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

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

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use sin^2(x) =1/2 * (1-cos2x) cos 2x - use def cosx from table
forgot to add that, it says to (Use the Identity 2sinxcosx = sin(2x)
f(x)=sin^2(0) f'(x)=2*sin(x)*cos(x)

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Other answers:

Ouch! seems easier to use:sin^2(x) =1/2 * (1-cos2x) - no multiplication...
therefore f'(x)=sin(2x) f''(x)=2cos(2x)
what is 2sinxcosx has to do with sin^2(x)...? did you mean in initial date: sin2x or sin^2(x)
f'''(x)=-4sin(2x)=-4*f'(x)
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.
putting it into maclaurin's series im getting 0+0+ 2x^2/2! + 0... am i doing something wrong
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
not using derivatives for Maclaurin series? How do you do that?
James Stuart Essential Calculus ch8... I'm trying to find ref on Internet - let you know
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.
thanks

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