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

##1-x^2+x^4/3-2/45 x^6+x^8/315+\cdots##

There are two methods.

1) Let ##f(x)=cos^2(x)## and use ##f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+\cdots##

We have, by the and/or ,

##f'(x)=-2cos(x)sin(x)##, ##f''(x)=2sin^2(x)-2cos^2(x)##

##f'''(x)=4sin(x)cos(x)+4cos(x)sin(x)=8cos(x)sin(x)##

##f''''(x)=-8sin^2(x)+8cos^2(x)##, ##f'''''(x)=\cdots=-32cos(x)sin(x)##,

##f''''''(x)=32sin^2(x)-32cos^2(x)##, etc...

Hence, ##f(0)=1##, ##f'(0)=0##, ##f''(0)=-2##, ##f'''(0)=0##, ##f''''(0)=8##, ##f'''''(0)=0##, ##f''''''(0)=-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^4/3-2/45 x^6+\cdots##

This does happen to converge for all ##x## and it does happen to equal ##cos^2(x)## 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+\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(x)=(1-x^2/2+x^4/24-x^6/720+\cdots)+\cdots##

Next, multiply ##-x^2/2## by everything in the series to get

##cos^2(x)=(1-x^2/2+x^4/24-x^6/720+\cdots)+(-x^2/2+x^4/4-x^6/48+x^8/1440+\cdots)+\cdots##

Then multiply ##x^4/24## by everything in the series to get

##cos^2(x)=(1-x^2/2+x^4/24-x^6/720+\cdots)+(-x^2/2+x^4/4-x^6/48+x^8/1440+\cdots)+(x^4/24-x^6/48+x^8/576-x^10/17280+\cdots)+\cdots##

etc...

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

##1-x^2+x^4/3-2/45 x^6+\cdots##

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