Chapter 9.3, Problem 44E

(a)

To determine

To obtain: Maclaurin series for ${\sin }^{2}x$ .

Answer to Problem 44E

Maclaurin series for ${\sin }^{2}x$ is $x^2-\frac{1}{3}{x}^4+\frac{2}{45}{x}^6-\dots -\frac{{\left(-1\right)}^{n-1}{2}^{2n-1}{x}^{2n}}{\left(2n\right)!}+\cdots$ .

Explanation of Solution

Given information: Function ${\sin }^{2}x$ along with the identity ${\sin }^{2}x=\frac{1}{2}\left(1-\cos \left(2x\right)\right)$ is given.

Formula used :

The Maclaurin series for the function $\cos x$ is given by,

  $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\cdots +{\left(-1\right)}^n\frac{x^{2n}}{\left(2n\right)!}+\cdots$

Calculation:

Replace $x$ with $2x$ in the Maclaurin series for the function $\cos x$ and simplify the right side of the resulting equation.

  $\begin{array}{c}\cos \left(2x\right)=1-\frac{{\left(2x\right)}^2}{2!}+\frac{{\left(2x\right)}^4}{4!}-\frac{{\left(2x\right)}^6}{6!}+\cdots +{\left(-1\right)}^n\frac{x^{2n}}{\left(2n\right)!}+\cdots \\ =1-\frac{4x^2}{2}+\frac{16x^4}{24}-\frac{64x^5}{720}+\cdots +{\left(-1\right)}^n\frac{{\left(2x\right)}^{2n}}{\left(2n\right)!}+\cdots \\ =1-2x^2+\frac{2}{3}{x}^4-\frac{4}{45}{x}^5+\cdots +{\left(-1\right)}^n\frac{{\left(2x\right)}^{2n}}{\left(2n\right)!}+\cdots \end{array}$

Since ${\sin }^{2}x=\frac{1}{2}\left(1-\cos \left(2x\right)\right)$ , substitute , $\cos \left(2x\right)=1-2x^2+\frac{2}{3}{x}^4-\frac{4}{45}{x}^6+\cdots +{\left(-1\right)}^n\frac{{\left(2x\right)}^{2n}}{\left(2n\right)!}+\cdots$ into ${\sin }^{2}x=\frac{1}{2}\left(1-\cos \left(2x\right)\right)$ and simplify to obtain the required result.

  $\begin{array}{c}{\sin }^{2}x=\frac{1}{2}\left(1-\left(1-2x^2+\frac{2}{3}{x}^4-\frac{4}{45}{x}^6+\cdots +{\left(-1\right)}^n\frac{{\left(2x\right)}^{2n}}{\left(2n\right)!}+\cdots \right)\right)\\ =\frac{1}{2}-\frac{1}{2}\left(1-2x^2+\frac{2}{3}{x}^4-\frac{4}{45}{x}^6\cdots +{\left(-1\right)}^n\frac{{\left(2x\right)}^{2n}}{\left(2n\right)!}+\cdots \right)\\ =\frac{1}{2}-\frac{1}{2}+x^2-\frac{1}{3}{x}^4+\frac{2}{45}{x}^6-\cdots +{\left(-1\right)}^n\frac{{\left(2x\right)}^{2n}}{2\left(2n\right)!}+\cdots \\ =x^2-\frac{1}{3}{x}^4+\frac{2}{45}{x}^6-\cdots +{\left(-1\right)}^n\frac{2^{2n-1}{x}^{2n}}{\left(2n\right)!}+\cdots \end{array}$

As a result, the Maclaurin series for ${\sin }^{2}x$ is ${\sin }^{2}x=x^2-\frac{1}{3}{x}^4+\frac{2}{45}{x}^6-\dots +\frac{{\left(-1\right)}^{n-1}{2}^{2n-1}{x}^{2n}}{\left(2n\right)!}+\cdots$

: (b).

To determine

To differentiate: The series ${\sin }^{2}x=x^2-\frac{1}{3}{x}^4+\frac{2}{45}{x}^6-\dots +\frac{{\left(-1\right)}^{n-1}{2}^{2n-1}{x}^{2n}}{\left(2n\right)!}+\cdots$ to obtain the Maclaurin series for $2\sin x\cos x$ .

Answer to Problem 44E

Maclaurin series for $2\sin x\cos x$ is $2x-\frac{4}{3}{x}^3+\frac{4}{15}{x}^5-\dots$ .

Explanation of Solution

Given information: Function ${\sin }^{2}x$ along with the identity ${\sin }^{2}x=\frac{1}{2}\left(1-\cos \left(2x\right)\right)$ is given.

Concept used:

Use the Maclaurin series ${\sin }^{2}x=x^2-\frac{1}{3}{x}^4+\frac{2}{45}{x}^6-\dots +\frac{{\left(-1\right)}^{n-1}{2}^{2n-1}{x}^{2n}}{\left(2n\right)!}+\cdots$

Calculation:

Differentiate ${\sin }^{2}x=x^2-\frac{1}{3}{x}^4+\frac{2}{45}{x}^6-\dots +\frac{{\left(-1\right)}^{n-1}{2}^{2n-1}{x}^{2n}}{\left(2n\right)!}+\cdots$ on both sides and simplify to obtain the required result.

  $\begin{array}{c}\frac{d}{dx}\left({\sin }^{2}x\right)=\frac{d}{dx}\left(x^2-\frac{1}{3}{x}^4+\frac{2}{45}{x}^6-\frac{1}{315}{x}^8+\cdots \right)\\ 2\sin x\frac{d}{dx}\left(\sin x\right)=2x-\frac{4x^3}{3}+\frac{4x^5}{15}-\frac{8x^7}{315}+\cdots \\ 2\sin x\cos x=2x-\frac{4}{3}{x}^3+\frac{4}{15}{x}^5-\frac{8}{315}{x}^7+\cdots \end{array}$

Therefore, the required Maclaurin series for $2\sin x\cos x$ is $2x-\frac{4}{3}{x}^3+\frac{4}{15}{x}^5-\frac{8}{315}{x}^7+\dots$ .

(c)

To determine

To verify: The obtained series in part (b) is the series for $\sin \left(2x\right)$ .

Explanation of Solution

Given information: Function ${\sin }^{2}x$ along with the identity ${\sin }^{2}x=\frac{1}{2}\left(1-\cos \left(2x\right)\right)$ is given.

Concept used :

The Maclaurin series for $\sin x$ is given by $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots +{\left(-1\right)}^n\frac{x^{2n}}{\left(2n\right)!}+\cdots$

Calculation:

Replace $x$ with $2x$ into $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots +{\left(-1\right)}^n\frac{x^{2n}}{\left(2n\right)!}+\cdots$ and simplify the obtained equation.

  $\begin{array}{c}\sin \left(2x\right)=2x-\frac{{\left(2x\right)}^3}{3!}+\frac{{\left(2x\right)}^5}{5!}-\frac{{\left(2x\right)}^7}{7!}+\cdots +{\left(-1\right)}^n\frac{{\left(2x\right)}^{2n}}{\left(2n\right)!}+\cdots \\ =2x-\frac{4}{3}{x}^3+\frac{4}{15}{x}^5-\frac{8}{315}{x}^7+\cdots \end{array}$

Substitute $2\sin x\cos x$ for $\sin \left(2x\right)$ into the left side of the obtained equation.

  $2\sin x\cos x=2x-\frac{4}{3}{x}^3+\frac{4}{15}{x}^5-\frac{8}{315}{x}^7+\cdots$

The obtained series is same as the one obtained in part (b).

This implies that the series of $\sin \left(2x\right)$ is the same as the series obtained in part (b).

Hence, the obtained series in part (b) is the series for $\sin \left(2x\right)$ .