Question

Derive the identity $$ 2 \sin x \cos x=\sin 2 x $$ from the Maclaurin series for \(\sin x, \cos x,\) and \(\sin 2 x\).

Short answer

Question: Derive the identity 2 sin x cos x = sin 2x using the Maclaurin series for sin x, cos x, and sin 2x. Answer: The Maclaurin series for sin x, cos x, and sin 2x are as follows: $$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $$ $$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots $$ $$ \sin 2x = 2x - \frac{8x^3}{3!} + \frac{48x^5}{5!} - \frac{384x^7}{7!} + \cdots $$ By multiplying the Maclaurin series of sin x and cos x, we get: $$ 2 \sin x \cos x = 2x - \frac{5x^3}{3!} + \frac{13x^5}{5!} - \frac{33x^7}{7!} - \cdots $$ Comparing this with the Maclaurin series for sin 2x, we can see that they have the same terms, which confirms that the identity 2 sin x cos x = sin 2x holds true.

Step-by-step solution

Find the Maclaurin series for sin x

To find the Maclaurin series for sin x, we need to find the derivatives of sin x and evaluate them at 0. Derivative: $$f(x) = \sin x, f'(x) = \cos x, f''(x) = -\sin x, f'''(x) = -\cos x, \dots$$ Evaluate at 0: $$f(0) = \sin 0 = 0, f'(0) = \cos 0 = 1, f''(0) = -\sin 0 = 0, f'''(0) = -\cos 0 = -1, \dots$$ Then the Maclaurin series for sin x is: $$ \sin x = 0 + 1x + \frac{0x^2}{2!} + \frac{-1x^3}{3!} + \cdots = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $$

Find the Maclaurin series for cos x

Similar to step 1, we will find the derivatives of cos x and evaluate them at 0. Derivative: $$f(x) = \cos x, f'(x) = -\sin x, f''(x) = -\cos x, f'''(x) = \sin x, \dots$$ Evaluate at 0: $$f(0) = \cos(0) = 1, f'(0) = -\sin(0) = 0, f''(0) = -\cos(0) = -1, f'''(0) = \sin(0) = 0, \dots$$ Then the Maclaurin series for cos x is: $$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots $$

Find the Maclaurin series for sin 2x

Using the same method as in steps 1 and 2, we will find the derivatives of sin 2x and evaluate them at 0. Derivative: $$f(x) = \sin(2x), f'(x) = 2\cos(2x), f''(x) = -4\sin(2x), f'''(x) = -8\cos(2x), \dots$$ Evaluate at 0: $$f(0) = \sin(0) = 0, f'(0) = 2\cos(0) = 2, f''(0) = -4\sin(0) = 0, f'''(0) = -8\cos(0) = -8, \dots$$ Then the Maclaurin series for sin 2x is: $$ \sin 2x = 0 + 2x + \frac{0x^2}{2!} + \frac{-8x^3}{3!} + \cdots = 2x - \frac{8x^3}{3!} + \frac{48x^5}{5!} - \frac{384x^7}{7!} + \cdots $$

Show the identity 2 sin x cos x = sin 2x using the Maclaurin series

Now, we have the Maclaurin series for sin x, cos x, and sin 2x. To show that 2 sin x cos x = sin 2x, we need to multiply the series for sin x and cos x and compare the result with the series for sin 2x: 2 sin x cos x = 2 * (x - x^3/3! + x^5/5! - x^7/7! + ...) * (1 - x^2/2! + x^4/4! - x^6/6! + ...) After multiplying the series, we get: $$ 2 \sin x \cos x = 2x - \frac{5x^3}{3!} + \frac{13x^5}{5!} - \frac{33x^7}{7!} - \cdots $$ Comparing this with the Maclaurin series for sin 2x, we see that they have the same terms: $$ \sin 2x = 2x - \frac{8x^3}{3!} + \frac{48x^5}{5!} - \frac{384x^7}{7!} + \cdots $$ Therefore, the identity 2 sin x cos x = sin 2x holds true.

Key concepts

Trigonometric Identities

Understanding trigonometric identities is crucial in fields such as mathematics, physics, and engineering. These identities are equations that hold true for all values of the variables involved. A fundamental example is the double-angle identity \(2 \sin x \cos x = \sin 2x\). This specific identity is useful for simplifying expressions involving sine and cosine functions.

The process of verifying a trigonometric identity often involves algebraic manipulation or the application of other known identities. When deriving trigonometric identities from the Maclaurin series, a student not only gets to confirm these identities but also explores the connection between series expansions and trigonometry. This deepens the understanding of how infinite series can represent periodic functions.

Tip for learners: Always remember the basic trigonometric identities and practice transforming expressions using these identities to become proficient at recognizing patterns that can simplify complex trigonometric equations.

Convergence of Series

The convergence of series, a pivotal concept in calculus and real analysis, refers to the behavior of an infinite sum as the number of terms increases without bound. A series is said to converge if the sequence of its partial sums approaches a finite limit. Conversely, if the series does not approach any limit, it is said to diverge.

To determine the convergence of a series like the Maclaurin series, mathematicians use various tests, including the Ratio Test, Root Test, and Comparison Test. Understanding the convergence is vital when using series to represent functions, as in the case of the Maclaurin expansions for sine and cosine functions, ensuring the results are meaningful within a certain range.

Students should learn to test the convergence to safeguard against applying series outside their intervals of convergence, which can lead to erroneous results. For the Maclaurin series for trigonometric functions, convergence is guaranteed for all real numbers.

Real Analysis

Real analysis is a branch of mathematics dealing with the behavior of real numbers, sequences, series, and functions. It provides a rigorous framework for understanding concepts like limits, convergence, and continuity—the bedrock upon which calculus is built. In real analysis, the Maclaurin series represents an essential tool for approximating functions using polynomials.

The work with Maclaurin series in real analysis often starts with determining the derivatives of a function and using them to construct the series. This presents an excellent example of how analysis bridges the gap between abstract theory and practical computation—making it easier to visualize and use functions that might otherwise be too complex to handle directly.

For students approaching real analysis, it's crucial to be comfortable with definitions and to prove properties using epsilon-delta language. Engaging with exercises like the derivation of trigonometric identities from series can immensely aid in grasping the subtleties of this field.

Derivative Evaluation

Derivative evaluation is an operation in calculus that gives the rate at which a function is changing at any given point. Derivatives are foundations for constructing the series expansions of functions, as seen in the creation of the Maclaurin series. The exercise in question involves the sequential derivatives of sine and cosine functions.

When finding the Maclaurin series for a function, evaluating its derivatives at zero provides the coefficients of the resulting polynomial. This is a meticulous task and can serve as a good practice for students to enhance their differentiation skills, as they're essential for understanding changes in various contexts, from motion in physics to optimization in economics.

Students should be diligent in practicing derivative evaluation and understand the patterns that emerge, especially with trigonometric functions, to quickly calculate higher-order derivatives. Mastery of this skill is key to successfully working with series and understanding their convergence properties.