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Question: If \( {2^{n - 1}}\left( {\cos \theta - \cos \dfrac{\pi }{n}} \right)\left( {\cos \theta - \cos \dfra...

If 2n−1(cos⁡θ−cos⁡πn)(cos⁡θ−cos⁡2πn)...(cos⁡θ−cos⁡n−1nπ){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. sin⁡θsin⁡(nθ)\dfrac{{\sin \theta }}{{\sin \left( {n\theta } \right)}}
B. sin⁡nθsin⁡θ\dfrac{{\sin n\theta }}{{\sin \theta }}
C. 11
D. −ncos⁡θ- n\cos \theta

Explanation

Solution

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 nn factors in the required product, we will use the equation z2n−1=0{z^{2n}} - 1 = 0 where zz is a complex number, and will further simplify.

Complete step by step solution:
We have to find the product of 2n−1(cos⁡θ−cos⁡πn)(cos⁡θ−cos⁡2πn)...(cos⁡θ−cos⁡n−1nπ){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 nn factors multiplied together.
The nn roots of unity are given as the solution of zn−1=0{z^n} - 1 = 0 where rth{r^{th}} root is zr=cos⁡2πrn+isin⁡2πrn{z_r} = \cos \dfrac{{2\pi r}}{n} + i\sin \dfrac{{2\pi r}}{n} and rr varies from 11 to nn .
We will use the equation z2n−1=0{z^{2n}} - 1 = 0 where zz is a complex number.
This equation will have 2n2n roots of unity where rth{r^{th}} root will be given as,
zr=cos⁡2πr2n+isin⁡2πr2n{z_r} = \cos \dfrac{{2\pi r}}{{2n}} + i\sin \dfrac{{2\pi r}}{{2n}} where rr varies from 11 to 2n2n .
⇒zr=cos⁡πrn+isin⁡πrn\Rightarrow {z_r} = \cos \dfrac{{\pi r}}{n} + i\sin \dfrac{{\pi r}}{n}
We have,
z1=cos⁡πn+isin⁡πn z2=cos⁡2πn+isin⁡2πn   {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,
zn−1=cos⁡(n−1)πn+isin⁡(n−1)πn zn=cos⁡nπn+isin⁡nπn=cos⁡π+isin⁡π=−1 zn+1=cos⁡(n+1)πn+isin⁡(n+1)πn=cos⁡(2π−(n−1)πn)+isin⁡(2π−(n−1)πn)=cos⁡(n−1)πn−isin⁡(n−1)πn   {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,
z2n−2=cos⁡(2n−2)πn+isin⁡(2n−2)πn=cos⁡(2π−2πn)+isin⁡(2π−2πn)=cos⁡2πn−isin⁡2πn z2n−1=cos⁡(2n−1)πn+isin⁡(2n−1)πn=cos⁡(2π−πn)+isin⁡(2π−πn)=cos⁡πn−isin⁡πn z2n=cos⁡2nπn+isin⁡2nπn=cos⁡2π+isin⁡2π=1   {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 z2n−1{z_{2n - 1}} is the conjugate of z1{z_1} , z2n−2{z_{2n - 2}} is the conjugate of z2{z_2} and zn+1{z_{n + 1}} is the conjugate of zn−1{z_{n - 1}} . Thus, for any rr , z2n−r{z_{2n - r}} is the conjugate of zr{z_r} . Also, zn=−1{z_n} = - 1 and z2n=1{z_{2n}} = 1 .
Since z1,z2,...,z2n−1,z2n{z_1},{z_2},...,{z_{2n - 1}},{z_{2n}} are the roots of the equation z2n−1=0{z^{2n}} - 1 = 0 , we can write,
z2n−1=(z−z1)(z−z2)...(z−z2n−1)(z−z2n){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,
⇒z2n−1=(z−z1)(z−z2n−1)(z−z2)(z−z2n−2)...(z−zn−1)(z−zn+1)(z−zn)(z−z2n)\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,
(z−z1)(z−z2n−1)=(z2−(z1+z2n−1)z+z1z2n−1) =(z2−(cos⁡πn+isin⁡πn+cos⁡πn−isin⁡πn)z+(cos⁡πn+isin⁡πn)(cos⁡πn−isin⁡πn)) =(z2−2zcos⁡πn+1)   \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,

⇒z2n−1=(z2−2zcos⁡πn+1)(z2−2zcos⁡2πn+1)...(z2−2zcos⁡(n−1)πn+1)(z+1)(z−1) ⇒z2n−1=(z2−2zcos⁡πn+1)(z2−2zcos⁡2πn+1)...(z2−2zcos⁡(n−1)πn+1)(z2−1) ⇒z2n−1zn=(z2−2zcos⁡πn+1)(z2−2zcos⁡2πn+1)...(z2−2zcos⁡(n−1)πn+1)(z2−1)zn ⇒zn−1zn=(z2−2zcos⁡πn+1z)(z2−2zcos⁡2πn+1z)...(z2−2zcos⁡(n−1)πn+1z)(z2−1z) ⇒zn−1zn=(z−2cos⁡πn+1z)(z−2cos⁡2πn+1z)...(z−2cos⁡(n−1)πn+1z)(z−1z)   \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=(cos⁡θ+isin⁡θ)z = \left( {\cos \theta + i\sin \theta } \right) . Then 1z=z−1=(cos⁡θ+isin⁡θ)−1=(cos⁡(−θ)+isin⁡(−θ))=(cos⁡θ−isin⁡θ)\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+1z=(cos⁡θ+isin⁡θ)+(cos⁡θ−isin⁡θ)=2cos⁡θz + \dfrac{1}{z} = \left( {\cos \theta + i\sin \theta } \right) + \left( {\cos \theta - i\sin \theta } \right) = 2\cos \theta
And, z−1z=(cos⁡θ+isin⁡θ)−(cos⁡θ−isin⁡θ)=2isin⁡θz - \dfrac{1}{z} = \left( {\cos \theta + i\sin \theta } \right) - \left( {\cos \theta - i\sin \theta } \right) = 2i\sin \theta
Similarly, zn=(cos⁡θ+isin⁡θ)n=(cos⁡nθ+isin⁡nθ){z^n} = {\left( {\cos \theta + i\sin \theta } \right)^n} = \left( {\cos n\theta + i\sin n\theta } \right)
And 1zn=z−n=(cos⁡θ+isin⁡θ)−n=(cos⁡(−nθ)+isin⁡(−nθ))=(cos⁡nθ−isin⁡nθ)\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, zn−1zn=(cos⁡nθ+isin⁡nθ)−(cos⁡nθ−isin⁡nθ)=2isin⁡nθ{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,

zn−1zn=(z−2cos⁡πn+1z)(z−2cos⁡2πn+1z)...(z−2cos⁡(n−1)πn+1z)(z−1z) ⇒2isin⁡nθ=(2cos⁡θ−2cos⁡πn)(2cos⁡θ−2cos⁡2πn)...(2cos⁡θ−2cos⁡(n−1)πn)(2isin⁡θ) ⇒sin⁡nθsin⁡θ=2n−1(cos⁡θ−cos⁡πn)(cos⁡θ−cos⁡2πn)...(cos⁡θ−cos⁡(n−1)πn)   {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 sin⁡nθsin⁡θ\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.