Question

The value of $\tan^{-1} \left[ \frac{1}{4\sqrt{3}} \sum_{n=0}^{10} \frac{1}{\cos(\frac{7\pi}{12}+\frac{n\pi}{2})\cos(\frac{7\pi}{12}+\frac{(n+1)\pi}{2})} \right]$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ equals

Answer 1

$\frac{\pi}{6}$

Answer 2

Let the given expression be $E$. $E = \tan^{-1} \left[ \frac{1}{4\sqrt{3}} \sum_{n=0}^{10} \frac{1}{\cos(\frac{7\pi}{12}+\frac{n\pi}{2})\cos(\frac{7\pi}{12}+\frac{(n+1)\pi}{2})} \right]$.

Let $\theta_n = \frac{7\pi}{12}+\frac{n\pi}{2}$. The general term of the sum is $\frac{1}{\cos(\theta_n)\cos(\theta_{n+1})}$. We have $\theta_{n+1} - \theta_n = (\frac{7\pi}{12}+\frac{(n+1)\pi}{2}) - (\frac{7\pi}{12}+\frac{n\pi}{2}) = \frac{\pi}{2}$.

Consider the identity $\tan(B) - \tan(A) = \frac{\sin(B-A)}{\cos(A)\cos(B)}$. Let $A = \theta_n$ and $B = \theta_{n+1}$. Then $B-A = \frac{\pi}{2}$. $\sin(B-A) = \sin(\frac{\pi}{2}) = 1$. So, $\tan(\theta_{n+1}) - \tan(\theta_n) = \frac{1}{\cos(\theta_n)\cos(\theta_{n+1})}$.

The sum is a telescoping series: $S = \sum_{n=0}^{10} \frac{1}{\cos(\theta_n)\cos(\theta_{n+1})} = \sum_{n=0}^{10} (\tan(\theta_{n+1}) - \tan(\theta_n))$. $S = (\tan(\theta_1) - \tan(\theta_0)) + (\tan(\theta_2) - \tan(\theta_1)) + \dots + (\tan(\theta_{11}) - \tan(\theta_{10}))$. $S = \tan(\theta_{11}) - \tan(\theta_0)$.

We need to calculate $\theta_0$ and $\theta_{11}$. $\theta_0 = \frac{7\pi}{12} + \frac{0\pi}{2} = \frac{7\pi}{12}$. $\theta_{11} = \frac{7\pi}{12} + \frac{11\pi}{2} = \frac{7\pi}{12} + \frac{66\pi}{12} = \frac{73\pi}{12}$.

Now we calculate $\tan(\theta_0)$ and $\tan(\theta_{11})$. $\tan(\theta_0) = \tan(\frac{7\pi}{12}) = \tan(\frac{3\pi}{12} + \frac{4\pi}{12}) = \tan(\frac{\pi}{4} + \frac{\pi}{3})$. Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$: $\tan(\frac{7\pi}{12}) = \frac{\tan(\pi/4) + \tan(\pi/3)}{1 - \tan(\pi/4)\tan(\pi/3)} = \frac{1 + \sqrt{3}}{1 - 1 \cdot \sqrt{3}} = \frac{1 + \sqrt{3}}{1 - \sqrt{3}}$. Multiply numerator and denominator by $1+\sqrt{3}$: $\tan(\frac{7\pi}{12}) = \frac{(1+\sqrt{3})^2}{(1-\sqrt{3})(1+\sqrt{3})} = \frac{1 + 3 + 2\sqrt{3}}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}$.

$\tan(\theta_{11}) = \tan(\frac{73\pi}{12})$. $\frac{73\pi}{12} = \frac{72\pi + \pi}{12} = 6\pi + \frac{\pi}{12}$. Using the property $\tan(n\pi + x) = \tan(x)$ for integer $n$: $\tan(\frac{73\pi}{12}) = \tan(6\pi + \frac{\pi}{12}) = \tan(\frac{\pi}{12})$. $\tan(\frac{\pi}{12}) = \tan(15^\circ) = \tan(45^\circ - 30^\circ)$. Using the formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$: $\tan(\frac{\pi}{12}) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ)\tan(30^\circ)} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$. Multiply numerator and denominator by $\sqrt{3}-1$: $\tan(\frac{\pi}{12}) = \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3 + 1 - 2\sqrt{3}}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.

The sum $S = \tan(\theta_{11}) - \tan(\theta_0) = (2 - \sqrt{3}) - (-2 - \sqrt{3}) = 2 - \sqrt{3} + 2 + \sqrt{3} = 4$.

The expression inside the inverse tangent is $\frac{1}{4\sqrt{3}} \cdot S = \frac{1}{4\sqrt{3}} \cdot 4 = \frac{1}{\sqrt{3}}$.

We need to find the value of $\tan^{-1}(\frac{1}{\sqrt{3}})$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. The principal value of $\tan^{-1}(\frac{1}{\sqrt{3}})$ is $\frac{\pi}{6}$. Since $-\frac{\pi}{2} < \frac{\pi}{6} < \frac{\pi}{2}$, the value is $\frac{\pi}{6}$.