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19-11-2016
Mathematics

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

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syed514 syed514

21-11-2016

use sin^2(x) =1/2 * (1-cos2x)

cos 2x - use def cosx from table

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

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

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

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

putting it into maclaurin's series

We get,

0+0+ 2x^2/2! + 0...

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