The positive integer value of \(n>3\) satisfying the equation \[ \frac{1}{\sin \left(\frac{\pi}{n}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{n}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{n}\right)} \] is

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Trigonometric Equations

The positive integer value of

n > 3

satisfying the equation

\frac{1}{sin ( \frac{π}{n} )} = \frac{1}{sin ( \frac{2 π}{n} )} + \frac{1}{sin ( \frac{3 π}{n} )}

Answer: 7

Rearranging:

\frac{1}{s i n \frac{π}{n}} - \frac{1}{s i n \frac{3 π}{n}} = \frac{1}{s i n \frac{2 π}{n}}

\frac{s i n \frac{3 π}{n} - s i n \frac{π}{n}}{s i n \frac{π}{n} s i n \frac{3 π}{n}} = \frac{1}{s i n \frac{2 π}{n}}

.

Using

sin A - sin B = 2 cos \frac{A + B}{2} sin \frac{A - B}{2}

sin \frac{3 π}{n} - sin \frac{π}{n} = 2 cos \frac{2 π}{n} sin \frac{π}{n}

.

Substituting:

\frac{2 c o s \frac{2 π}{n} s i n \frac{π}{n}}{s i n \frac{π}{n} s i n \frac{3 π}{n}} = \frac{1}{s i n \frac{2 π}{n}}

.

Simplifying:

\frac{2 c o s \frac{2 π}{n}}{s i n \frac{3 π}{n}} = \frac{1}{s i n \frac{2 π}{n}}

.

Cross-multiplying:

2 cos \frac{2 π}{n} sin \frac{2 π}{n} = sin \frac{3 π}{n}

.

Using

2 sin A cos A = sin 2 A

sin \frac{4 π}{n} = sin \frac{3 π}{n}

.

This gives

\frac{4 π}{n} = π - \frac{3 π}{n}

(taking the supplementary angle), so

\frac{4 π}{n} + \frac{3 π}{n} = π

, giving

n = 7

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Integer answer