Question: 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

Answer: 
If you know $AD = 12$ already, then $DF = 12/3$. This is because $F$ is the centroid of $\triangle ABC$, and the centroid of a triangle divides each median into $1:2$.
F is the intersection of all three medians of the triangle. You can know the length of AD (which is a median) with <http://en.wikipedia.org/wiki/Apollonius%27_theorem> if you have the length of the 3 sides (in ABC), or <http://en.wikipedia.org/wiki/Law_of_cosines> if you have 2 angles and 1 length (in ACD), which you seem to have.
After that, it is a result that the length of DF is equal to the third of the length of AD.

EDIT : I just saw that ABC is rectangle. That simplifies the problem of knowing the length of AD if you know that if ABC is rectangle in A, BC is the diameter of a circle passing by A, B and C. What relation can you deduce between AD and BD ?