Question

Let $n$ be a positive integer such that $\sin \dfrac{\pi }{{2n}} + \cos \dfrac{\pi }{{2n}} = \dfrac{{\sqrt n }}{2}$ . Then
A. $6 \leqslant n \leqslant 8$
B. $4 < n \leqslant 8$
C. $4 \leqslant n \leqslant 8$
D. $4 < n < 8$

Answer 1

Hint : In this question, we have to find the value of $n$ that satisfy the given an inequality $\sin \dfrac{\pi }{{2n}} + \cos \dfrac{\pi }{{2n}} = \dfrac{{\sqrt n }}{2}$ . We first proceed by simplifying the given inequality and reducing it in terms of either sine or cosine. Then we will use the fact that the value of sine or cosine always lies between $- 1$ and $1$ . Thus we then equate the simplified inequality to lie between $- 1$ and $1$ , and solve to get the required value of $n$ .

Complete step-by-step answer :
The given question is based on application of trigonometry ratio. Trigonometric ratio is the ratio of the side of the right angled triangle. The various ratios are sine, cosine, tangent, cotangent, secant and cosecant. The value of trigonometric ratio Sine and Cosine always lies between $- 1$ and $1$ .
Consider the given question,
The given inequality is: $\sin \dfrac{\pi }{{2n}} + \cos \dfrac{\pi }{{2n}} = \dfrac{{\sqrt n }}{2}$
On squaring both sides, we have
$\Rightarrow {\left( {\sin \dfrac{\pi }{{2n}} + \cos \dfrac{\pi }{{2n}}} \right)^2} = {\left( {\dfrac{{\sqrt n }}{2}} \right)^2}$
$\Rightarrow {\sin ^2}\dfrac{\pi }{{2n}} + {\cos ^2}\dfrac{\pi }{{2n}} + 2\sin \dfrac{\pi }{{2n}}\cos \dfrac{\pi }{{2n}} = \dfrac{n}{4}$
We know that, ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and $2\sin \theta \cos \theta = \sin 2\theta$ .
Taking $\theta = \dfrac{\pi }{{2n}}$ , we have ${\sin ^2}\dfrac{\pi }{{2n}} + {\cos ^2}\dfrac{\pi }{{2n}} = 1$ and $2\sin \dfrac{\pi }{{2n}}\cos \dfrac{\pi }{{2n}} = \sin 2\left( {\dfrac{\pi }{{2n}}} \right)$
Therefore, we have
$\Rightarrow 1 + \sin 2\left( {\dfrac{\pi }{{2n}}} \right) = \dfrac{n}{4}$
On subtracting 1 from both side, we get
$\Rightarrow \sin \left( {\dfrac{\pi }{n}} \right) = \dfrac{n}{4} - 1$ ………………………….. eq(i)
Now we know that the value of sine lies between $- 1$ and $1$ .
i.e. $- 1 \leqslant \sin \theta \leqslant 1$
Taking $\theta = \dfrac{\pi }{n}$ , we have $- 1 \leqslant \sin \left( {\dfrac{\pi }{n}} \right) \leqslant 1$
Since we are given that $n$ is positive integer, therefore we have
$\Rightarrow 0 \leqslant \sin \left( {\dfrac{\pi }{n}} \right) \leqslant 1$
From eq(i), we have
$\Rightarrow 0 \leqslant \dfrac{n}{4} - 1 \leqslant 1$
Adding $1$ to each part of inequality, we get
$\Rightarrow 1 \leqslant \dfrac{n}{4} \leqslant 2$
Multiplying $4$ to each part of inequality, we get
$\Rightarrow 4 \leqslant n \leqslant 8$
Hence, Option $(C)$ is correct.
So, the correct answer is “Option C”.

Note : To find the square of ${\left( {\sin \dfrac{\pi }{{2n}} + \cos \dfrac{\pi }{{2n}}} \right)^2}$ , we use the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
Taking $a = \sin \dfrac{\pi }{{2n}}$ and $b = \cos \dfrac{\pi }{{2n}}$ into the formula, we have
$\Rightarrow {\left( {\sin \dfrac{\pi }{{2n}} + \cos \dfrac{\pi }{{2n}}} \right)^2} = {\left( {\sin \dfrac{\pi }{{2n}}} \right)^2} + {\left( {\cos \dfrac{\pi }{{2n}}} \right)^2} + 2\left( {\sin \dfrac{\pi }{{2n}}} \right)\left( {\cos \dfrac{\pi }{{2n}}} \right)$
On solving, we get
$\Rightarrow {\left( {\sin \dfrac{\pi }{{2n}} + \cos \dfrac{\pi }{{2n}}} \right)^2} = {\sin ^2}\dfrac{\pi }{{2n}} + {\cos ^2}\dfrac{\pi }{{2n}} + 2\sin \dfrac{\pi }{{2n}}\cos \dfrac{\pi }{{2n}}$
Product of two same numbers in root is the number itself.
i.e. $\sqrt 2 \times \sqrt 2 = 2$ or ${\left( {\sqrt 2 } \right)^2} = 2$ .