What do you need help with?
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)
<span>putting it into maclaurin's series</span>
<span>We get,</span>
<span> 0+0+ 2x^2/2! + 0...</span>
What number do u need help on?
Answer:
(B)
Step-by-step explanation:
we can solve by using formula (1/2) x base x height to get the answer.