Prove that: $$\dfrac{1}{\cos x+\cos {3x}} + \dfrac{1}{\cos x+ \cos {5x}}+\dots+\dfrac{1}{\cos x+ \cos {(2n+1)x}} \\= \frac{1}{2}\csc x \,[ \tan{(n+1)x}-\tan{x}]$$

I tried to prove this using the regular formulas. But failed. Please help me.


$$\sum_{k=1}^n\frac{1}{\cos{x}+\cos(2k+1)x}=\sum_{k=1}^n\frac{1}{2\cos{kx}\cos(k+1)x}=$$ $$=\frac{1}{2\sin{x}}\sum_{k=1}^n\left(\tan(k+1)x-\tan kx\right)=\frac{1}{2\sin{x}}\left(\tan(n+1)x-\tan x\right)$$ and we are done!

I used the following reasoning. $$\tan\alpha-\tan\beta=\frac{\sin\alpha}{\cos\alpha}-\frac{\sin\beta}{\cos\beta}=\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\cos\alpha\cos\beta}=\frac{\sin(\alpha-\beta)}{\cos\alpha\cos\beta}.$$

  • $\begingroup$ Seems to me it might take a few steps to get from the second sum to the third expression. $\endgroup$ – Gerry Myerson Nov 18 '17 at 12:35
  • 1
    $\begingroup$ I added something. See now. $\endgroup$ – Michael Rozenberg Nov 18 '17 at 12:39

In the case $n=0$ there is no summand on the left-hand side and the right-hand side is $0$, so the base case of the induction holds.

Suppose it holds for $n-1$; then the left-hand side with $n$ can be written, by the induction hypothesis, $$ \frac{1}{2\sin x}\bigl(\tan nx-\tan x\bigr)+\frac{1}{\cos x+\cos(2n+1)x} $$ and you want to prove this equals $$ \frac{1}{2\sin x}\bigl(\tan(n+1)x-\tan x\bigr) $$ which is equivalent to $$ \frac{\tan nx}{2\sin x}+\frac{1}{\cos x+\cos(2n+1)x}= \frac{\tan(n+1)x}{2\sin x} $$ By the sum-to-product formulas, this becomes $$ \frac{\sin nx}{2\sin x\cos nx}+\frac{1}{2\cos nx\cos(n+1)x}= \frac{\sin(n+1)x}{2\sin x\cos(n+1)x} $$ Reduce to the same denominator and conclude the equality holds.

The relation $$\sin nx\cos(n+1)x+\sin x=\sin(n+1)x\cos nx$$ is true, because it is equivalent to $$\sin x=\sin(n+1)x\cos nx-\cos(n+1)x\sin nx$$


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