Determine the first four terms of the Maclaurin series for sin 2x (a) by using the definition of

Reggie

Reggie

Answered question

2021-03-11

For sin2x, find the first four terms in the Maclaurin series.
(a) by using the definition of Maclaurin series.
(b) by replacing x by 2x in the series for sin2x.
(c) by multiplying 2 by the series for sinx by the series for  cosx, because sin2x=2 sinxcosx

Answer & Explanation

toroztatG

toroztatG

Skilled2021-03-12Added 98 answers

(a)
Calculate the derivative for f(x)=sin(2x) and the the value for the derivatives till 8th order.
f(x)=sin2x
f1(x)=2cos2x
f2(x)=4sin2x
f3(x)=8cos2x
f4(x)=16sin(2x)
f5(x)=32cos(2x)
f6(x)=64sin(2x)
f7(x)=128cos(2x)
f8(x)=256sin(2x)
At x=0, f(0)=sin(2×0)=0
f1(x)=2cos(2×0)=2
f2(x)=4sin(2×0)=0
f3(x)=8cos(2×0)=8
f4(x)=16sin(2×0)=0
f5(x)=32cos(2×0)=32
f6(x)=64sin(2×0)=0
f7(x)=128cos(2×0)=128
f8(x)=256sin(2×0)=0
Now, consider the formula for the maclaurin's series and express f(x) as Maclaurin's series.
Maclaurin's series expansion,
f(x)=f(0)+f1(0)x1!+f2(0)x22!+f3(0)x33!+f4(0)x44!+f5(0)x55!+f6(0)x66!+f7(0)x77!+f8(0)x88!+f9(0)x99!
=0+f1(0)x1!+0+f3(0)x33!+0+f5(0)x55!+0+f7(0)x77!+0
=2x1!+(8)x33!+(32)x55!+(128)x77!
So, f(x)=2x4x33+415x58315x7
Hence, sin2x=2x4x33+415x58315x7
(b) Now, replace x by 2x and evaluate the series for sin4x.

Jeffrey Jordon

Jeffrey Jordon

Expert2021-12-16Added 2605 answers

Answer is given below (on video)

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