Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The value of $\sum\limits_{k = 1}^{13} {\dfrac{1}{{\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})}}}$ is equal to ${\text{(a) }}3 - \sqrt 3 \\ {\text{(b) }}2(3 - \sqrt 3 ) \\ {\text{(c) }}2(\sqrt 3 - 1) \\ {\text{(d) 2(2 + }}\sqrt 3 ) \\$

Last updated date: 17th Jun 2024
Total views: 403.2k
Views today: 12.03k
Verified
403.2k+ views
Hint: In this trigonometrical problem, please note that denominator is the part of expansion of compound formula. Multiply the common factor for compound formula in both numerator and denominator to get simplified expression for the problem.

Firstly, write the expression given in the question as,
$\sum\limits_{k = 1}^{13} {\dfrac{1}{{\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})}}}$,
Now multiply both numerator and denominator by $\sin \dfrac{\pi }{6}$. The $\sin \dfrac{\pi }{6}$ can be factored in the following way, $\sin \dfrac{\pi }{6} = \sin [(\dfrac{\pi }{4} + \dfrac{{k\pi }}{6}) - (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})] \\ \\$. Therefore the expression will become as
$\Rightarrow \sum\limits_{k = 1}^{13} {\dfrac{{\sin [(\dfrac{\pi }{4} + \dfrac{{k\pi }}{6}) - (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})]}}{{\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})\sin \dfrac{\pi }{6}}}}$
Now, expand the numerator by using compound formula, $\sin (A - B) = \sin A\cos B - \cos A\sin B$. The expression will be
$\Rightarrow \sum\limits_{k = 1}^{13} {\dfrac{{\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})\cos (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6}) - \cos (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})}}{{\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})\sin \dfrac{\pi }{6}}}}$
As we know that$\sin \dfrac{\pi }{6} = \dfrac{1}{2}$, hence put this value and take 2 out of summation. Therefore,
$\Rightarrow 2\sum\limits_{k = 1}^{13} {\dfrac{{\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})\cos (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6}) - \cos (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})}}{{\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})}}}$
Now, simplify the above expression by separating it and dividing them by common denominator $\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})$, we get$\Rightarrow 2\sum\limits_{k = 1}^{13} {[\dfrac{{\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})\cos (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})}}{{\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})}}} - \dfrac{{\cos (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})}}{{\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})}}]$
Nullify the same terms in numerators and denominators, we get
$\Rightarrow 2\sum\limits_{k = 1}^{13} {[\dfrac{{\cos (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})}}{{\sin (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})}}} - \dfrac{{\cos (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})}}{{\sin (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})}}]$
As we know that$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$. With this identity, simplify the above expression, we get
$\Rightarrow 2\sum\limits_{k = 1}^{13} {[\cot (\dfrac{\pi }{4} + \dfrac{{(k - 1)\pi }}{6})} - \cot (\dfrac{\pi }{4} + \dfrac{{k\pi }}{6})]$
Now, if we expand the above equation, we find out that due to negative signs between the terms, all the terms will be cancelled out and only first and last term will remain in the equation. Since we know this fact, we can simply put the value of k in the above expression, then
$\Rightarrow 2[\cot (\dfrac{\pi }{4} + \dfrac{{12\pi }}{6}) - \cot (\dfrac{\pi }{4} + \dfrac{{13\pi }}{6})]$
As we know that $\cot \dfrac{\pi }{4} = 1$and by taking the LCM in the second term, we will get the following expression,
$\Rightarrow 2[\cot (2\pi + \dfrac{\pi }{4}) - \cot (\dfrac{{3\pi + 26\pi }}{{12}})]$
After, simplifying, we will get
$\Rightarrow 2[1 - \cot \dfrac{{29\pi }}{{12}}] \\ \Rightarrow 2[1 - \cot (2\pi + \dfrac{{5\pi }}{{12}})] \\ \Rightarrow 2[1 - \cot \dfrac{{5\pi }}{{12}}] \\$
Since we know that $\cot \dfrac{{5\pi }}{{12}} = 2 - \sqrt 3$, so put this value in the above expression, we will get
$\Rightarrow 2[1 - (2 - \sqrt 3 ] \\ \Rightarrow 2(\sqrt 3 - 1) \\$

Therefore, the answer to the problem is option (c).

Note: It is very common practice to solve these types of trigonometrical problems with the help of compound formula. You shall have to identify which compound formula can be used in particular problem. Remembering all compound formulas will help you a lot to solve such problems. Some formulae are shown below:
$\begin{array}{*{20}{l}} \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B \\ \sin \left( {A-B} \right) = \sin A\cos B-\cos A\sin B \\ \cos \left( {A + B} \right) = \cos A\cos B-\sin A\cos B \\ \cos \left( {A-B} \right) = \cos A\cos B + \sin A\cos B \\ \tan \left( {A + B} \right) = \left[ {\tan A{\text{ }} + \tan B} \right]/\left[ {1-\tan A\tan B} \right] \\ \tan \left( {A-B} \right) = \left[ {\tan A-\tan B} \right]/\left[ {1 + \tan A\tan B} \right] \\ \sin \left( {A + B} \right)\sin \left( {A-B} \right) = {\sin ^2}A-{\sin ^2}B = {\cos ^2}B-{\cos ^2}A \\ \cos \left( {A + B} \right)\cos \left( {A-B} \right) = {\cos ^2}A{\text{ }}-{\sin ^2}A-{\sin ^2}B = {\cos ^2}B-{\sin ^2}A \\ \\ \end{array}$