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anonymous
 5 years ago
Find the Maclaurin series for f(x) = sin^2(x)
anonymous
 5 years ago
Find the Maclaurin series for f(x) = sin^2(x)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use sin^2(x) =1/2 * (1cos2x) cos 2x  use def cosx from table

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0forgot to add that, it says to (Use the Identity 2sinxcosx = sin(2x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(x)=sin^2(0) f'(x)=2*sin(x)*cos(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ouch! seems easier to use:sin^2(x) =1/2 * (1cos2x)  no multiplication...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0therefore f'(x)=sin(2x) f''(x)=2cos(2x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is 2sinxcosx has to do with sin^2(x)...? did you mean in initial date: sin2x or sin^2(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f'''(x)=4sin(2x)=4*f'(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0inik, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0putting it into maclaurin's series im getting 0+0+ 2x^2/2! + 0... am i doing something wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0got 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not using derivatives for Maclaurin series? How do you do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0James Stuart Essential Calculus ch8... I'm trying to find ref on Internet  let you know

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0huykma, 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.
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