Evaluate $(\sum_{k=1}^{7} \tan^2(\frac{k\pi}{16})) - \tan^2(\frac{4\pi}{16})$

Question

The question is:

Evaluate $$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$ The given answer:$34$

What I've tried:

$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$ $$=\tan^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+\tan^2\left(\frac{3\pi}{16}\right)+\tan^2\left(\frac{5\pi}{16}\right)+\tan^2\left(\frac{6\pi}{16}\right)+\tan^2\left(\frac{7\pi}{16}\right)$$ $$=\tan^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+...+\tan^2\left(\frac{\pi}{2}-\frac{2\pi}{16}\right)+\tan^2\left(\frac{\pi}{2}-\frac{\pi}{16}\right)$$ $$=\tan^2\left(\frac{\pi}{16}\right)+\cot^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+\cot^2\left(\frac{2\pi}{16}\right)+\tan^2\left(\frac{3\pi}{16}\right)+\cot^2\left(\frac{3\pi}{16}\right)$$

How do I proceed from here?

https://math.stackexchange.com/questions/951522/trig-sum-tan-21-circ-tan-22-circ-tan2-89-circ — lab bhattacharjee, May 15 '19 at 06:44
@BarryCipra but I've just looked at the source again, there is no $\tan^2 (4\pi /6)$ term given — QuasarChaser, May 15 '19 at 06:51
@labbhattacharjee I've spend time reading your answer in the linked post, but I'm not sure about how I should apply that here. Any suggestions? — QuasarChaser, May 15 '19 at 06:53
@ExtremeRaider, Then, see https://math.stackexchange.com/questions/1610812/how-do-you-find-the-value-of-sum-r-044-tan22r1 — lab bhattacharjee, May 15 '19 at 06:54
@ExtremeRaider, if you write out all the terms explicitly instead of using "$+...+$" you should see that $k=4$ gives a $\tan^2(4\pi/16)$ that doesn't pair with anything. — Barry Cipra, May 15 '19 at 06:55
@BarryCipra now I see the misunderstanding, I'm going to edit the question. I apologize for not making it clear earlier. — QuasarChaser, May 15 '19 at 06:56

DXT · Accepted Answer · 2019-05-15T07:13:57.467

2

Using $\displaystyle \tan^2(x)+\cot^2(x)=\frac{\sin^4(x)+\cos^4(x)}{\sin^2(x)\cdot \cos^2(x)}=\frac{1-2\sin^2 x\cos^2 x}{\sin^2 x\cos^2 x}.$

So we have $\displaystyle =4\csc^2(2x)-2=2+4\cot^2(2x).$

So our expression is

$\displaystyle 2+4\cot^2(\pi/8)+2+4\cot^2(\pi/4)+2+4\cot^2(3\pi/8)$

$\displaystyle 6+4+4\bigg[\cot^2(\pi/8)+\tan^2(\pi/8)\bigg]$

$\displaystyle =10+4\bigg[2+4\cot^2(\pi/4)\bigg]$

$=10 + 4\bigg[2+4\bigg]=34.$

edited May 15 '19 at 07:13

answered May 15 '19 at 06:54

DXT

11,241

1

Thank you for pointing out the expression for $\tan^2x + \cot^2x$. – QuasarChaser May 15 '19 at 07:12

score 1 · Answer 2 · answered May 15 '19 at 07:22

1

enter image description here

The summation is 35 and subtract -1 because tan pi/4 is 1

Finally the result is 34

answered May 15 '19 at 07:22

ashok knv

103

Thank you for the effort. – QuasarChaser May 15 '19 at 07:26

score 1 · Answer 3 · answered May 15 '19 at 18:44

Using Expansion of $\tan(nx)$ in powers of $\tan(x)$,

$\tan\dfrac{k\pi}{2n},0\le k<2n,k\ne n$ are the roots of

$$\binom{2n}1t-\binom{2n}3t^3+\cdots+(-1)^{n-1}\binom{2n}{2n-1}t^{2n-1}=0$$

$$\implies \binom{2n}{2n-1}t^{2n-2}-\binom{2n}{2n-3}t^{2n-3}+\cdots+(-1)^{n-1}\binom{2n}1=0$$ will have the roots $\tan\dfrac{k\pi}n, 1\le k<2n;k\ne n$

So, the roots of $$2nc^{n-1}-\binom{2n}3c^{n-2}+\cdots=0$$ will be $c_k=\tan^2\dfrac{k\pi}n,1\le k<n$

$$\sum_{k=1}^{n-1}c_k=\dfrac{\binom{2n}3}{2n}=?$$

Here $2n=16$

Evaluate $(\sum_{k=1}^{7} \tan^2(\frac{k\pi}{16})) - \tan^2(\frac{4\pi}{16})$

3 Answers3

Linked