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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

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  • $\begingroup$ Yes @hardmath...thats why i didnt ask about E $\endgroup$
    – user220382
    Oct 26, 2015 at 17:50
  • $\begingroup$ @hardmath easier said than done atleast for me... $\endgroup$
    – user220382
    Oct 26, 2015 at 17:55
  • $\begingroup$ Do the coefficients "one at a time". In other words, whether $T$ has terms containing $A$ can be worked out just by collecting those terms in $T$. $\endgroup$
    – hardmath
    Oct 26, 2015 at 18:00
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    $\begingroup$ Would it be tedious if doing one such "collecting terms" allows you to do all of them? $\endgroup$
    – hardmath
    Oct 26, 2015 at 18:15
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    $\begingroup$ I looked up the question paper and the options given make it really easy, you have to only check for $B$. $\endgroup$
    – najayaz
    Oct 28, 2015 at 9:30

3 Answers 3

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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 $$f\left(\frac{2\pi}{5}\right)+f\left(\frac{4\pi}{5}\right) = 2\left[...\right]$$

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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.

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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.

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