# 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