Question

Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.

Hint:

For trigonometric sums, look for telescoping series using identities like \( \cot a - \cot b \).

Answer 1

We use the identity $\frac{1}{\sin A \sin B} = \frac{1}{\sin(B-A)} (\cot A - \cot B)$. Here, $B - A = 1^\circ$, so $\sin(B-A) = \sin 1^\circ$. Thus, \[ \frac{1}{\sin n^\circ \sin (n+1)^\circ} = \frac{\cot n^\circ - \cot (n+1)^\circ}{\sin 1^\circ}. \] The sum $\alpha$ becomes a telescoping sum: \begin{align*} \alpha &= \sum_{n=60}^{119} \frac{1}{\sin n^\circ \sin (n+1)^\circ}
&= \sum_{n=60}^{119} \frac{\cot n^\circ - \cot (n+1)^\circ}{\sin 1^\circ}
&= \frac{1}{\sin 1^\circ} \left[ (\cot 60^\circ - \cot 61^\circ) + (\cot 61^\circ - \cot 62^\circ) + \cdots + (\cot 119^\circ - \cot 120^\circ) \right]
&= \frac{1}{\sin 1^\circ} (\cot 60^\circ - \cot 120^\circ). \end{align*} We know that $\cot 60^\circ = \frac{1}{\sqrt{3}}$ and $\cot 120^\circ = \cot (180^\circ - 60^\circ) = -\cot 60^\circ = -\frac{1}{\sqrt{3}}$. Therefore, \[ \alpha = \frac{1}{\sin 1^\circ} \left( \frac{1}{\sqrt{3}} - \left( -\frac{1}{\sqrt{3}} \right) \right) = \frac{1}{\sin 1^\circ} \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} \right) = \frac{1}{\sin 1^\circ} \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3} \sin 1^\circ}. \] We need to find the value of $\left( \frac{\csc 1^\circ}{\alpha} \right)^2$. We know that $\csc 1^\circ = \frac{1}{\sin 1^\circ}$. \[ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 = \left( \frac{1/\sin 1^\circ}{2/(\sqrt{3} \sin 1^\circ)} \right)^2 = \left( \frac{1}{\sin 1^\circ} \cdot \frac{\sqrt{3} \sin 1^\circ}{2} \right)^2 = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4}. \]