Which of the constants A,B,C,D does T depend on? Let $f(x)=cos(5x)+Acos(4x)+Bcos(3x)+Ccos(2x)+Dcos(x)+E$ and $T=f(0)-f(\pi/5)+f(2\pi/5)-f(3\pi/5)+..-f(9\pi/5)$.Then out of A,B,C,D which does T depend on?
Hints please!
P.S:KVPY 2011 question
 A: HINT: You can write $\frac{9\pi}{5}=2\pi-\frac{\pi}{5}$,$\frac{8\pi}{5}=2\pi-\frac{2\pi}{5}$,$\frac{7\pi}{5}=2\pi-\frac{3\pi}{5}$,$\frac{6\pi}{5}=2\pi-\frac{4\pi}{5}$ and see if it helps.
A: There might be more elegant ways, but here's the 'dumb' approach: Start by factoring out the variables and you might start to see a pattern pretty soon.
$$\begin{align}f(x)=&\cos(0)-\cos(\pi)+\cos(2\pi)-\ldots\\
&+A\left(\cos(0)-\cos\left(\frac{4}{5}\pi\right)+\cos\left(\frac{4}{5}2\pi\right)-\cos\left(\frac{4}{5}3\pi\right)+\ldots\right)\\
&+B\left(\cos(0)-\cos\left(\frac{3}{5}\pi\right)+\cos\left(\frac{3}{5}2\pi\right)-\cos\left(\frac{3}{5}3\pi\right)+\ldots\right)\\
&+C(\ldots)\\
&+D(\ldots)\end{align}$$
Alternatively written in a more compact way:
$$\begin{align}f(x)=&\sum_{k=0}^9 (-1)^k \cos(k\pi)\\&+A\sum_{k=0}^9 (-1)^k\cos\left(\frac{4}{5}k\pi\right)\\&+B\sum_{k=0}^9(-1)^k\cos\left(\frac{3}{5}k\pi\right)\\&+C\sum_{k=0}^9(-1)^k\cos\left(\frac{2}{5}k\pi\right)\\&+D\sum_{k=0}^9(-1)^k\cos\left(\frac{1}{5}k\pi\right)\end{align}.$$
Now you need some knowledge of how $\cos$ behaves, count angles and you should be done.
A: Given $$f(x) = \cos 5x+A\cos 4x+B\cos 3x+C\cos 2x+d\cos x+E$$
Using $$f(\pi-x) = f(\pi+x)\Rightarrow f(x) = f(2\pi-x).$$
So we get $$\displaystyle f\left(\frac{\pi}{5}\right)=f\left(\frac{9\pi}{5}\right)\;\;\;,\;\;\; \displaystyle f\left(\frac{2\pi}{5}\right)=f\left(\frac{8\pi}{5}\right)$$
and $$\displaystyle f\left(\frac{3\pi}{5}\right)=f\left(\frac{7\pi}{5}\right)\;\;\;,\;\;\; \displaystyle f\left(\frac{4\pi}{5}\right)=f\left(\frac{6\pi}{5}\right)$$
Now $$T=f(0)-2\left[\displaystyle f\left(\frac{\pi}{5}\right)+f\left(\frac{3\pi}{5}\right)\right]+2\left[\displaystyle f\left(\frac{2\pi}{5}\right)+f\left(\frac{4\pi}{5}\right)\right]-f(\pi)$$
Now $$f(0)-f(\pi) = 2\left[...\right]$$
and $$f\left(\frac{\pi}{5}\right)+f\left(\frac{3\pi}{5}\right) = 2\left[...\right]$$
and $$f\left(\frac{2\pi}{5}\right)+f\left(\frac{4\pi}{5}\right) = 2\left[...\right]$$
