Finding $\log_{2b – a}(2a – b)$, where $a=\sum_{r=1}^{11}\tan^2(\frac{r\pi}{24})$ and $b=\sum_{r=1}^{11}(-1)^{r-1}\tan^2(\frac{r\pi}{24})$

Let $a = \sum\limits_{r = 1}^{11} {{{\tan }^2}\left( {\frac{{r\pi }}{{24}}} \right)} $ and $b = \sum\limits_{r = 1}^{11} {{{\left( { - 1} \right)}^{r - 1}}{{\tan }^2}\left( {\frac{{r\pi }}{{24}}} \right)} $, then find the value of $\log_{(2b – a)}(2a – b)$.

My approach to solve this is as follows: $\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} \Rightarrow 1 - \frac{{2\tan \theta }}{{\tan 2\theta }} = {\tan ^2}\theta.$

Solution 1:

Some symmetries and trig identities help to simplify the sums.

The most important,

$$\cos{x+\pi}=-\cos{x}$$ $$\frac{1}{1+\cos{x}}+\frac{1}{1-\cos{x}}=\frac{4}{1-\cos{2x}}$$

Then:

$$a=\sum_{k=1}^{11}\tan^2{\frac{k\pi}{24}}=\sum_{k=1}^{11}(\sec^2{\frac{k\pi}{24}}-1)=-11+\sum_{k=1}^{11}\frac{2}{1+\cos{\frac{k\pi}{12}}}$$

$$2a=-22+2\sum_{k-1}^{11}(\frac{1}{1+\cos{\frac{k\pi}{12}}}+\frac{1}{1+\cos{\frac{(k+12)\pi}{12}}})=-22+2\sum_{k-1}^{11}(\frac{1}{1+\cos{\frac{k\pi}{12}}}+\frac{1}{1-\cos{\frac{k\pi}{12}}})$$

$$a=-11+4\sum_{k=1}^{11}\frac{1}{1-\cos{\frac{k\pi}{6}}}=-9+4\sum_{k=1}^{5}(\frac{1}{1-\cos{\frac{k\pi}{6}}}+\frac{1}{1-\cos{\frac{(k+6)\pi}{6}}})$$

$$a=-9+16\sum_{k=1}^{5}\frac{1}{1-\cos{\frac{k\pi}{3}}}$$

$$a=-1+16\sum_{k=1}^2\frac{2}{\sin^2{\frac{k\pi}{3}}}=-1+32(\frac{1}{3/4}+\frac{1}{3/4})=-1+256/3=253/3$$

So $a=253/3$.

Instead of calculating $b$, it's easier to calculate $a-b$. This cancels terms in the sum for odd values of $k$.

$$a-b=\sum_{k=1}^52(\sec^2{\frac{k\pi}{12}}-1)=-10+4\sum_{k=1}^5\frac{1}{1+\cos{\frac{k\pi}{6}}}$$

$$2(a-b)=-20+4\sum_{k=1}^5\frac{1}{1+\cos{k\pi/6}}+\frac{1}{1-\cos{k\pi/6}}=-20+16\sum_{k=1}^5\frac{1}{1-\cos{k\pi/3}}$$

$$a-b=-10+8\sum_{k=1}^5\frac{1}{1-\cos{k\pi/3}}$$

$a+9=2(a-b+10)\implies b=(a+11)/2=143/3$

Since $a=253/3$ and $b=143/3$,

$2a-b=(506-143)/3=121$

$2b-a=(286-253)/3=11$

ans: $\ln{121}/\ln{11}=2$

Solution 2:

The goal of my answer is to give you a series of hints and let you work out the exact calculations (I know, I'm lazy).

The main idea is that you can compute $a$ and $b$ by looking at these as the sums of roots of certain polynomials. As you know, the sum of the roots of a polynomial is related to its constant coefficient.

Let $n=12$. For $a$ you need to compute $$\sum_{r=0}^{n-1}\tan^2\left(\frac {k\pi}{2n}\right) $$ So you'd like to find a polynomial whose roots are those squared tangents.

Start with $P(X)=(1+X)^{2n}-(1-X)^{2n}$. Then any root $x$ of $P$ verifies $$(1+x)^{2n} = (1-x)^{2n}$$ or equivalently, $$\left( \frac{1+x}{1-x}\right)^{2n} = 1$$ which means that the roots $x_r$ verify $\frac{1+x_r}{1-x_r}=e^{i\frac{r\pi}n}$ for $r=0,...,2n-1$. In other words, the roots of $P$ are $$x_r=\frac{e^{i\frac{r\pi}n} - 1}{e^{i\frac{r\pi}n} + 1}=\frac{e^{i\frac{r\pi}{2n}}-e^{-i\frac{r\pi}{2n}}}{e^{i\frac{r\pi}{2n}} +e^{-i\frac{r\pi}{2n}}}=\tan \left(\frac {r\pi}{2n}\right)$$

Now the first sum, $a$, uses the squares of the tangents, so $P(X)$ is not quite what we need. But $P$ only has terms of even degree, so you can write $P(X) = Q(X^2)$, and the roots of $Q$ are, $$y_r = \tan^2\left(\frac {r\pi}{2n}\right)$$

So by looking at the constant coefficient for $Q$, you should be able to compute $a$.

As for $b$, you can notice that you can separate the sum into two sums (the terms for odd $r$'s, and the terms with even $r$'s). The sum with odd $r$'s has the same form as the one we had for $a$, except that you need to replace $n$ with $n/2$.

The second sum (even $r$'s) can be deduced from the first one (odd $r$'s) and $a$ itself.