Question

The Fourier cosine series for an even function f(x) given by $f\left(x\right)=a_0+{\sum }_{n=1}^{\infty }{a}_n\cos n\left(x\right)$ The value of the coefficient $a_2$ for the function $f\left(x\right)={\cos }^2\left(x\right)in[0,\pi ]$ is

• A. -0.5
• B. 0.0
• C. 0.5
• D. 1.0

Answer 1

The correct option is C 0.5

Fourier cosine series is
$f\left(x\right)=\frac{a_0}{2}+\sum a_n\cos n\left(x\right)....\left(i\right)$
where $a_n=\frac{1}{\pi }{\int }_0^{\pi }f\left(x\right)\cos nxdx$

Now by (i) we have
$f\left(x\right)=\frac{a_0}{2}+a_1\cos x+a_2\cos 2x+a_3\cos 3x+...$
${\cos }^{2}x=\frac{a_0}{2}+a_1\cos x+a_{2}cos2x+a_3\cos 3x+....$
$\frac{1}{2}+\frac{\cos 2x}{2}=\frac{a_0}{2}+a_1\cos x+a_2\cos 2x+...$
$\Rightarrow a_0=1,a_1=0,a_2=\frac{1}{2},a_3=0$ and so on...