Let f(x)=cos 5 x+A cos 4 x+B cos 3 x +C cos 2 | vedclass

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Question

(c)

We have,

$f(x)=\cos 5 x+A \cos 4 x+B \cos 3 x $

$+C \cos 2 x+D \cos x+E $

We know that,

$\cos x =\cos (2 \pi-x)  f(x) =f(2 \pi

Vedclass · Accepted Answer

depends on $B, D$, but independent of $A, C, E$

Vedclass · Answer

(c)

We have,

$f(x)=\cos 5 x+A \cos 4 x+B \cos 3 x $

$+C \cos 2 x+D \cos x+E $

We know that,

$\cos x =\cos (2 \pi-x)  f(x) =f(2 \pi-x)$

$[\because f(x)$ contains only cosines terms]

$f\left(\frac{\pi}{5}\right)=f\left(2 \pi-\frac{\pi}{5}\right)=f\left(\frac{9 \pi}{5}\right)$

Similarly, $f\left(\frac{2 \pi}{5}\right)=f\left(\frac{8 \pi}{5}\right), f\left(\frac{3 \pi}{5}\right)=f\left(\frac{7 \pi}{5}\right)$

$f\left(\frac{4 \pi}{5}\right)=f\left(\frac{6 \pi}{5}\right)$

Let $T=f(0)-f\left(\frac{\pi}{5}\right)+f\left(\frac{2 \pi}{5}\right)-f\left(\frac{3 \pi}{5}\right)$

$T=f(0)-2\left[f\left(\frac{\pi}{5}\right)+f\left(\frac{3 \pi}{5}\right)\right]$

$\quad+2\left[f\left(\frac{2 \pi}{5}\right)+f\left(\frac{4 \pi}{5}\right)\right]-f(\pi)$

Now, $f(0)=1+A+B+C+D+E$

$f(\pi)=-1+A-B+C-D+E$

$\because f(0)-f(\pi)=2(1+B+D)$

$f\left(\frac{\pi}{5}\right) +f\left(\frac{3 \pi}{5}\right)$

$=2\left(1+B \cos \frac{3 \pi}{5}+D \cos \frac{\pi}{5}\right)$

$f\left(\frac{2 \pi}{5}\right) +f\left(\frac{4 \pi}{5}\right)$

$=2\left(1+B \cos \frac{6 \pi}{5}+D \cos \frac{2 \pi}{5}\right)$

Clearly, $T$ contains only $B$ and $D$ terms.

Let $f(x)=\cos 5 x+A \cos 4 x+B \cos 3 x$ $+C \cos 2 x+D \cos x+E$, and

$T=f(0)-f\left(\frac{\pi}{5}\right)+f\left(\frac{2 \pi}{5}\right)-f\left(\frac{3 \pi}{5}\right)+\ldots+f\left(\frac{8 \pi}{5}\right)-f\left(\frac{9 \pi}{5}\right) \text {. }$Then, $T$

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Let $f(x)=\cos 5 x+A \cos 4 x+B \cos 3 x$ $+C \cos 2 x+D \cos x+E$, and $T=f(0)-f\left(\frac{\pi}{5}\right)+f\left(\frac{2 \pi}{5}\right)-f\left(\frac{3 \pi}{5}\right)+\ldots+f\left(\frac{8 \pi}{5}\right)-f\left(\frac{9 \pi}{5}\right) \text {. }$Then, $T$