Question

If $R={\cos }^2\left(\frac{n-1}{n}\right)\pi +{\cos }^2\left(\frac{n-2}{n}\right)\pi +\cdots +{\cos }^2\left(\frac{2\pi }{n}\right)+{\cos }^2\left(\frac{\pi }{n}\right)$, then value of $R$ is

• A. $\frac{n(n+1)}{2}$
• B. $\frac{n}{2}$
• C. $\frac{n-1}{2}$
• D. $\frac{n-2}{2}$

Answer 1

The correct option is D $\frac{n-2}{2}$

Step 1: Given:

$R={\cos }^2\left(\frac{n-1}{n}\right)\pi +{\cos }^2\left(\frac{n-2}{n}\right)\pi +{\cos }^2\left(\frac{n-3}{n}\right)\pi +\cdots +{\cos }^2\left(\frac{\pi }{n}\right)$

Step 2: To find:

The value of $R$.

Step 3: Assumption:

$n\in N$

Step 4: Concepts/Formulas used:

(1) $\cos 2A=2{\cos }^{2}A-1$

(2) $2\cos A\sin B=\sin (A+B)-\sin (A-B)$

Step 5: Solution:

From given, we find out

$R={\cos }^2\left(\frac{n-1}{n}\right)\pi +{\cos }^2\left(\frac{n-2}{n}\right)\pi +{\cos }^2\left(\frac{n-3}{n}\right)\pi +\cdots +{\cos }^2\left(\frac{\pi }{n}\right)$ ...(i)

Using formula (1) provide above in step 4, we get

$R=\frac{1}{2}(1-\cos 2\left(\frac{n-1}{n}\right)\pi +1-\cos 2\left(\frac{n-2}{n}\right)\pi +1-\cos 2\left(\frac{n-3}{n}\right)\pi +\cdots +1-\cos 2\left(\frac{\pi }{n}\right))$

$\Rightarrow R=\frac{1}{2}(n-1+[\cos 2\left(\frac{n-1}{n}\right)\pi +\cos 2\left(\frac{n-2}{n}\right)\pi +\cos 2\left(\frac{n-3}{n}\right)\pi +\cdots +\cos 2\left(\frac{\pi }{n}\right)])$ ...(ii)

Step 6:

Now wonder the term in square braces

$S=\cos 2\left(\frac{n-1}{n}\right)\pi +\cos 2\left(\frac{n-2}{n}\right)\pi +\cos 2\left(\frac{n-3}{n}\right)\pi +\cdots +\cos 2\left(\frac{\pi }{n}\right)$

Multiply both sides by $2\sin \frac{\pi }{n}$, we get

$2S\sin \frac{\pi }{n}=2\sin \frac{\pi }{n}(\cos 2\left(\frac{n-1}{n}\right)\pi +\cos 2\left(\frac{n-2}{n}\right)\pi +\cos 2\left(\frac{n-3}{n}\right)\pi +\cdots +\cos 2\left(\frac{\pi }{n}\right))$ ...(iii)

Step 7:

Using the formula(2) in the result of step 6, we get

$2S\sin \frac{\pi }{n}=\sum _{i=1}^{n-1}(\sin \left(\frac{2i+1}{n}\right)\pi -\sin \left(\frac{2i-1}{n}\right)\pi )$ ....(iv)

Step 8:

Using the method of telescopic sum, we get

$2S\sin \frac{\pi }{n}=\sin \frac{3\pi }{n}-\sin \frac{\pi }{n}+\sin \frac{5\pi }{n}-\sin \frac{3\pi }{n}+..\sin \frac{2n-1}{n}\pi -\sin \frac{2n-3}{n}\pi$

Solving it, we get,

$2S\sin \frac{\pi }{n}=\sin \frac{2n-1}{n}\pi -\sin \frac{\pi }{n}$

$\Rightarrow 2S\sin \frac{\pi }{n}=-\sin \frac{\pi }{n}-\sin \frac{\pi }{n}$

$\Rightarrow 2S\sin \frac{\pi }{n}-2\sin \frac{\pi }{n}$

$\Rightarrow S=-1$

Step 9:

Substituting the value of $S$ in (ii), we get

$R=\frac{1}{2}(n-1+(-1\left)\right)$

$\Rightarrow R=\frac{n-2}{2}$

Step 10: Result:

$R=\frac{n-2}{2}$

Hence, option D.