The Problem Says it All!

The problem is essentially based on the fact that $\sin {r \theta } = 2^{r-1} \prod_{k=0}^{r-1} \sin {(\theta + \frac {k\pi}{r})}$

This is turn can be proven as follows:

Consider the $n^{th}$ roots of the complex number $e^{2ni\theta}$ .

Let these roots be $z_{0},z_{2}...z_{n-1}$

It's fairly obvious that $z_{r}$ can be expressed as: $z_{r}=e^{i(2\theta +\frac{2\pi r}{n})}$

Now, using the fact that these are the roots of the equation $z^n=e^{2ni\theta}$ , we can write:

$z^n-e^{2ni\theta}=(z-z_{0})(z-z_{1})(z-z_{2})...(z-z_{n-1})$

Note that this equation is true for any value of $z$ . For the purpose of this question, make the substitution, $z=1$ to get: $1-e^{2ni\theta}=(1-z_{0})(1-z_{1})(1-z_{2})...(1-z_{n-1})$

Take the modulus of both sides. Consider the general term: $|1- e^{i(2\theta +\frac{2\pi r}{n})}|=|e^{i(\theta +\frac{\pi r}{n})}||e^{-i(\theta +\frac{\pi r}{n})}-e^{i(\theta +\frac{\pi r}{n})}|\\=|e^{-i(\theta +\frac{\pi r}{n})}-e^{i(\theta +\frac{\pi r}{n})}|\\=|2sin(\theta+\frac{r\pi}{n})|$

Do this for every term on the LHS and RHS and the required result is obtained.

In general,

$\sin {r \theta } = 2^{r-1} \prod_{k=0}^{r-1} \sin {(\theta + \frac {k\pi}{r})}$

After this the problem is very easy.

Therefore, $T(r,\theta)=2^{r-1}$

$S(n,\theta)=2^r-1$

Therefore,

S(6)=63

T(7)=64

Thus,Answer=127.

Note:What is interesting is that the answer is independent of $\theta$

EDIT:I don't know the proof to the identity.I started from r=1,2,3,4 and found it while searching the web for general formula of $\sin{r\theta}$