Forsin⁡2x, find the first four terms in the Maclaurin series.(a) by using the definition of...

(a)
Calculate the derivative for $f(x)=\sin (2x)$ and the the value for the derivatives till 8th order.
$f(x)=\sin 2x$
$f_1(x)=2\cos 2x$
$f_2(x)=-4\sin 2x$
$f_3(x)=-8\cos 2x$
$f_4(x)=16\sin (2x)$
$f_5(x)=32\cos (2x)$
$f_6(x)=-64\sin (2x)$
$f_7(x)=-128\cos (2x)$
$f_8(x)=256\sin (2x)$
At $x=0$, $f(0)=\sin (2\times 0)=0$
$f_1(x)=2\cos (2\times 0)=2$
$f_2(x)=-4\sin (2\times 0)=0$
$f_3(x)=-8\cos (2\times 0)=-8$
$f_4(x)=16\sin (2\times 0)=0$
$f_5(x)=32\cos (2\times 0)=32$
$f_6(x)=-64\sin (2\times 0)=0$
$f_7(x)=-128\cos (2\times 0)=-128$
$f_8(x)=256\sin (2\times 0)=0$
Now, consider the formula for the maclaurin's series and express f(x) as Maclaurin's series.
Maclaurin's series expansion,
$f(x)=f(0)+\frac{f_1(0)x}{1!}+\frac{f_2(0)x^2}{2!}+\frac{f_3(0)x^3}{3!}+\frac{f_4(0)x^4}{4!}+\frac{f_5(0)x^5}{5!}+\frac{f_6(0)x^6}{6!}+\frac{f_7(0)x^7}{7!}+\frac{f_8(0)x^8}{8!}+\frac{f_9(0)x^9}{9!}$
$=0+\frac{f_1(0)x}{1!}+0+\frac{f_3(0)x^3}{3!}+0+\frac{f_5(0)x^5}{5!}+0+\frac{f_7(0)x^7}{7!}+0$
$=\frac{2x}{1!}+\frac{(-8)x^3}{3!}+\frac{(32)x^5}{5!}+\frac{(-128)x^7}{7!}$
So, $f(x)=2x-\frac{4x^3}{3}+\frac{4}{15}{x}^5-\frac{8}{315}{x}^7$
Hence, $\sin 2x=2x-\frac{4x^3}{3}+\frac{4}{15}{x}^5-\frac{8}{315}{x}^7$
(b) Now, replace x by 2x and evaluate the series for $\sin 4x$.