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Question

A series expansion for the function sin θ is

The correct answer is \(\theta - \frac{{{\theta ^3}}}{{3!}} + \frac{{{\theta^5}}}{{5!}} - \ldots\)

Understanding Series Expansions for Functions

A series expansion, like the Taylor or Maclaurin series, represents a function as an infinite sum of terms calculated from the values of its derivatives at a single point. The Maclaurin series is a special case of the Taylor series centered at zero (\(\theta = 0\)). For a function \(f(\theta)\) that has derivatives of all orders at \(\theta = 0\), the Maclaurin series is given by:

\(\displaystyle f(\theta) = f(0) + f'(0)\theta + \frac{f''(0)}{2!}\theta^2 + \frac{f'''(0)}{3!}\theta^3 + \ldots + \frac{f^{(n)}(0)}{n!}\theta^n + \ldots\)

Finding the Series Expansion for sin θ

To find the Maclaurin series for the function \(f(\theta) = \sin \theta\), we need to evaluate the function and its derivatives at \(\theta = 0\). Let's calculate the first few terms:

  • \(f(\theta) = \sin \theta \Rightarrow f(0) = \sin(0) = 0\)
  • \(f'(\theta) = \cos \theta \Rightarrow f'(0) = \cos(0) = 1\)
  • \(f''(\theta) = -\sin \theta \Rightarrow f''(0) = -\sin(0) = 0\)
  • \(f'''(\theta) = -\cos \theta \Rightarrow f'''(0) = -\cos(0) = -1\)
  • \(f^{(4)}(\theta) = \sin \theta \Rightarrow f^{(4)}(0) = \sin(0) = 0\)
  • \(f^{(5)}(\theta) = \cos \theta \Rightarrow f^{(5)}(0) = \cos(0) = 1\)

We can see a pattern emerging for the values of the derivatives at \(\theta = 0\): 0, 1, 0, -1, 0, 1, 0, -1, ... This pattern repeats every four terms.

Constructing the Maclaurin Series for sin θ

Now, substitute these values into the Maclaurin series formula:

\(\displaystyle \sin \theta = f(0) + f'(0)\theta + \frac{f''(0)}{2!}\theta^2 + \frac{f'''(0)}{3!}\theta^3 + \frac{f^{(4)}(0)}{4!}\theta^4 + \frac{f^{(5)}(0)}{5!}\theta^5 + \ldots\)

Substitute the calculated values:

\(\displaystyle \sin \theta = 0 + (1)\theta + \frac{0}{2!}\theta^2 + \frac{-1}{3!}\theta^3 + \frac{0}{4!}\theta^4 + \frac{1}{5!}\theta^5 + \ldots\)

Simplifying the terms, we get:

\(\displaystyle \sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \ldots\)

This is the series expansion for sin θ.

Comparing with Options

Let's compare the derived series expansion with the given options:

  • Option 1: \(1 - \frac{{\theta ^2}}{{2!}} + \frac{{\theta^4}}{{4!}} - \ldots\) (This is the series expansion for \(\cos \theta\))
  • Option 2: \(\theta - \frac{{\theta ^3}}{{3!}} + \frac{{\theta^5}}{{5!}} - \ldots\) (This matches our derived series)
  • Option 3: \(1 + \theta + \frac{{\theta ^2}}{{2!}} + \frac{{\theta^3}}{{3!}} + \ldots\) (This is the series expansion for \(e^{\theta}\))
  • Option 4: \(\theta + \frac{{\theta ^3}}{{3!}} + \frac{{\theta^5}}{{5!}} + \ldots\) (This is the series expansion for \(\sinh \theta\))

Therefore, the series expansion for sin θ is given by option 2.

Function Maclaurin Series Expansion
\(e^\theta\) \(1 + \theta + \frac{{\theta ^2}}{{2!}} + \frac{{\theta^3}}{{3!}} + \ldots\)
\(\sin \theta\) \(\theta - \frac{{\theta ^3}}{{3!}} + \frac{{\theta^5}}{{5!}} - \frac{{\theta^7}}{{7!}} + \ldots\)
\(\cos \theta\) \(1 - \frac{{\theta ^2}}{{2!}} + \frac{{\theta^4}}{{4!}} - \frac{{\theta^6}}{{6!}} + \ldots\)
\(\sinh \theta\) \(\theta + \frac{{\theta ^3}}{{3!}} + \frac{{\theta^5}}{{5!}} + \frac{{\theta^7}}{{7!}} + \ldots\)

Revision Table: Trigonometric Series Expansions

Here's a quick summary of common series expansions to remember:

Function Series Expansion
\(\sin \theta\) \(\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\theta^{2n+1} = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \ldots\)
\(\cos \theta\) \(\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\theta^{2n} = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \ldots\)
\(e^\theta\) \(\sum_{n=0}^{\infty} \frac{1}{n!}\theta^n = 1 + \theta + \frac{\theta^2}{2!} + \ldots\)

Additional Information: Properties of sin θ Series

The Maclaurin series for sin θ has several notable properties:

  • It includes only odd powers of \(\theta\). This is because sin θ is an odd function.
  • The signs of the terms alternate (+, -, +, -, ...).
  • The denominators are the factorials of the corresponding odd powers (3!, 5!, 7!, ...).
  • This series converges for all real values of \(\theta\).

Understanding the properties of the function (like being odd or even) can help recall or verify its series expansion.

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Important Questions from Mean Value Theorem

  1. If f is the derivative of some function on [a, b], then there exists a number c in (a, b) such that Integral of f with respect to x =

  2. Which condition is not required in checking for Taylor's theorem?

  3. What is the interval of Taylor series expansion of tan(x)?
  4. According to the Mean Value Theorem, for a continuous function f(x) in the interval [a, b], there exists a value ξ in this interval such that \(\mathop \smallint \limits_a^b f\left( x \right)dx =\)

  5. Let f(x) = x2 - 2x + 2 be a continuous function defined on x ∈ [1, 3]. The point x at which the tangent of f(x) becomes parallel to the straight line joining f(1) and f(3) is

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