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Find the largest possible value of

$$\sin(a_1)\cos(a_2) + \sin(a_2)\cos(a_3) + \cdots + \sin(a_{2014})\cos(a_1)$$

Since the range of the $\sin$ and $\cos$ function is between $1$ and $-1$, shouldn't the answer be $2014$?

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    $\begingroup$ You cannot have $\sin (a_1)=\cos (a_1)=1$ so the maximum is not 2014. $\endgroup$ Aug 8, 2018 at 8:24
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    $\begingroup$ Fairly obviously it won't exceed $2014$ but it does not necessarily achieve it either. $\endgroup$
    – badjohn
    Aug 8, 2018 at 8:24

2 Answers 2

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Let $a_{2015}=a_1.$

Thus, by AM-GM $$\sum_{k=1}^{2014}\sin{a_k}\cos{a_{k+1}} \leq\sum_{k=1}^{2014}|\sin{a_k}||\cos{a_{k+1}}| \le$$ $$\leq\sum_{k=1}^{2014}\frac{\sin^2a_k+\cos^2a_{k+1}}{2}=\sum_{k=1}^{2014}\frac{\sin^2a_k+\cos^2a_k}{2}=\frac{2014}{2}=1007.$$

The equality occurs for $a_i=45^{\circ},$ which says that $1007$ is a maximal value.

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  • $\begingroup$ You mention AM-GM, which makes me expect to see a big sum turn into a big product at some point. But it never happens. Something clever is clearly going on in the third inequality -- but what? Could you expand this just a bit more? $\endgroup$ Aug 8, 2018 at 13:19
  • $\begingroup$ @Daniel Wagner I used $ab\leq\frac{a^2+b^2}{2}.$ $\endgroup$ Aug 8, 2018 at 13:32
  • $\begingroup$ Ah! Very simple -- I kept trying to figure out where the calculation of a 2014th root went. Thanks for the clarification. $\endgroup$ Aug 8, 2018 at 13:35
  • $\begingroup$ You are welcome! $\endgroup$ Aug 8, 2018 at 13:37
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From

$$ f_n(a) = \sum_{k=1}^n \sin a_k \cos a_{k+1} $$

with $a_{n+1} = a_1$

the stationary points are located at the solutions for

$$ \frac{\partial }{\partial a_k}f_n(a) = -\sin a_{k-1}\sin a_k + \cos a_k \cos a_{k+1} = 0 $$

and then

$$ \tan a_n\tan a_{n-1}\cdots\tan a_{2} = \cot a_1 $$

or

$$ \tan a_n\tan a_{n-1}\cdots\tan a_{2}\tan a_1 = 1 $$

or

$$ \prod_k\sin a_k = \prod_k\cos a_k $$

which is obtained for $a_k = \frac{\pi}{4}$ when

$$ f_n(a) = \frac n2 $$

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  • $\begingroup$ Nice answer. The calculus is easier than it looks at first. Is it obvious that $a_k=\frac{\pi}{4}$ is the absolute maximum? The $\tan a_n\tan a_{n-1}\cdots\tan a_{2}\tan a_1 = 1$ may have many solutions but I guess they are not all absolute maximum... $\endgroup$
    – Taladris
    Aug 8, 2018 at 13:49
  • $\begingroup$ @Taladris There are many relative extrema. For instance for $n = 3$ we have: $\left\{-\frac{3}{2},\frac{3}{2},-\sqrt{2},\sqrt{2}\right\}$ but with some additional considerations like symmetry... $\endgroup$
    – Cesareo
    Aug 8, 2018 at 14:44

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