Question

If $ {2^{n - 1}}\left( {\cos \theta - \cos \dfrac{\pi }{n}} \right)\left( {\cos \theta - \cos \dfrac{{2\pi }}{n}} \right)...\left( {\cos \theta - \cos \dfrac{{n - 1}}{n}\pi } \right) $ equals
A. $ \dfrac{{\sin \theta }}{{\sin \left( {n\theta } \right)}} $
B. $ \dfrac{{\sin n\theta }}{{\sin \theta }} $
C. $ 1 $
D. $ - n\cos \theta $

Answer 1

Hint: We have been given to find the product of the expression containing trigonometric terms. We can observe that the second term of each expression is in series. We can use the properties of the complex number to simplify and arrive at the required product. Since there are $ n $ factors in the required product, we will use the equation $ {z^{2n}} - 1 = 0 $ where $ z $ is a complex number, and will further simplify.

Complete step by step solution:
We have to find the product of $ {2^{n - 1}}\left( {\cos \theta - \cos \dfrac{\pi }{n}} \right)\left( {\cos \theta - \cos \dfrac{{2\pi }}{n}} \right)...\left( {\cos \theta - \cos \dfrac{{n - 1}}{n}\pi } \right) $ .
It has $ n $ factors multiplied together.
The $ n $ roots of unity are given as the solution of $ {z^n} - 1 = 0 $ where $ {r^{th}} $ root is $ {z_r} = \cos \dfrac{{2\pi r}}{n} + i\sin \dfrac{{2\pi r}}{n} $ and $ r $ varies from $ 1 $ to $ n $ .
We will use the equation $ {z^{2n}} - 1 = 0 $ where $ z $ is a complex number.
This equation will have $ 2n $ roots of unity where $ {r^{th}} $ root will be given as,
 $ {z_r} = \cos \dfrac{{2\pi r}}{{2n}} + i\sin \dfrac{{2\pi r}}{{2n}} $ where $ r $ varies from $ 1 $ to $ 2n $ .
\[ \Rightarrow {z_r} = \cos \dfrac{{\pi r}}{n} + i\sin \dfrac{{\pi r}}{n}\]
We have,
 $
  {z_1} = \cos \dfrac{\pi }{n} + i\sin \dfrac{\pi }{n} \\
  {z_2} = \cos \dfrac{{2\pi }}{n} + i\sin \dfrac{{2\pi }}{n} \;
  $
Similarly,
 $
  {z_{n - 1}} = \cos \dfrac{{\left( {n - 1} \right)\pi }}{n} + i\sin \dfrac{{\left( {n - 1} \right)\pi }}{n} \\
  {z_n} = \cos \dfrac{{n\pi }}{n} + i\sin \dfrac{{n\pi }}{n} = \cos \pi + i\sin \pi = - 1 \\
  {z_{n + 1}} = \cos \dfrac{{\left( {n + 1} \right)\pi }}{n} + i\sin \dfrac{{\left( {n + 1} \right)\pi }}{n} = \cos \left( {2\pi - \dfrac{{\left( {n - 1} \right)\pi }}{n}} \right) + i\sin \left( {2\pi - \dfrac{{\left( {n - 1} \right)\pi }}{n}} \right) = \cos \dfrac{{\left( {n - 1} \right)\pi }}{n} - i\sin \dfrac{{\left( {n - 1} \right)\pi }}{n} \;
  $
And,
 $
  {z_{2n - 2}} = \cos \dfrac{{\left( {2n - 2} \right)\pi }}{n} + i\sin \dfrac{{\left( {2n - 2} \right)\pi }}{n} = \cos \left( {2\pi - \dfrac{{2\pi }}{n}} \right) + i\sin \left( {2\pi - \dfrac{{2\pi }}{n}} \right) = \cos \dfrac{{2\pi }}{n} - i\sin \dfrac{{2\pi }}{n} \\
  {z_{2n - 1}} = \cos \dfrac{{\left( {2n - 1} \right)\pi }}{n} + i\sin \dfrac{{\left( {2n - 1} \right)\pi }}{n} = \cos \left( {2\pi - \dfrac{\pi }{n}} \right) + i\sin \left( {2\pi - \dfrac{\pi }{n}} \right) = \cos \dfrac{\pi }{n} - i\sin \dfrac{\pi }{n} \\
  {z_{2n}} = \cos \dfrac{{2n\pi }}{n} + i\sin \dfrac{{2n\pi }}{n} = \cos 2\pi + i\sin 2\pi = 1 \;
  $
We can observe that $ {z_{2n - 1}} $ is the conjugate of $ {z_1} $ , $ {z_{2n - 2}} $ is the conjugate of $ {z_2} $ and $ {z_{n + 1}} $ is the conjugate of $ {z_{n - 1}} $ . Thus, for any $ r $ , $ {z_{2n - r}} $ is the conjugate of $ {z_r} $ . Also, $ {z_n} = - 1 $ and $ {z_{2n}} = 1 $ .
Since $ {z_1},{z_2},...,{z_{2n - 1}},{z_{2n}} $ are the roots of the equation $ {z^{2n}} - 1 = 0 $ , we can write,
 $ {z^{2n}} - 1 = \left( {z - {z_1}} \right)\left( {z - {z_2}} \right)...\left( {z - {z_{2n - 1}}} \right)\left( {z - {z_{2n}}} \right) $
We can rearrange this to write conjugate terms together as,
 $ \Rightarrow {z^{2n}} - 1 = \left( {z - {z_1}} \right)\left( {z - {z_{2n - 1}}} \right)\left( {z - {z_2}} \right)\left( {z - {z_{2n - 2}}} \right)...\left( {z - {z_{n - 1}}} \right)\left( {z - {z_{n + 1}}} \right)\left( {z - {z_n}} \right)\left( {z - {z_{2n}}} \right) $
We can simplify one term for reference,
 $
  \left( {z - {z_1}} \right)\left( {z - {z_{2n - 1}}} \right) = \left( {{z^2} - \left( {{z_1} + {z_{2n - 1}}} \right)z + {z_1}{z_{2n - 1}}} \right) \\
   = \left( {{z^2} - \left( {\cos \dfrac{\pi }{n} + i\sin \dfrac{\pi }{n} + \cos \dfrac{\pi }{n} - i\sin \dfrac{\pi }{n}} \right)z + \left( {\cos \dfrac{\pi }{n} + i\sin \dfrac{\pi }{n}} \right)\left( {\cos \dfrac{\pi }{n} - i\sin \dfrac{\pi }{n}} \right)} \right) \\
   = \left( {{z^2} - 2z\cos \dfrac{\pi }{n} + 1} \right) \;
  $
Similarly we can simplify other terms to write,
\[
   \Rightarrow {z^{2n}} - 1 = \left( {{z^2} - 2z\cos \dfrac{\pi }{n} + 1} \right)\left( {{z^2} - 2z\cos \dfrac{{2\pi }}{n} + 1} \right)...\left( {{z^2} - 2z\cos \dfrac{{\left( {n - 1} \right)\pi }}{n} + 1} \right)\left( {z + 1} \right)\left( {z - 1} \right) \\
   \Rightarrow {z^{2n}} - 1 = \left( {{z^2} - 2z\cos \dfrac{\pi }{n} + 1} \right)\left( {{z^2} - 2z\cos \dfrac{{2\pi }}{n} + 1} \right)...\left( {{z^2} - 2z\cos \dfrac{{\left( {n - 1} \right)\pi }}{n} + 1} \right)\left( {{z^2} - 1} \right) \\
   \Rightarrow \dfrac{{{z^{2n}} - 1}}{{{z^n}}} = \dfrac{{\left( {{z^2} - 2z\cos \dfrac{\pi }{n} + 1} \right)\left( {{z^2} - 2z\cos \dfrac{{2\pi }}{n} + 1} \right)...\left( {{z^2} - 2z\cos \dfrac{{\left( {n - 1} \right)\pi }}{n} + 1} \right)\left( {{z^2} - 1} \right)}}{{{z^n}}} \\
   \Rightarrow {z^n} - \dfrac{1}{{{z^n}}} = \left( {\dfrac{{{z^2} - 2z\cos \dfrac{\pi }{n} + 1}}{z}} \right)\left( {\dfrac{{{z^2} - 2z\cos \dfrac{{2\pi }}{n} + 1}}{z}} \right)...\left( {\dfrac{{{z^2} - 2z\cos \dfrac{{\left( {n - 1} \right)\pi }}{n} + 1}}{z}} \right)\left( {\dfrac{{{z^2} - 1}}{z}} \right) \\
   \Rightarrow {z^n} - \dfrac{1}{{{z^n}}} = \left( {z - 2\cos \dfrac{\pi }{n} + \dfrac{1}{z}} \right)\left( {z - 2\cos \dfrac{{2\pi }}{n} + \dfrac{1}{z}} \right)...\left( {z - 2\cos \dfrac{{\left( {n - 1} \right)\pi }}{n} + \dfrac{1}{z}} \right)\left( {z - \dfrac{1}{z}} \right) \;
 \]
Now let us assume $ z = \left( {\cos \theta + i\sin \theta } \right) $ . Then $ \dfrac{1}{z} = {z^{ - 1}} = {\left( {\cos \theta + i\sin \theta } \right)^{ - 1}} = \left( {\cos \left( { - \theta } \right) + i\sin \left( { - \theta } \right)} \right) = \left( {\cos \theta - i\sin \theta } \right) $
Thus, $ z + \dfrac{1}{z} = \left( {\cos \theta + i\sin \theta } \right) + \left( {\cos \theta - i\sin \theta } \right) = 2\cos \theta $
And, $ z - \dfrac{1}{z} = \left( {\cos \theta + i\sin \theta } \right) - \left( {\cos \theta - i\sin \theta } \right) = 2i\sin \theta $
Similarly, $ {z^n} = {\left( {\cos \theta + i\sin \theta } \right)^n} = \left( {\cos n\theta + i\sin n\theta } \right) $
And $ \dfrac{1}{{{z^n}}} = {z^{ - n}} = {\left( {\cos \theta + i\sin \theta } \right)^{ - n}} = \left( {\cos \left( { - n\theta } \right) + i\sin \left( { - n\theta } \right)} \right) = \left( {\cos n\theta - i\sin n\theta } \right) $
Thus, \[{z^n} - \dfrac{1}{{{z^n}}} = \left( {\cos n\theta + i\sin n\theta } \right) - \left( {\cos n\theta - i\sin n\theta } \right) = 2i\sin n\theta \]
Thus we have,
\[
  {z^n} - \dfrac{1}{{{z^n}}} = \left( {z - 2\cos \dfrac{\pi }{n} + \dfrac{1}{z}} \right)\left( {z - 2\cos \dfrac{{2\pi }}{n} + \dfrac{1}{z}} \right)...\left( {z - 2\cos \dfrac{{\left( {n - 1} \right)\pi }}{n} + \dfrac{1}{z}} \right)\left( {z - \dfrac{1}{z}} \right) \\
   \Rightarrow 2i\sin n\theta = \left( {2\cos \theta - 2\cos \dfrac{\pi }{n}} \right)\left( {2\cos \theta - 2\cos \dfrac{{2\pi }}{n}} \right)...\left( {2\cos \theta - 2\cos \dfrac{{\left( {n - 1} \right)\pi }}{n}} \right)\left( {2i\sin \theta } \right) \\
   \Rightarrow \dfrac{{\sin n\theta }}{{\sin \theta }} = {2^{n - 1}}\left( {\cos \theta - \cos \dfrac{\pi }{n}} \right)\left( {\cos \theta - \cos \dfrac{{2\pi }}{n}} \right)...\left( {\cos \theta - \cos \dfrac{{\left( {n - 1} \right)\pi }}{n}} \right) \;
 \]
We can see that the RHS is the expression whose value we have to find. The required product is equal to \[\dfrac{{\sin n\theta }}{{\sin \theta }}\].
Hence, option (B) is the correct answer.
So, the correct answer is “Option B”.

Note: The approach in such questions can be confusing as nothing related to complex numbers is given in the question. We can only practice more and more questions to have the aptitude of approach. The problems of trigonometric series are often solved easily using complex numbers. Calculation error in such problems requiring messy calculations can be common and one should very carefully carry out the steps.