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19-11-2016
Mathematics

Answered

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

Verified Answer :

syed514 syed514

20-11-2016

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

cos 2x - use def cosx from table

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

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

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

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

putting it into maclaurin's series

We get,

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

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