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I'm reading a solution right now, and I am stuck.

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

How could these two equations be equal.

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Do you have your own thought? –  Shuchang Sep 15 '13 at 1:40
    
@Shuchang I posted this because I don't have any thought. –  therexists Sep 15 '13 at 1:51
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@therexists you still should post what you have tried, other wise you might end up with hints like: "use trigonometrical properties", and also, whay did you tag this in abstract algebra? –  Ana Galois Sep 15 '13 at 2:08
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@Dre You edited my post wrongly.... sin(theta/2) should be in the denominator of LHS. –  therexists Sep 15 '13 at 2:12
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I also think $\cos(n+1/2)\theta$ in the LHS should be actually $\cos\{(n+1)\theta/2\}$. Because then the RHS will be exactly equal to the LHS. –  Abishanka Saha Sep 15 '13 at 3:03

1 Answer 1

Not sure about the accuracy/format of the problem as

using $\displaystyle \cos C-\cos D=2\sin\frac{C+D}2\sin\frac{D-C}2$

$$\cos\frac\theta2-\cos\left(n+\frac12\right)\theta=2\sin\frac{n\theta}2\sin\frac{(n+1)\theta}2 $$

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