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I am stuck understanding why $\frac{\cos(n\theta)-\cos(n+1)\theta)}{2-2\cos\theta}=\frac{\sin(n+1/2)\theta}{2\sin(\theta/2)}$. I have tried $2-2\cos \theta=4\sin^2(\theta/2)$ and standard trig identities to maniplate the expression but I don't get it.

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Hint. One may recall that $$ \cos \alpha -\cos \beta=2\sin \frac{\alpha+\beta}2\:\sin \frac{\beta-\alpha}2. $$

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  • $\begingroup$ The form $\;-2\sin \dfrac{\alpha+\beta}2\:\sin \dfrac{\alpha-\beta}2$ is better to easily memorise it. $\endgroup$ – Bernard Feb 4 '18 at 17:59
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hint

put $n+1/2=m$ then

$$\cos ((n+1)t)=\cos ((n+1/2)t+t/2) $$

$$=\cos (mt)\cos (t/2)-\sin (mt)\sin (t/2) $$

and

$$\cos (nt)=\cos (mt-t/2)=$$ $$\cos (mt)\cos (t/2)+\sin (mt)\sin (t/2) $$

done.

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use that $$\cos(n\theta)-\cos((n+1)\theta)=-2 \sin \left(\frac{\theta n}{2}-\frac{1}{2} \theta (n+1)\right) \sin \left(\frac{\theta n}{2}+\frac{1}{2} \theta (n+1)\right)$$

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