Question

Find the Maclaurin series for f(x) = sin^2(x)

Answer 1

use sin^2(x) =1/2 * (1-cos2x)
cos 2x - use def cosx from table

Answer 2

forgot to add that, it says to (Use the Identity 2sinxcosx = sin(2x)

Answer 3

f(x)=sin^2(0)
f'(x)=2*sin(x)*cos(x)

Answer 4

Ouch! seems easier to use:sin^2(x) =1/2 * (1-cos2x) - no multiplication...

Answer 5

therefore f'(x)=sin(2x)
f''(x)=2cos(2x)

Answer 6

what is 2sinxcosx has to do with sin^2(x)...?
did you mean in initial date: sin2x or sin^2(x)

Answer 7

f'''(x)=-4sin(2x)=-4*f'(x)

Answer 8

inik, it's because the first derivative of the function is 2cos(x)sin(x) which is sin(2x), according to the identity given by asker.

Answer 9

putting it into maclaurin's series im getting 0+0+ 2x^2/2! + 0...
am i doing something wrong

Answer 10

got that... I thought he asked for Maclaurin series representation...I don't use derivatives. there is known presentation of cos & sin - why not to use it? it's simple

Answer 11

not using derivatives for Maclaurin series? How do you do that?

Answer 12

James Stuart Essential Calculus ch8... I'm trying to find ref on Internet - let you know

Answer 13

huykma, i think you are doing it right, when you get to f''''(0) the term would be something times f''(0) and the cycle goes on and on.

Answer 14

thanks