(a)

Calculate the derivative for f(x) = sin(2x) and the the value for the derivatives till 8th order.

\(f(x)=\sin2x\)

\(f_1(x)=2\cos2x\)

\(f_2(x)=-4\sin2x\)

\(f_3(x)=-8\cos2x\)

\(f_4(x)=16\sin(2x)\)

\(f_5(x)=32\cos(2x)\)

\(f_6(x)=-64\sin(2x)\)

\(f_7(x)=-128\cos(2x)\)

\(f_8(x)=256\sin(2x)\)

At x=0, \(f(0)=\sin(2\times0)=0\)

\(f_1(x)=2\cos(2\times0)=2\)

\(f_2(x)=-4\sin(2\times0)=0\)

\(f_3(x)=-8\cos(2\times0)=-8\)

\(f_4(x)=16\sin(2\times0)=0\)

\(f_5(x)=32\cos(2\times0)=32\)

\(f_6(x)=-64\sin(2\times0)=0\)

\(f_7(x)=-128\cos(2\times0)=-128\)

\(f_8(x)=256\sin(2\times0)=0\)

Now, consider the formula for the maclaurin's series and express f(x) as Maclaurin's series.

Maclaurin's series expansion,

\(f(x)=f(0)+\frac{f_1(0)x}{1!}+\frac{f_2(0)x^2}{2!}+\frac{f_3(0)x^3}{3!}+\frac{f_4(0)x^4}{4!}+\frac{f_5(0)x^5}{5!}+\frac{f_6(0)x^6}{6!}+\frac{f_7(0)x^7}{7!}+\frac{f_8(0)x^8}{8!}+\frac{f_9(0)x^9}{9!}\)

\(=0+\frac{f_1(0)x}{1!}+0+\frac{f_3(0)x^3}{3!}+0+\frac{f_5(0)x^5}{5!}+0+\frac{f_7(0)x^7}{7!}+0\)

\(=\frac{2x}{1!}+\frac{(-8)x^3}{3!}+\frac{(32)x^5}{5!}+\frac{(-128)x^7}{7!}\)

So, \(f(x)=2x-\frac{4x^3}{3}+\frac{4}{15}x^5-\frac{8}{315}x^7\)

Hence, \(\sin2x=2x-\frac{4x^3}{3}+\frac{4}{15}x^5-\frac{8}{315}x^7\)

(b) Now, replace x by 2x and evaluate the series for \(\sin4x\).

\(\sin2x=2x-\frac{4x^3}{3}+\frac{4}{15}x^5-\frac{8}{315}x^7\)

\(\sin(2\times2x)=(2\times2x)-\frac{4(2x)^3}{3}+\frac{4}{15}(2x)^5-\frac{8}{315}(2x)^7\)

\(\sin4x=4x-\frac{8x^3}{3}+\frac{4\times32x^5}{15}-\frac{8\times128x^7}{315}\)

\(\sin4x=4x-\frac{32x^3}{3}+\frac{128x^5}{15}-\frac{1024x^7}{315}\)

Hence, \(\sin4x=4x-\frac{32x^3}{3}+\frac{128x^5}{15}-\frac{1024x^7}{315}\)

(c)

Now, determine the product for the 2, maclaurin's series fro sin x and cos x and compare it with the maclaurin's series for sin2x.

The series is same for both the expressions.

\(\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}\)

\(\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}\)

\(2\sin x\cos x=2\times(x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040})\times(1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720})\)

\(=2(x-\frac{2x^3}{3}+\frac{2x^5}{15}-\frac{4x^7}{315})\)

\(=2x-\frac{4x^3}{3}+\frac{4x^5}{15}-\frac{8x^7}{315}\)

\(\sin2x=2x-\frac{4x^3}{3}+\frac{4x^5}{15}-\frac{8x^7}{315}\)

Hence, \(2\sin2x\cos x=2x-\frac{4x^3}{3}+\frac{4x^5}{15}-\frac{8x^7}{315}\)

Calculate the derivative for f(x) = sin(2x) and the the value for the derivatives till 8th order.

\(f(x)=\sin2x\)

\(f_1(x)=2\cos2x\)

\(f_2(x)=-4\sin2x\)

\(f_3(x)=-8\cos2x\)

\(f_4(x)=16\sin(2x)\)

\(f_5(x)=32\cos(2x)\)

\(f_6(x)=-64\sin(2x)\)

\(f_7(x)=-128\cos(2x)\)

\(f_8(x)=256\sin(2x)\)

At x=0, \(f(0)=\sin(2\times0)=0\)

\(f_1(x)=2\cos(2\times0)=2\)

\(f_2(x)=-4\sin(2\times0)=0\)

\(f_3(x)=-8\cos(2\times0)=-8\)

\(f_4(x)=16\sin(2\times0)=0\)

\(f_5(x)=32\cos(2\times0)=32\)

\(f_6(x)=-64\sin(2\times0)=0\)

\(f_7(x)=-128\cos(2\times0)=-128\)

\(f_8(x)=256\sin(2\times0)=0\)

Now, consider the formula for the maclaurin's series and express f(x) as Maclaurin's series.

Maclaurin's series expansion,

\(f(x)=f(0)+\frac{f_1(0)x}{1!}+\frac{f_2(0)x^2}{2!}+\frac{f_3(0)x^3}{3!}+\frac{f_4(0)x^4}{4!}+\frac{f_5(0)x^5}{5!}+\frac{f_6(0)x^6}{6!}+\frac{f_7(0)x^7}{7!}+\frac{f_8(0)x^8}{8!}+\frac{f_9(0)x^9}{9!}\)

\(=0+\frac{f_1(0)x}{1!}+0+\frac{f_3(0)x^3}{3!}+0+\frac{f_5(0)x^5}{5!}+0+\frac{f_7(0)x^7}{7!}+0\)

\(=\frac{2x}{1!}+\frac{(-8)x^3}{3!}+\frac{(32)x^5}{5!}+\frac{(-128)x^7}{7!}\)

So, \(f(x)=2x-\frac{4x^3}{3}+\frac{4}{15}x^5-\frac{8}{315}x^7\)

Hence, \(\sin2x=2x-\frac{4x^3}{3}+\frac{4}{15}x^5-\frac{8}{315}x^7\)

(b) Now, replace x by 2x and evaluate the series for \(\sin4x\).

\(\sin2x=2x-\frac{4x^3}{3}+\frac{4}{15}x^5-\frac{8}{315}x^7\)

\(\sin(2\times2x)=(2\times2x)-\frac{4(2x)^3}{3}+\frac{4}{15}(2x)^5-\frac{8}{315}(2x)^7\)

\(\sin4x=4x-\frac{8x^3}{3}+\frac{4\times32x^5}{15}-\frac{8\times128x^7}{315}\)

\(\sin4x=4x-\frac{32x^3}{3}+\frac{128x^5}{15}-\frac{1024x^7}{315}\)

Hence, \(\sin4x=4x-\frac{32x^3}{3}+\frac{128x^5}{15}-\frac{1024x^7}{315}\)

(c)

Now, determine the product for the 2, maclaurin's series fro sin x and cos x and compare it with the maclaurin's series for sin2x.

The series is same for both the expressions.

\(\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}\)

\(\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}\)

\(2\sin x\cos x=2\times(x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040})\times(1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720})\)

\(=2(x-\frac{2x^3}{3}+\frac{2x^5}{15}-\frac{4x^7}{315})\)

\(=2x-\frac{4x^3}{3}+\frac{4x^5}{15}-\frac{8x^7}{315}\)

\(\sin2x=2x-\frac{4x^3}{3}+\frac{4x^5}{15}-\frac{8x^7}{315}\)

Hence, \(2\sin2x\cos x=2x-\frac{4x^3}{3}+\frac{4x^5}{15}-\frac{8x^7}{315}\)