Question ID: #491
If $\sum_{r=1}^{13}\left|\frac{1}{\sin(\frac{\pi}{4}+(r-1)\frac{\pi}{6})\sin(\frac{\pi}{4}+\frac{r\pi}{6})}\right|=a\sqrt{3}+b$, where $a, b\in Z$, then $a^{2}+b^{2}$ is equal to :
- (1) 10
- (2) 2
- (3) 8
- (4) 4
Solution:
Let the term be $T_r$. The series is telescoping in nature.
$$T_r = \frac{1}{\sin((\frac{\pi}{4}+(r-1)\frac{\pi}{6}))\sin(\frac{\pi}{4}+\frac{r\pi}{6})}$$
Multiplying and dividing by $\sin(\frac{\pi}{6})$ (which is the difference of angles):
$$T_r = \frac{1}{\sin(\frac{\pi}{6})} \left[ \cot\left(\frac{\pi}{4}+(r-1)\frac{\pi}{6}\right) – \cot\left(\frac{\pi}{4}+\frac{r\pi}{6}\right) \right]$$
Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$, we have $\frac{1}{\sin(\pi/6)} = 2$.
Let $S = \sum_{r=1}^{13} |T_r|$.
Since the terms inside the modulus are positive for the given range or handled by the modulus at the end:
$$S = 2 \left| \sum_{r=1}^{13} \left[ \cot\left(\frac{\pi}{4}+(r-1)\frac{\pi}{6}\right) – \cot\left(\frac{\pi}{4}+\frac{r\pi}{6}\right) \right] \right|$$
This is a telescoping sum. The first and last terms remain:
$$S = 2 \left| \cot\left(\frac{\pi}{4}\right) – \cot\left(\frac{\pi}{4} + \frac{13\pi}{6}\right) \right|$$
$$S = 2 \left| 1 – \cot\left(2\pi + \frac{\pi}{6} + \frac{\pi}{4}\right) \right|$$
Actually, $\frac{\pi}{4} + \frac{13\pi}{6} = \frac{3\pi + 26\pi}{12} = \frac{29\pi}{12} = 2\pi + \frac{5\pi}{12}$.
$$S = 2 \left| 1 – \cot\left(\frac{5\pi}{12}\right) \right|$$
We know that $\cot(75^\circ) = 2 – \sqrt{3}$.
$$S = 2 \left| 1 – (2 – \sqrt{3}) \right|$$
$$S = 2 |-1 + \sqrt{3}| = 2(\sqrt{3} – 1) = 2\sqrt{3} – 2$$
Comparing with $a\sqrt{3} + b$:
$a = 2$ and $b = -2$.
We need $a^2 + b^2$:
$$a^2 + b^2 = (2)^2 + (-2)^2 = 4 + 4 = 8$$
Ans. (3)
Was this solution helpful?
YesNo