The positive integer value of $n > 3$ satisfying the equation $\frac{1}{\sin(\pi/n)} = \frac{1}{\sin(2\pi/n)} + \frac{1}{\sin(3\pi/n)}$ is ____.

### Analyzing the Setup We are tasked with solving the trigonometric equation: $$\frac{1}{\sin(\pi/n)} = \frac{1}{\sin(2\pi/n)} + \frac{1}{\sin(3\pi/n)}$$ To simplify our perspective, we define $\theta = \frac{\pi}{n}$. This substitution transforms the equation into: $$\frac{1}{\sin \theta} = \frac{1}{\sin 2\theta} + \frac{1}{\sin 3\theta}$$ ### The Master Equation Observe the symmetry between $\theta$ and $3\theta$, whose average is exactly $2\theta$. We rearrange the terms to group $\theta$ and $3\theta$ together: $$\frac{1}{\sin \theta} - \frac{1}{\sin 3\theta} = \frac{1}{\sin 2\theta}$$ Taking the common denominator on the left side, we obtain: $$\frac{\sin 3\theta - \sin \theta}{\sin \theta \sin 3\theta} = \frac{1}{\sin 2\theta}$$ ### Applying Trigonometric Identities We apply the sum-to-product identity, $\sin 3\theta - \sin \theta = 2 \cos(2\theta) \sin(\theta)$, to the numerator: $$\frac{2 \cos 2\theta \sin \theta}{\sin \theta \sin 3\theta} = \frac{1}{\sin 2\theta}$$ The $\sin \theta$ terms cancel out, simplifying the expression to: $$\frac{2 \cos 2\theta}{\sin 3\theta} = \frac{1}{\sin 2\theta}$$ ### Final Calculation Cross-multiplying the terms yields: $$2 \sin 2\theta \cos 2\theta = \sin 3\theta$$ Using the double-angle identity $2 \sin A \cos A = \sin 2A$, we recognize that $2 \sin 2\theta \cos 2\theta = \sin 4\theta$. Thus, the equation reduces to: $$\sin 4\theta = \sin 3\theta$$ Since $\theta = \pi/n$, we have $4\theta = 4\pi/n$ and $3\theta = 3\pi/n$. For the sine values to be equal, we consider the general solution $4\theta = \pi - 3\theta$ (as $4\theta = 3\theta$ leads to a trivial solution): $$7\theta = \pi \implies 7\left(\frac{\pi}{n}\right) = \pi$$ Solving for $n$, we find the final result: **$n = 7$**

Wait, why do we reject the case where $4\theta = 3\theta$? Isn't that a valid solution for $\sin A = \sin B$?

That is a fantastic observation! You are absolutely right that $\sin A = \sin B$ implies $A = B$ or $A = \pi - B$. However, if $4\theta = 3\theta$, then $\theta = 0$. Since $\theta = \frac{\pi}{n}$, this would imply $n$ is infinite, which contradicts our constraint that $n > 3$. Always check your boundary conditions—it's the difference between a good student and a topper!

JEE Advanced 2011

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: The positive integer value of $n > 3$ satisfying the equation $\frac{1}{s i n ( π / n )} = \frac{1}{s i n ( 2 π / n )} + \frac{1}{s i n ( 3 π / n )}$ is ____.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Look closely at the angles involved:

\frac{π}{n}

\frac{2 π}{n}

, and

\frac{3 π}{n}

. Do you see the arithmetic progression? Before you start calculating, visualize the symmetry. The middle term,

\frac{2 π}{n}

, is the average of the other two. This is a massive hint that grouping the outer terms will yield something related to the middle term.

Hint 2

Hint 2: Don't rush to expand everything into sines and cosines immediately. That is a trap that leads to a swamp of algebra. Instead, rearrange the equation to isolate terms that share a common structure. Specifically, try moving the

\frac{1}{s i n ( 3 π / n )}

term to the left side to pair it with

\frac{1}{s i n ( π / n )}

Hint 3

Hint 3: Once you have

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

, use the common denominator on the left. The numerator will involve

sin 3 θ - sin θ

. Apply the sum-to-product identity

sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})

to unlock the path forward.

Visualized Solution

Substitution: Let $θ = \frac{π}{n}$

Let $θ = \frac{π}{n}$ where $n > 3$ .
The given equation becomes:
$\frac{1}{s i n θ} = \frac{1}{s i n 2 θ} + \frac{1}{s i n 3 θ}$

Rearranging the Terms

Rearrange the equation to group terms with $θ$ and $3 θ$ :
$\frac{1}{s i n θ} - \frac{1}{s i n 3 θ} = \frac{1}{s i n 2 θ}$

Taking the Common Denominator (LCM)

Take LCM on the LHS:
$\frac{s i n 3 θ - s i n θ}{s i n θ s i n 3 θ} = \frac{1}{s i n 2 θ}$

Applying $sin C - sin D$ Formula

Apply $sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})$ :
$\frac{2 c o s 2 θ s i n θ}{s i n θ s i n 3 θ} = \frac{1}{s i n 2 θ}$

Simplifying the Fraction

Cancel $sin θ$ (since $θ \neq = 0$ ):
$\frac{2 c o s 2 θ}{s i n 3 θ} = \frac{1}{s i n 2 θ}$

Cross-Multiplication

Cross-multiply to eliminate fractions:
$2 sin 2 θ cos 2 θ = sin 3 θ$

Applying the Double-Angle Identity

Using $2 sin A cos A = sin 2 A$ :
$sin 4 θ = sin 3 θ$

Solving the Trigonometric Equation

Since $sin 4 θ = sin 3 θ$ , we have:
$4 θ = π - 3 θ$ (as $4 θ = 3 θ ⟹ θ = 0$ , not possible)
$7 θ = π$

Solving for $n$

Substitute $θ = \frac{π}{n}$ :
$7 (\frac{π}{n}) = π \Rightarrow n = 7$
Final Answer: 7

00:00 / 00:00

The Sigma Insight: Trigonometric Ratios and Identities

Solution Diagram

Analyzing the Setup

We are tasked with solving the trigonometric equation:

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

To simplify our perspective, we define

θ = \frac{π}{n}

. This substitution transforms the equation into:

\frac{1}{sin θ} = \frac{1}{sin 2 θ} + \frac{1}{sin 3 θ}

The Master Equation

Observe the symmetry between

θ

and

3 θ

, whose average is exactly

2 θ

. We rearrange the terms to group

θ

and

3 θ

together:

\frac{1}{sin θ} - \frac{1}{sin 3 θ} = \frac{1}{sin 2 θ}

Taking the common denominator on the left side, we obtain:

\frac{sin 3 θ - sin θ}{sin θ sin 3 θ} = \frac{1}{sin 2 θ}

Applying Trigonometric Identities

We apply the sum-to-product identity,

sin 3 θ - sin θ = 2 cos (2 θ) sin (θ)

, to the numerator:

\frac{2 cos 2 θ sin θ}{sin θ sin 3 θ} = \frac{1}{sin 2 θ}

The

sin θ

terms cancel out, simplifying the expression to:

\frac{2 cos 2 θ}{sin 3 θ} = \frac{1}{sin 2 θ}

Final Calculation

Cross-multiplying the terms yields:

2 sin 2 θ cos 2 θ = sin 3 θ

Using the double-angle identity

2 sin A cos A = sin 2 A

, we recognize that

2 sin 2 θ cos 2 θ = sin 4 θ

. Thus, the equation reduces to:

sin 4 θ = sin 3 θ

Since

θ = π / n

, we have

4 θ = 4 π / n

and

3 θ = 3 π / n

. For the sine values to be equal, we consider the general solution

4 θ = π - 3 θ

(as

4 θ = 3 θ

leads to a trivial solution):

7 θ = π ⟹ 7 (\frac{π}{n}) = π

Solving for

n

, we find the final result:

$n = 7$

Common Traps & FAQs

Wait, why do we reject the case where

4 θ = 3 θ

? Isn't that a valid solution for

sin A = sin B

That is a fantastic observation! You are absolutely right that

sin A = sin B

implies

A = B

A = π - B

. However, if

4 θ = 3 θ

, then

θ = 0

. Since

θ = \frac{π}{n}

, this would imply

n

is infinite, which contradicts our constraint that

n > 3

. Always check your boundary conditions—it's the difference between a good student and a topper!

Why can't we just cross-multiply everything from the start? It seems like the most direct way to get rid of fractions.

You *could* do that, but you would end up with a nightmare of product terms like

sin 2 θ sin 3 θ = sin θ sin 3 θ + sin θ sin 2 θ

. Expanding these into sums would create a massive algebraic expression that is very easy to mess up. By grouping terms first, we use the structure of the equation to simplify it *before* the algebra gets heavy. Work smarter, not harder!

Is there a geometric way to solve this? It feels like it relates to polygons.

You have a great intuition! This equation actually appears in the study of regular polygons. Specifically, it relates to the lengths of chords in a circle. While the algebraic path is the standard JEE approach, seeing the geometric connection is a sign of a deep mathematical mind. Keep nurturing that curiosity!

Animated Solution for Mathematics - Trigonometry: The positive integer value of $n > 3$ satisfying the equation $\frac{1}{s i n ( π / n )} = \frac{1}{s i n ( 2 π / n )} + \frac{1}{s i n ( 3 π / n )}$ is ____.

Enter Numerical Value:

Visualized Solution

Substitution: Let $θ = \frac{π}{n}$

Rearranging the Terms

Taking the Common Denominator (LCM)

Applying $sin C - sin D$ Formula

Simplifying the Fraction

Cross-Multiplication

Applying the Double-Angle Identity

Solving the Trigonometric Equation

Solving for $n$

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Master Equation

Applying Trigonometric Identities

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The positive integer value of n>3 satisfying the equation sin(π/n)1​=sin(2π/n)1​+sin(3π/n)1​ is ____.

Enter Numerical Value:

Visualized Solution

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Rearranging the Terms

Taking the Common Denominator (LCM)

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Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

The Sigma Insight: Trigonometric Ratios and Identities

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Animated Solution for Mathematics - Trigonometry: The positive integer value of $n > 3$ satisfying the equation $\frac{1}{s i n ( π / n )} = \frac{1}{s i n ( 2 π / n )} + \frac{1}{s i n ( 3 π / n )}$ is ____.

Substitution: Let $θ = \frac{π}{n}$

Applying $sin C - sin D$ Formula

Solving for $n$

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If for $x \in (0, \frac{π}{2})$ , $lo g_{10} sin x + lo g_{10} cos x = - 1$ and $lo g_{10} (sin x + cos x) = \frac{1}{2} (lo g_{10} n - 1)$ , then the value of $n$ is equal to :

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