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

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...

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

rx and sx

step-by-step explanation: