Question

By comparing the first four terms, show that the Maclaurin series for \(\sin ^{2} x\) can be found (a) by squaring the Maclaurin series for \(\sin x,\) (b) by using the identity \(\sin ^{2} x=\frac{1-\cos 2 x}{2},\) or \((c)\) by computing the coefficients using the definition.

Short answer

Question: Show that the first four terms of the Maclaurin series for \(\sin^2 x\) are the same when obtained by squaring the Maclaurin series of \(\sin x\), using the identity \(\sin ^{2} x=\frac{1-\cos 2 x}{2}\), and computing the coefficients directly using the definition. Answer: All three methods (squaring the Maclaurin series for \(\sin x\), using the identity, and computing the coefficients directly using the definition) result in the following first four terms of the Maclaurin series for \(\sin^2 x\): \(\sin^2 x = x^2 - 2x^2\frac{x^2}{3!} + (\frac{x^2}{3!})^2 + \cdots\)

Step-by-step solution

Determine the Maclaurin series for \(\sin x\)

Recall that the Maclaurin series for \(\sin x\) is given by: \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\)

Square the Maclaurin series for \(\sin x\)

To find the Maclaurin series for \(\sin^2 x\), we square the Maclaurin series for \(\sin x\): \(\sin^2 x = (x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots)^2\) Expanding only the first four terms, we get: \(\sin^2 x = x^2 - 2x^2\frac{x^2}{3!} + (\frac{x^2}{3!})^2 + \cdots\) (b) Using the identity \(\sin ^{2} x=\frac{1-\cos 2 x}{2}\)

Determine the Maclaurin series for \(\cos 2x\)

Recall that the Maclaurin series for \(\cos x\) is given by: \(\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\) Thus, the Maclaurin series for \(\cos 2x\) is: \(\cos 2x = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \cdots\)

Apply the identity to find the Maclaurin series for \(\sin^2 x\)

Using the identity, we get: \(\sin^2 x = \frac{1 - \cos 2x}{2} = \frac{1}{2} - \frac{1}{2}\cos 2x\) Substitute the Maclaurin series for \(\cos 2x\): \(\sin^2 x = \frac{1}{2} - \frac{1}{2}(1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \cdots) = x^2 - 2x^2\frac{x^2}{3!} + (\frac{x^2}{3!})^2 + \cdots\) (c) Computing the coefficients using the definition

Find the general form of the Maclaurin series for \(\sin^2 x\)

The general form of the Maclaurin series for any function \(f(x)\) is given by: \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\) For \(\sin^2 x\), we need to find its derivatives and evaluate them at \(0\).

Compute the derivatives and substitute the values

We have: \(f(x) = \sin^2 x\) \(f'(x) = 2\sin x \cos x\) \(f''(x) = 2(\cos^2 x - \sin^2 x)\) \(f'''(x) = -8\sin x \cos x\) \(f''''(x) = 8(2\sin^2 x - \cos^2 x)\) Now evaluate the derivatives at \(0\), keeping the first 4 derivatives: \(f(0) = \sin^2 0 = 0\) \(f'(0) = 2\sin 0 \cos 0 = 0\) \(f''(0) = 2(\cos^2 0 - \sin^2 0) = 2\) \(f'''(0) = -8\sin 0 \cos 0 = 0\) \(f''''(0) = 8(2\sin^2 0 - \cos^2 0) = 0\) Substitute the values in the general form of the Maclaurin series: \(\sin^2 x = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = 0 + 0x + \frac{2}{2!}x^2 + 0x^3 + \cdots = x^2 - 2x^2\frac{x^2}{3!} + (\frac{x^2}{3!})^2 + \cdots\) This confirms that all three methods give the same first four terms for the Maclaurin series of \(\sin^2 x\).

Key concepts

Sin Function

The sine function, denoted as \(\sin x\), is a fundamental trigonometric function that represents the y-coordinate of a unit circle at a given angle \(x\) measured from the horizontal axis. It's periodic with a period of \(2\pi\), meaning it repeats its pattern every \(2\pi\) radians.
In terms of Taylor or Maclaurin series expansions, \(\sin x\) is expressed as an infinite series:

• \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\)

This series is centered around \(x = 0\), known as a Maclaurin series. Each term involves increasing odd powers of \(x\), alternating signs, and factorial in the denominators.
This expansion is particularly useful for approximations in scientific calculations when \(x\) is small. Understanding this series is key when dealing with complex expressions like \(\sin^2 x\), as we can manipulate it using different mathematical techniques to find other useful series expressions.

Cosine Function

The cosine function, represented as \(\cos x\), measures the x-coordinate of a point on a unit circle corresponding to an angle \(x\). Like the sine function, \(\cos x\) is periodic with a period of \(2\pi\). Its series expansion provides a tool for calculating the function's values for small \(x\) and is defined by:

• \(\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\)

This expansion is a Maclaurin series, involving only even powers of \(x\) with alternating signs and factorial divisions.
By substituting \(2x\) into the cosine series, we can find \(\cos 2x\), a crucial step in further expansions for expressions like \(\sin^2 x\). The identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\) elegantly ties \(\sin^2 x\) to the cosine function, allowing us to shift between these trigonometric identities and related series expansions seamlessly.

Series Expansion

A series expansion is a way to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For Maclaurin series, the expansion occurs around the point \(x = 0\). Each term contributes more precision to the approximation of the function.
These expansions are particularly powerful in calculus and analysis, allowing for:

• Approximation of complex functions with simpler polynomial forms
• Simplifying calculations, especially for small values of \(x\)
• Analyzing function behavior through its polynomial form

In our example, the series for \(\sin x\) and \(\cos x\) provide the basis for constructing the series for \(\sin^2 x\). By either directly squaring its components, using trigonometric identities, or calculating coefficients from derivatives, we can find consistent series expressions. Each method provides valuable insights into the structure and interrelations of trigonometric functions.

Derivative Calculation

Derivative calculation is fundamental to constructing series expansions. The derivative of a function measures its rate of change at any point. For Maclaurin series, derivatives up to the desired term order are evaluated at \(x = 0\).
To expand \(\sin^2 x\) using its definition, we compute derivatives like:

• \(f(x) = \sin^2 x\)
• \(f'(x) = 2\sin x \cos x\)
• \(f''(x) = 2(\cos^2 x - \sin^2 x)\)
• \(f'''(x) = -8\sin x \cos x\)
• \(f''''(x) = 8(2\sin^2 x - \cos^2 x)\)

After evaluating these derivatives at \(x = 0\), they provide coefficients for the respective powers of \(x\) in the series expansion. This systematic calculation can be repetitive yet unveils the beauty of how functions can be deconstructed and approximated through polynomial forms.