If _r=1^131 (( )/(4)+(r-1)( )/(6)) (( )/(4)+(r )/(6))=a√3+b,a,b Z, then a^2+b^2 is equal t…

Question

If $\sum _{r=1}^{13}{\frac{1}{\sin (\frac{\pi }{4}+(r-1)\frac{\pi }{6})\sin (\frac{\pi }{4}+\frac{r\pi }{6})}}=a\sqrt{3}+b,a,b\in Z$, then ${a}^{2}+{b}^{2}$ is equal to :

Accepted Answer

Using $\dfrac1{\sin X\sin Y} = \dfrac{\cot X-\cot Y}{\sin(Y-X)}$ with $Y-X = \dfrac\pi6$ (so $\sin(Y-X)=\tfrac12$):

$$\frac1{\sin\left(\tfrac\pi4+(r-1)\tfrac\pi6\right)\sin\left(\tfrac\pi4+r\tfrac\pi6\right)} = 2\left[\cot\left(\tfrac\pi4+(r-1)\tfrac\pi6\right) - \cot\left(\tfrac\pi4+r\tfrac\pi6\right)\right]$$

This telescopes:

$$\sum_{r=1}^{13}(\cdots) = 2\left[\cot\tfrac\pi4 - \cot\left(\tfrac\pi4+13\cdot\tfrac\pi6\right)\right] = 2\left[1-\cot\tfrac{29\pi}{12}\right]$$

Since $\cot\tfrac{29\pi}{12} = \cot\left(\tfrac{29\pi}{12}-2\pi\right) = \cot\tfrac{5\pi}{12} = \cot75^\circ = 2-\sqrt3$:

$$\text{Sum} = 2[1-(2-\sqrt3)] = 2(\sqrt3-1) = 2\sqrt3-2$$

So $a=2,\ b=-2$, giving $a^2+b^2 = 4+4=8$.

Solutions · 1