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I have to prove the following trigonometric identities. However, since I can't prove them, I am starting to think they are not correctly stated.

Problem 1:

$$\cos\frac{2π}{3}+\cos\frac{4π}{9}+\cos\frac{6π}{3}+\cos\frac{8π}{9}=-\frac{1}{2}$$

Problem 2:

$$\frac{\cot^2\frac{α}{2}-\cot\frac{3α}{2}}{\cos^2\frac{α}{2}\cosα(1+\cot^2\frac{3α}{2})}=8$$

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  • $\begingroup$ I suppose that the first problem is $$\cos\frac{2π}{9}+\cos\frac{4π}{9}+\cos\frac{6π}{9}+\cos\frac{8π}{9}=-\frac{1}{2}$$ Typo's probably. The second is not true; typo's again ! $\endgroup$ – Claude Leibovici Mar 29 '15 at 13:29
  • $\begingroup$ @parkhyeyoo, See math.stackexchange.com/questions/117114/… $\endgroup$ – lab bhattacharjee Mar 29 '15 at 18:11
  • $\begingroup$ Why the second is not correct? Can you exaplain?? $\endgroup$ – parkhyeyoo Apr 17 '15 at 16:10
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The first formula is correct, if we assume it is a typo.

Problem 1:

$$\cos\frac{2π}{3}+\cos\frac{4π}{3}+\cos\frac{6π}{3}+\cos\frac{8π}{3}=-\frac{1}{2}$$

Sum of $n$ angles in $ AP == \dfrac{\sin n \beta/2}{\sin \beta/2} \cos ( Average Angle)$. Here angle is in degrees.

$$ \dfrac{\sin(4\cdot 120/2)}{\sin (120/2)}\cos{300} = -\frac{1}2 $$

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