Question

Maclaurin expansion series of sin^2 x

Answer 1

Use an identity.

Which I identity will I use?

Answer 2

sin^2(x) = 1/2 - 1/2 cos(2x)

Answer 3

http://www.math.com/tables/trig/identities.htm

Answer 4

Would that be easier?

Answer 5

yes, because now all you need to do is find taylor expansion of cos(x)

Answer 6

why cos x? not cos 2x?

Answer 7

you actually want cos(2x), but it is easier to find series for cos(x) then replace every x with 2x

Answer 8

Would that work?

Answer 9

yes

Answer 10

What would f^(n) be?

Answer 11

what series expansion of cos(x)

Answer 12

maclaurin. a=0

Answer 13

yes

Answer 14

I find it hard to get the f^n given 1/2-1/2cos 2x
How do you do that?

Answer 15

I would only try to find series expansion of cos(x)

Answer 16

thanks!

Answer 17

Did you find it?

Answer 18

yes.

cos(0)+ -sin(0)/1 (x) + -cos(0)/2! (x)^2

1 -x^2/2! + x^4/4!

Answer 19

that's

cos(x)=
\[\sum _{n=0}^{\infty } \frac{\left((-1)^nx^{2 n}\right)}{(2k)!}\]

Answer 20

now replace x with 2x

Answer 21

cos(2x)=\[\sum _{n=0}^{\infty } \frac{\left((-1)^n(2x)^{2 n}\right)}{(2n)!}\]

Answer 22

1/2 - 1/2 cos (2x)

replace cos(2x) that series

Answer 23

1/2 - 1/2 (\[\sum _{n=0}^{\infty } \frac{\left((-1)^n(2x)^{2 n}\right)}{(2n)!}\])

Answer 24

This is the maclaurin expansion? COOL! :D Thanks @imranmeah91!

Answer 25

wait. is this a=0?

Answer 26

yes, it is

Answer 27

THAAAAAAAAAAAANKS! :)

Answer 28

you are welcome