# Determine the first four terms of the Maclaurin series for sin 2x (a) by using the definition of Maclaurin series. (b) by replacing x by 2x in the series for sin 2x. (c) by multiplying 2 by the series for sin x by the series for cos x, because sin 2x = 2 sin x cos x

Question
Series
Determine the first four terms of the Maclaurin series for sin 2x
(a) by using the definition of Maclaurin series.
(b) by replacing x by 2x in the series for sin 2x.
(c) by multiplying 2 by the series for sin x by the series for cos x, because sin 2x = 2 sin x cos x

2021-03-12
(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, $$2\sin2x\cos x=2x-\frac{4x^3}{3}+\frac{4x^5}{15}-\frac{8x^7}{315}$$

### Relevant Questions

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