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I'm trying to solve a problem related to the Gaussian quadrature.

The first step of the problem is to prove that the following claim holds:

enter image description here

I was able to prove the $\theta = \pi k$ part very easily by using some simple trigonometric identities, but I got stuck trying to prove the rest for $\theta \neq \pi k$.

The question suggested using proof by induction on the sum of the cosine function, so I followed it and got to the point where I get the following equality:

$sin(2n\theta))/(2sin\theta) - 2sin(2n\theta)sin\theta$

The left part of what I got is what I intended to get eventually, but the part on the right does not go to $0$ because of the way $\theta$ is configured.

Could somebody please help me find my mistake in this?

Thanks!!!

RandyMarsh
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1 Answers1

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You should have $$ \sum_{j=0}^n\cos((2j+1)\theta)=\frac{\sin(2n\theta)}{2\sin\theta}+\cos((2n+1)\theta) =\frac{\sin(2n\theta)+2\cos((2n+1)\theta)\sin\theta}{2\sin\theta} $$ then $2\cos((2n+1)\theta)\sin\theta=\sin((2n+2)\theta)-\sin(2n\theta)$.

user10354138
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