Question

Find cosine series for f(x)= sin x in (0,π)​.​

Answer 1

Step-by-step explanation:

What is the Fourier cosine series expansion of f(x) =sin(x) on 0≤x≤π?

Using my previous answer:

We wish to set c0=1L∫L0f(t)dt and cn=2L∫L0f(t)cos(πntL)dt for n⩾1 , and then have f(t)=∑∞n=0cncos(πntL) .

Since L=π here, we have

c0=1π∫π0f(t)dt

cn=2π∫π0f(t)cos(nt)dt for n⩾1

and then f(t)=∑∞n=0cncos(nt) .

With that, c0=1π∫π0sintdt=1π(−cost)∣∣π0=2π . Moreover, for n⩾1 ,

cn=2π∫π0f(t)cos(nt)dt=2π∫π0sintcos(nt)dt=2π∫π012(sin((n+1)t)−sin((n−1)t))dt=1π∫π0(sin((n+1)t)−sin((n−1)t))dt