Fourier Cosine Series Coefficient Calculation

The problem asks us to find the value of the coefficient \(a_2\) for the function \(f(x) = \cos^2(x)\) in the interval \([0, \pi]\), given its Fourier cosine series.

Understanding Fourier Cosine Series

For an even function \(f(x)\) defined on the interval \([0, \pi]\), its Fourier cosine series is given by:

\[f(x) = {a_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos (nx)}\]

The coefficients \(a_0\) and \(a_n\) are determined using the following integral formulas:

• For the constant term \(a_0\): \[a_0 = \frac{1}{\pi} \int_0^\pi f(x) \, dx\]
• For the cosine coefficients \(a_n\) (where \(n = 1, 2, 3, \dots\)): \[a_n = \frac{2}{\pi} \int_0^\pi f(x) \cos(nx) \, dx\]

Given Function and Target Coefficient

We are given the function \(f(x) = \cos^2(x)\) and asked to find the value of the specific coefficient \(a_2\).

To find \(a_2\), we will use the formula for \(a_n\) with \(n=2\):

\[a_2 = \frac{2}{\pi} \int_0^\pi f(x) \cos(2x) \, dx\]

Substitute \(f(x) = \cos^2(x)\) into the formula:

\[a_2 = \frac{2}{\pi} \int_0^\pi \cos^2(x) \cos(2x) \, dx\]

Step-by-Step Calculation of \(a_2\)

To evaluate this integral, it's helpful to use trigonometric identities to simplify the integrand. We know the identity: \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\).

Applying this identity to \(\cos^2(x)\):

\[\cos^2(x) = \frac{1 + \cos(2x)}{2}\]

Now substitute this into the integral for \(a_2\):

\[a_2 = \frac{2}{\pi} \int_0^\pi \left( \frac{1 + \cos(2x)}{2} \right) \cos(2x) \, dx\]

Take the constant \(\frac{1}{2}\) out of the integral and multiply the terms inside:

\[a_2 = \frac{2}{\pi} \cdot \frac{1}{2} \int_0^\pi (1 + \cos(2x))\cos(2x) \, dx\]

\[a_2 = \frac{1}{\pi} \int_0^\pi (\cos(2x) + \cos^2(2x)) \, dx\]

We encounter \(\cos^2(2x)\) again. Use the same identity: \(\cos^2(A) = \frac{1 + \cos(2A)}{2}\). Here, \(A = 2x\), so \(2A = 4x\).

\[\cos^2(2x) = \frac{1 + \cos(4x)}{2}\]

Substitute this back into the integral:

\[a_2 = \frac{1}{\pi} \int_0^\pi \left( \cos(2x) + \frac{1 + \cos(4x)}{2} \right) \, dx\]

Separate the terms for integration:

\[a_2 = \frac{1}{\pi} \int_0^\pi \left( \cos(2x) + \frac{1}{2} + \frac{\cos(4x)}{2} \right) \, dx\]

Now, perform the integration:

\[a_2 = \frac{1}{\pi} \left[ \frac{\sin(2x)}{2} + \frac{x}{2} + \frac{\sin(4x)}{8} \right]_0^\pi\]

Evaluate the definite integral by plugging in the upper limit (\(\pi\)) and the lower limit (\(0\)):

• At the upper limit \(x = \pi\): \[\frac{\sin(2\pi)}{2} + \frac{\pi}{2} + \frac{\sin(4\pi)}{8} = \frac{0}{2} + \frac{\pi}{2} + \frac{0}{8} = \frac{\pi}{2}\]
• At the lower limit \(x = 0\): \[\frac{\sin(0)}{2} + \frac{0}{2} + \frac{\sin(0)}{8} = \frac{0}{2} + 0 + \frac{0}{8} = 0\]

Substitute these values back:

\[a_2 = \frac{1}{\pi} \left( \frac{\pi}{2} - 0 \right)\]

\[a_2 = \frac{1}{\pi} \cdot \frac{\pi}{2}\]

\[a_2 = \frac{1}{2}\]

Therefore, the value of the coefficient \(a_2\) is \(0.5\).