}{3}$, etc..., you'll eventually come to the same answer as above:

$1 - {x}^{2} + {x}^{4} / 3 - \frac{2}{45} {x}^{6} + \setminus \cdots$