Question

The positive integer value of \[n > 3\] satisfying the equation \[\dfrac{1}{{\sin \left( {\dfrac{\pi }{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}} + \dfrac{1}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}}\]is
A. 7
B. 6
C. 4
D. 5

Answer 1

Hint: We shift one of the values from RHS to LHS and take LCM. Use the formula of \[\sin A - \sin B\] in the numerator. Cancel possible terms from numerator and denominator and bring out all the terms with the same angle on one side of the equation. Take inverse function on both sides and equate the angles.
* \[\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\]

Complete step by step answer:
We are given the equation \[\dfrac{1}{{\sin \left( {\dfrac{\pi }{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}} + \dfrac{1}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}}\]
We shift second term in RHS to LHS of the equation
\[ \Rightarrow \dfrac{1}{{\sin \left( {\dfrac{\pi }{n}} \right)}} - \dfrac{1}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}}\]
Take LCM on left hand side of the equation
\[ \Rightarrow \dfrac{{\sin \left( {\dfrac{{3\pi }}{n}} \right) - \sin \left( {\dfrac{\pi }{n}} \right)}}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)\sin \left( {\dfrac{\pi }{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}}\]
Use formula of \[\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\] in numerator
\[ \Rightarrow \dfrac{{2\cos \left( {\dfrac{{\dfrac{{3\pi }}{n} + \dfrac{\pi }{n}}}{2}} \right)\sin \left( {\dfrac{{\dfrac{{3\pi }}{n} - \dfrac{\pi }{n}}}{2}} \right)}}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)\sin \left( {\dfrac{\pi }{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}}\]
\[ \Rightarrow \dfrac{{2\cos \left( {\dfrac{{\dfrac{{4\pi }}{n}}}{2}} \right)\sin \left( {\dfrac{{\dfrac{{2\pi }}{n}}}{2}} \right)}}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)\sin \left( {\dfrac{\pi }{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}}\]
Cancel same factors from numerator and denominator in numerator of LHS
\[ \Rightarrow \dfrac{{2\cos \left( {\dfrac{{2\pi }}{n}} \right)\sin \left( {\dfrac{\pi }{n}} \right)}}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)\sin \left( {\dfrac{\pi }{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}}\]
Cancel same factors from numerator and denominator of LHS
\[ \Rightarrow \dfrac{{2\cos \left( {\dfrac{{2\pi }}{n}} \right)}}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}}\]
Cross multiply the terms from RHS to LHS
\[ \Rightarrow 2\cos \left( {\dfrac{{2\pi }}{n}} \right)\sin \left( {\dfrac{{2\pi }}{n}} \right) = \sin \left( {\dfrac{{3\pi }}{n}} \right)\]
Use formula of \[\sin 2x = 2\sin x\cos x\] in LHS of the equation
\[ \Rightarrow \sin \left( {\dfrac{{4\pi }}{n}} \right) = \sin \left( {\dfrac{{3\pi }}{n}} \right)\] … (1)
We know the general solution for \[\sin \theta = \sin \alpha \]is given by \[\theta = p\pi + {( - 1)^p}\alpha \] where p is an integer value.
Then general solution of equation (1) is
\[\dfrac{{4\pi }}{n} = p\pi + {( - 1)^p}\dfrac{{3\pi }}{n}\]
Since p is an integer it can either be even \[(p = 2n)\] or p can be odd\[(p = 2n + 1)\]
If p is even:
\[ \Rightarrow \dfrac{{4\pi }}{n} = 2k\pi + {( - 1)^{2k}}\dfrac{{3\pi }}{n}\]
\[ \Rightarrow \dfrac{{4\pi }}{n} = 2k\pi + \dfrac{{3\pi }}{n}\]
Shift fraction term to LHS
\[ \Rightarrow \dfrac{{4\pi }}{n} - \dfrac{{3\pi }}{n} = 2k\pi \]
\[ \Rightarrow \dfrac{\pi }{n} = 2k\pi \]
Cancel same factors from both sides of the equation
\[ \Rightarrow \dfrac{1}{n} = 2k\]
Since n was an even number, so, this is a contradiction.
If p is odd:
\[ \Rightarrow \dfrac{{4\pi }}{n} = (2k + 1)\pi + {( - 1)^{2k + 1}}\dfrac{{3\pi }}{n}\]
\[ \Rightarrow \dfrac{{4\pi }}{n} = (2k + 1)\pi - \dfrac{{3\pi }}{n}\]
Shift fraction term to LHS
\[ \Rightarrow \dfrac{{4\pi }}{n} + \dfrac{{3\pi }}{n} = (2k + 1)\pi \]
\[ \Rightarrow \dfrac{{7\pi }}{n} = (2k + 1)\pi \]
Cancel same factors from both sides of the equation
\[ \Rightarrow \dfrac{7}{n} = 2k + 1\]
Of we put \[k = 0\]
\[ \Rightarrow \dfrac{7}{n} = 1\]
\[ \Rightarrow n = 7\] which is greater than 3
\[\therefore \] Positive integer satisfying equation is 7

\[\therefore \]Option A is correct.

Note: Many students try to solve this question by directly substituting the values of ‘n’ given in the options, keep in mind we will have to solve the equation using trigonometric formulas for each of the given values which will take a lot of time. Instead we should solve generally for the solution.