Let $n$ be a positive integer such that $\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} = \frac{\sqrt{n}}{2}$. Then Options: (A) $6 \le n \le 8$ (B) $4 < n \le 8$ (C) $4 \le n \le 8$ (D) $4 < n < 8$

### Analyzing the Setup We are given the equation: $$\sin \left( \frac{\pi}{2n} \right) + \cos \left( \frac{\pi}{2n} \right) = \frac{\sqrt{n}}{2}$$ To simplify our approach, we define $\theta = \frac{\pi}{2n}$. The equation now transforms into: $$\sin \theta + \cos \theta = \frac{\sqrt{n}}{2}$$ ### The Power of Squaring To eliminate the square root and simplify the trigonometric terms, we square both sides of the equation: $$(\sin \theta + \cos \theta)^2 = \left( \frac{\sqrt{n}}{2} \right)^2$$ Expanding the left side using the identity $(a+b)^2 = a^2 + b^2 + 2ab$, we obtain: $$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = \frac{n}{4}$$ Applying the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ and the double angle identity $2 \sin \theta \cos \theta = \sin(2\theta)$, the equation collapses into: $$1 + \sin(2\theta) = \frac{n}{4}$$ ### The Bridge to the Variable We now substitute $\theta = \frac{\pi}{2n}$ back into the equation, which implies $2\theta = \frac{\pi}{n}$. The equation becomes: $$1 + \sin \left( \frac{\pi}{n} \right) = \frac{n}{4}$$ Rearranging the terms to isolate the trigonometric function, we get: $$\sin \left( \frac{\pi}{n} \right) = \frac{n}{4} - 1 = \frac{n-4}{4}$$ ### The Final Constraint For any positive integer $n > 2$, the angle $\frac{\pi}{n}$ lies in the interval $(0, \frac{\pi}{2}]$. In this domain, the sine function is strictly positive and less than or equal to $1$. Given the structure of our equation, we must satisfy the inequality: $$0 < \frac{n-4}{4} < 1$$ Multiplying the entire inequality by $4$, we obtain: $$0 < n - 4 < 4$$ Adding $4$ to all parts of the inequality yields the range: $$4 < n < 8$$ Since $n$ must be an integer, the possible values for $n$ are **$5, 6,$ and $7$**.

Why is the range $0 < \sin(\frac{\pi}{n}) < 1$? Can't it be 1?

Great observation. If $\sin(\frac{\pi}{n}) = 1$, then $\frac{\pi}{n} = \frac{\pi}{2}$, which implies $n=2$. If you plug $n=2$ back into the original equation, you get $\sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$. But the right side of the original equation is $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Since $\sqrt{2} \neq \frac{1}{\sqrt{2}}$, $n=2$ is not a solution. Thus, the strict inequality holds.

JEE Advanced 1994

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: Let $n$ be a positive integer such that $sin \frac{π}{2 n} + cos \frac{π}{2 n} = \frac{n}{2}$ . Then

Q: Wait, why do we square both sides? Couldn't we just use the harmonic addition formula $\sqrt{2} \sin(\theta + \pi/4)$?

You absolutely could! Using $\sqrt{2} \sin(\theta + \pi/4) = \frac{\sqrt{n}}{2}$ is a brilliant alternative. However, squaring is often the path of least resistance in JEE problems because it immediately eliminates the square root and invokes the Pythagorean identity. Both methods lead to the same result—mathematics is a garden with many paths!

Q: Why is the range $0 < \sin(\frac{\pi}{n}) < 1$? Can't it be 1?

Great observation. If $\sin(\frac{\pi}{n}) = 1$, then $\frac{\pi}{n} = \frac{\pi}{2}$, which implies $n=2$. If you plug $n=2$ back into the original equation, you get $\sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$. But the right side of the original equation is $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Since $\sqrt{2} \neq \frac{1}{\sqrt{2}}$, $n=2$ is not a solution. Thus, the strict inequality holds.

Q: Why do we only consider $n > 2$?

The problem states $n$ is a positive integer. If $n=1$, $\frac{\pi}{2n} = \pi/2$, $\sin(\pi/2) + \cos(\pi/2) = 1 + 0 = 1$. The RHS is $\frac{\sqrt{1}}{2} = 0.5$. $1 \neq 0.5$. If $n=2$, we already checked it. For $n \ge 3$, the angle $\frac{\pi}{n}$ is in $(0, \pi/3]$, which keeps the sine value strictly between 0 and 1.

Select Answer:

Visualized Solution

Define the Angle $θ$

Let $θ = \frac{π}{2 n}$
The given equation becomes: $sin θ + cos θ = \frac{n}{2}$

Squaring Both Sides

To eliminate the square root, square both sides.
$(sin θ + cos θ)^{2} = (\frac{n}{2})^{2}$

Expanding the Expression

Expand using $(a + b)^{2} = a^{2} + b^{2} + 2 ab$ :
$sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = \frac{n}{4}$

Applying Trigonometric Identities

Use the Pythagorean identity: $sin^{2} θ + cos^{2} θ = 1$
Use the double angle identity: $2 sin θ cos θ = sin (2 θ)$
$1 + sin (2 θ) = \frac{n}{4}$

Substituting $θ$ Back

Recall that $θ = \frac{π}{2 n}$
Therefore, $2 θ = 2 (\frac{π}{2 n}) = \frac{π}{n}$
$1 + sin (\frac{π}{n}) = \frac{n}{4}$

Isolating the Sine Term

Rearrange the equation to solve for the sine term:
$sin (\frac{π}{n}) = \frac{n}{4} - 1$
$sin (\frac{π}{n}) = \frac{n - 4}{4}$

Analyzing the Range of Sine

For any positive integer $n > 2$ , the angle $\frac{π}{n}$ lies in $(0, \frac{π}{2}]$ .
In this interval, the sine function is strictly positive and less than or equal to $1$ .
For $n = 2$ , $sin (\frac{π}{2}) = 1$ , but $\frac{2 - 4}{4} = - \frac{1}{2} \neq = 1$ .
Thus, for valid $n$ , $0 < sin (\frac{π}{n}) < 1$ .

Setting Up the Inequality

Since $0 < sin (\frac{π}{n}) < 1$
Substitute $sin (\frac{π}{n}) = \frac{n - 4}{4}$
$0 < \frac{n - 4}{4} < 1$

Solving the Inequality

Multiply the entire inequality by $4$ :
$0 < n - 4 < 4$
Add $4$ to all parts of the inequality:
$4 < n < 8$

Final Conclusion

The possible integer values for $n$ are $5, 6, 7$ .
The range is $4 < n < 8$ .
This matches the given option.

00:00 / 00:00

The Sigma Insight: Trigonometric Ratios and Identities

Solution Diagram

Analyzing the Setup

We are given the equation:

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

To simplify our approach, we define

θ = \frac{π}{2 n}

. The equation now transforms into:

sin θ + cos θ = \frac{n}{2}

The Power of Squaring

To eliminate the square root and simplify the trigonometric terms, we square both sides of the equation:

(sin θ + cos θ)^{2} = (\frac{n}{2})^{2}

Expanding the left side using the identity

(a + b)^{2} = a^{2} + b^{2} + 2 ab

, we obtain:

sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = \frac{n}{4}

Applying the Pythagorean identity

sin^{2} θ + cos^{2} θ = 1

and the double angle identity

2 sin θ cos θ = sin (2 θ)

, the equation collapses into:

1 + sin (2 θ) = \frac{n}{4}

The Bridge to the Variable

We now substitute

θ = \frac{π}{2 n}

back into the equation, which implies

2 θ = \frac{π}{n}

. The equation becomes:

1 + sin (\frac{π}{n}) = \frac{n}{4}

Rearranging the terms to isolate the trigonometric function, we get:

sin (\frac{π}{n}) = \frac{n}{4} - 1 = \frac{n - 4}{4}

The Final Constraint

For any positive integer

n > 2

, the angle

\frac{π}{n}

lies in the interval

(0, \frac{π}{2}]

. In this domain, the sine function is strictly positive and less than or equal to

1

Given the structure of our equation, we must satisfy the inequality:

0 < \frac{n - 4}{4} < 1

Multiplying the entire inequality by

4

, we obtain:

0 < n - 4 < 4

Adding

4

to all parts of the inequality yields the range:

4 < n < 8

Since

n

must be an integer, the possible values for

n

are $5, 6,$ and $7$ .

Common Traps & FAQs

Wait, why do we square both sides? Couldn't we just use the harmonic addition formula

2 sin (θ + π /4)

You absolutely could! Using

2 sin (θ + π /4) = \frac{n}{2}

is a brilliant alternative. However, squaring is often the path of least resistance in JEE problems because it immediately eliminates the square root and invokes the Pythagorean identity. Both methods lead to the same result—mathematics is a garden with many paths!

Why is the range

0 < sin (\frac{π}{n}) < 1

? Can't it be 1?

Great observation. If

sin (\frac{π}{n}) = 1

, then

\frac{π}{n} = \frac{π}{2}

, which implies

n = 2

. If you plug

n = 2

back into the original equation, you get

sin (π /4) + cos (π /4) = \frac{1}{2} + \frac{1}{2} = 2

. But the right side of the original equation is

\frac{2}{2} = \frac{1}{2}

. Since

2 \neq = \frac{1}{2}

n = 2

is not a solution. Thus, the strict inequality holds.

Why do we only consider

n > 2

The problem states

n

is a positive integer. If

n = 1

\frac{π}{2 n} = π /2

sin (π /2) + cos (π /2) = 1 + 0 = 1

. The RHS is

\frac{1}{2} = 0.5

1 \neq = 0.5

. If

n = 2

, we already checked it. For

n \geq 3

, the angle

\frac{π}{n}

is in

(0, π /3]

, which keeps the sine value strictly between 0 and 1.

Animated Solution for Mathematics - Trigonometry: Let $n$ be a positive integer such that $sin \frac{π}{2 n} + cos \frac{π}{2 n} = \frac{n}{2}$ . Then

Select Answer:

Visualized Solution

Define the Angle $θ$

Squaring Both Sides

Expanding the Expression

Applying Trigonometric Identities

Substituting $θ$ Back

Isolating the Sine Term

Analyzing the Range of Sine

Setting Up the Inequality

Solving the Inequality

Final Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Power of Squaring

The Bridge to the Variable

The Final Constraint

Similar Questions

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 ____.

For a positive integer $n$ , let $f_{n} (θ) = (tan \frac{θ}{2}) (1 + sec θ) (1 + sec 2 θ) (1 + sec 4 θ) \dots (1 + sec 2^{n} θ)$ . Then

For $α, β \in (0, \frac{π}{2})$ , let $3 sin (α + β) = 2 sin (α - β)$ and a real number $k$ be such that $tan α = k tan β$ . Then the value of $k$ is equal to :

For $0 < θ < π /2$ , the solution(s) of $\sum_{m = 1}^{6} csc (θ + \frac{( m - 1 ) π}{4}) csc (θ + \frac{mπ}{4}) = 42$ is (are)

If $0 < x, y < π$ and $cos x + cos y - cos (x + y) = \frac{3}{2}$ , then $sin x + cos y$ is equal to:

If $15 sin^{4} α + 10 cos^{4} α = 6$ , for some $α \in R$ , then the value of $27 sec^{6} α + 8 cosec^{6} α$ is equal to:

If $\frac{s i n ^{4} x}{2} + \frac{c o s ^{4} x}{3} = \frac{1}{5}$ , then

If $exp {(sin^{2} x + sin^{4} x + sin^{6} x + \dots \infty) ln 2}$ satisfies the equation $x^{2} - 9 x + 8 = 0$ , find the value of $\frac{c o s x}{c o s x + s i n x}, 0 < x < π /2$ .

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 :

Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n be a positive integer such that sin2nπ​+cos2nπ​=2n​​. Then

Select Answer:

Visualized Solution

Define the Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Squaring Both Sides

Expanding the Expression

Applying Trigonometric Identities

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ Back

Isolating the Sine Term

Analyzing the Range of Sine

Setting Up the Inequality

Solving the Inequality

Final Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Power of Squaring

The Bridge to the Variable

The Final Constraint

Similar Questions

.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 ____.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a positive integer n, let fn​(θ)=(tan2θ​)(1+secθ)(1+sec2θ)(1+sec4θ)…(1+sec2nθ). Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For α,β∈(0,2π​), let 3sin(α+β)=2sin(α−β) and a real number k be such that tanα=ktanβ. Then the value of k is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For 0<θ<π/2, the solution(s) of ∑m=16​csc(θ+4(m−1)π​)csc(θ+4mπ​)=42​ is (are)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 0<x,y<π and cosx+cosy−cos(x+y)=23​, then sinx+cosy is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 15sin4α+10cos4α=6, for some α∈R, then the value of 27sec6α+8cosec6α is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 2sin4x​+3cos4x​=51​, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If exp{(sin2x+sin4x+sin6x+…∞)ln2} satisfies the equation x2−9x+8=0, find the value of cosx+sinxcosx​,0<x<π/2.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for x∈(0,2π​), log10​sinx+log10​cosx=−1 and log10​(sinx+cosx)=21​(log10​n−1), then the value of n is equal to :

Animated Solution for Mathematics - Trigonometry: Let $n$ be a positive integer such that $sin \frac{π}{2 n} + cos \frac{π}{2 n} = \frac{n}{2}$ . Then

Define the Angle $θ$

Substituting $θ$ Back

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 ____.

For a positive integer $n$ , let $f_{n} (θ) = (tan \frac{θ}{2}) (1 + sec θ) (1 + sec 2 θ) (1 + sec 4 θ) \dots (1 + sec 2^{n} θ)$ . Then

For $α, β \in (0, \frac{π}{2})$ , let $3 sin (α + β) = 2 sin (α - β)$ and a real number $k$ be such that $tan α = k tan β$ . Then the value of $k$ is equal to :

For $0 < θ < π /2$ , the solution(s) of $\sum_{m = 1}^{6} csc (θ + \frac{( m - 1 ) π}{4}) csc (θ + \frac{mπ}{4}) = 42$ is (are)

If $0 < x, y < π$ and $cos x + cos y - cos (x + y) = \frac{3}{2}$ , then $sin x + cos y$ is equal to:

If $15 sin^{4} α + 10 cos^{4} α = 6$ , for some $α \in R$ , then the value of $27 sec^{6} α + 8 cosec^{6} α$ is equal to:

If $\frac{s i n ^{4} x}{2} + \frac{c o s ^{4} x}{3} = \frac{1}{5}$ , then

If $exp {(sin^{2} x + sin^{4} x + sin^{6} x + \dots \infty) ln 2}$ satisfies the equation $x^{2} - 9 x + 8 = 0$ , find the value of $\frac{c o s x}{c o s x + s i n x}, 0 < x < π /2$ .

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 :