Closed form of \(\displaystyle{\sum_{{{n}={1}}}^{{N}}}{{\cos}^{{2}}{\left({n}{x}\right)}}\) From the Lagrange identity

Elisha Frost

Elisha Frost

Answered question

2022-03-19

Closed form of n=1Ncos2(nx)
From the Lagrange identity we have:
n=0Ncos(nx)=12+sin[(2N+1)x2]2sinx2
Mathematica tells me that it is:
n=1Ncos2(nx)=N2+sinNxcos(Nx+x)2sinx
I actually want to see the process. If you could include the derivation of the original identity, that would be appreciated!
Also, am I correct to say that:
k=0ncos(kx)=k=1ncos(kx)+1

Answer & Explanation

Brennan Summers

Brennan Summers

Beginner2022-03-20Added 5 answers

Hint 1: (Telescoping)
2sinx2n=0Ncos(nx)=
=n=0Nsin(nx+12x)n=0Nsin(nx12x)
Hint 2:
cos2(nx)=12+12cos(2nx)

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