# How find this $\sum_{i=0}^{5}\frac{1}{2+\cos{\left(x+\frac{i\pi}{3}\right)}}\cdot \frac{1}{2+\cos{\left(x+\frac{(i+1)\pi}{3}\right)}}$

Find this follow function $f(x)$ range ,where $x\in R$, $$f(x)=\sum_{i=0}^{5}\dfrac{1}{2+\cos{\left(x+\dfrac{i\pi}{3}\right)}}\cdot \dfrac{1}{2+\cos{\left(x+\dfrac{(i+1)\pi}{3}\right)}}$$ or $$f(x)=\dfrac{1}{2+\cos{x}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{2\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{2\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{3\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{3\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{4\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{4\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{5\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{5\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{6\pi}{3}\right)}}$$

I think $f(x)$ have simple form,becasue this is exam problem

I think maybe can use $$f(2\pi-x)+f(x)=?$$ or maybe have use this $$\cos{x}\cos{y}=\dfrac{1}{2}[\cos{(x-y)}+\cos{(x+y)}]$$ I use this two idea all solve this problem,

Thank you

• One observation : If $\displaystyle\cos6A=\cos6x, 6A=2n\pi\pm6x$ where $n$ is any integer $$\displaystyle A=\frac{n\pi}3+x\text{ where }0\le n\le5\ \ \ \ (1)$$ Again, $$\displaystyle\cos6A=2\cos^23A-1=2(4\cos^3A-3\cos A)^2-1=32\cos^6A-48\cos^4A+18\cos^2A-1$$ So, $(1)$ is the set of roots of $$\displaystyle32c^6-48c^4+18c^2-(1+\cos6A)=0$$ – lab bhattacharjee Jan 12 '14 at 10:45
• Mathematica gives $\frac{2880}{1351-\cos(6x)}$. – Jack D'Aurizio Jan 13 '14 at 3:35

## 2 Answers

$$\text{Let }y_r=\left(2+\cos\left(x+\frac{r\pi}3\right)\right)\left(2+\cos\left(x+\dfrac{(r+1)\pi}3\right)\right)$$

$$y_r=4+2\left[\cos\left(x+\frac{r\pi}3\right)+\cos\left(x+\dfrac{(r+1)\pi}3\right)\right]+\cos\left(x+\frac{r\pi}3\right)\cos\left(x+\dfrac{(r+1)\pi}3\right)$$

Applying $2\cos A\cos B$ and $\cos C+\cos D$ formula,

$$2y_r=8+8\cos\left(x+\frac{(2r+1)\pi}6\right)\cos\frac\pi6+\left[\cos\frac\pi3+\cos\left(2x+\frac{(2r+1)\pi}3\right)\right]$$

$$=8+4\sqrt3c+\frac12+2c^2-1\text{ where }c=\cos\left(x+\frac{(2r+1)\pi}6\right)$$

$$\implies 4y_r=4c^2+8\sqrt3c+15\ \ \ \ (1)$$

So, we need to find $\displaystyle\sum_{r=0}^5\frac1{y_r}=\frac{\sum_{\text{cyc}} 5 y_r\text{-s at a time}}{\prod _{r=0}^5 y_r}$ which can be easily managed by Vieta's formula

Now, $\displaystyle \cos6\left(x+\frac{(2r+1)\pi}6\right)=\cos\left(6x+\overline{2r+1}\pi\right)=-\cos6x$

So if $\displaystyle \cos6A=-\cos6x=\cos(\pi+6x),$

$\displaystyle\implies6A=2n\pi\pm(\pi+6x)$ where $n$ is any integer

$\displaystyle\implies A=x+\frac{(2n+1)\pi}6$ where $0\le n\le 5\ \ \ \ (2)$

So, as I've explained in the comment,

$(2)$ is the set of roots of $\displaystyle32c^6−48c^4+18c^2+\cos6A-1=0\ \ \ \ (3)$

Now from $\displaystyle (1),c^2=\frac{4y_r-8\sqrt3c-15}{15}\ \ \ \ (4)$

Squaring we get $\displaystyle c^4=\frac{16y_r^2+192c^2+225-(64\sqrt3y_r)c-120y_r+240\sqrt3c}{225} \ \ \ \ (5)$

Put the value of $c^2$ from $(4)$ in $(5)$

Now multiply $(4),(5)$ to get the value of $c^6$

Put the values of $c^6,c^4,c^2$ in $(3)$ to find $c$ in terms of $y_r$

Now put the values of $c,c^2$ in $(4)$ to form a Sextic equation in $y_r$ where we need to apply Vieta's formula mentioned above

HINT

I should start simplifying the sum of the first and the sixth terms (result = A), then the sum of the second and fifth terms (result = B), then the sum of the third and fourth terms (result = C). Now, I should simplify A + B (result = D) and finally simplify C + D.

You will arrive to a surprizingly simple result.

• I believe because $\frac{2880}{1351-\cos(6x)}$ is not surprisingly simple. – Jack D'Aurizio Jan 13 '14 at 3:21