
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.$8$
B. $6$
C. $5$
D. $7$
Answer
232.8k+ views
Hint: Start solving the given equation step by step using trigonometric formula until only one term is left on both sides. Then equate both the sides and check whether that term is giving the value of $n$ or not. If not, then try to change the angle without changing the function and solve.
Formula Used:
Trigonometric formula –
$\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\sin 2A = 2\sin A\cos A$
Complete step by step solution:
Given that,
$\dfrac{1}{{\sin \left( {\dfrac{\pi }{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}} + \dfrac{1}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}}$
$\dfrac{1}{{\sin \left( {\dfrac{\pi }{n}} \right)}} - \dfrac{1}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}}$
$\dfrac{{\sin \left( {\dfrac{{3\pi }}{n}} \right) - \sin \left( {\dfrac{\pi }{n}} \right)}}{{\sin \left( {\dfrac{\pi }{n}} \right)\sin \left( {\dfrac{{3\pi }}{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}}$
$\left[ {\dfrac{{2\cos \left( {\dfrac{{2\pi }}{n}} \right)\sin \left( {\dfrac{\pi }{n}} \right)}}{{\sin \left( {\dfrac{\pi }{n}} \right)\sin \left( {\dfrac{{3\pi }}{n}} \right)}}} \right]\sin \left( {\dfrac{{2\pi }}{n}} \right) = 1$
$\dfrac{{2\sin \left( {\dfrac{{2\pi }}{n}} \right)\cos \left( {\dfrac{{2\pi }}{n}} \right)}}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}} = 1$
$\dfrac{{\sin \left( {\dfrac{{4\pi }}{n}} \right)}}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}} = 1$
$\sin \left( {\dfrac{{4\pi }}{n}} \right) = \sin \left( {\dfrac{{3\pi }}{n}} \right)$
$\sin \left( {\pi - \dfrac{{4\pi }}{n}} \right) = \sin \left( {\dfrac{{3\pi }}{n}} \right)$
$\pi - \dfrac{{4\pi }}{n} = \dfrac{{3\pi }}{n}$
$n = 7$
Option ‘D’ is correct
Note: The key concept involved in solving this problem is the good knowledge of Trigonometry formula, ratio, and identities. Students must remember that while applying any trigonometric formula do focus on its angle also.
Formula Used:
Trigonometric formula –
$\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\sin 2A = 2\sin A\cos A$
Complete step by step solution:
Given that,
$\dfrac{1}{{\sin \left( {\dfrac{\pi }{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}} + \dfrac{1}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}}$
$\dfrac{1}{{\sin \left( {\dfrac{\pi }{n}} \right)}} - \dfrac{1}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}}$
$\dfrac{{\sin \left( {\dfrac{{3\pi }}{n}} \right) - \sin \left( {\dfrac{\pi }{n}} \right)}}{{\sin \left( {\dfrac{\pi }{n}} \right)\sin \left( {\dfrac{{3\pi }}{n}} \right)}} = \dfrac{1}{{\sin \left( {\dfrac{{2\pi }}{n}} \right)}}$
$\left[ {\dfrac{{2\cos \left( {\dfrac{{2\pi }}{n}} \right)\sin \left( {\dfrac{\pi }{n}} \right)}}{{\sin \left( {\dfrac{\pi }{n}} \right)\sin \left( {\dfrac{{3\pi }}{n}} \right)}}} \right]\sin \left( {\dfrac{{2\pi }}{n}} \right) = 1$
$\dfrac{{2\sin \left( {\dfrac{{2\pi }}{n}} \right)\cos \left( {\dfrac{{2\pi }}{n}} \right)}}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}} = 1$
$\dfrac{{\sin \left( {\dfrac{{4\pi }}{n}} \right)}}{{\sin \left( {\dfrac{{3\pi }}{n}} \right)}} = 1$
$\sin \left( {\dfrac{{4\pi }}{n}} \right) = \sin \left( {\dfrac{{3\pi }}{n}} \right)$
$\sin \left( {\pi - \dfrac{{4\pi }}{n}} \right) = \sin \left( {\dfrac{{3\pi }}{n}} \right)$
$\pi - \dfrac{{4\pi }}{n} = \dfrac{{3\pi }}{n}$
$n = 7$
Option ‘D’ is correct
Note: The key concept involved in solving this problem is the good knowledge of Trigonometry formula, ratio, and identities. Students must remember that while applying any trigonometric formula do focus on its angle also.
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