Question

If $\theta =\dfrac{\pi }{{{2}^{n}}-1}$, then $\cos \theta \cos 2\theta \cos {{2}^{2}}\theta ...\cos {{2}^{n-1}}\theta =\dfrac{1}{{{a}^{n}}}$. Find the value of a.

Answer 1

To solve the above question, we first find the value of ${{2}^{n}}\theta$ and that can be found from the term of $\theta =\dfrac{\pi }{{{2}^{n}}-1}$ and after that we place the value in the equation ${{2}^{n}}\cos \theta \cos 2\theta \cos {{2}^{2}}\theta ...\cos {{2}^{n-1}}\theta =1$ and then solve both the RHS and LHS to prove the question.

Complete step by step solution:
To find the value of a, we first need to convert the equation from $\cos \theta \cos 2\theta \cos {{2}^{2}}\theta ...\cos {{2}^{n-1}}\theta$ to ${{2}^{n}}\cos \theta \cos 2\theta \cos {{2}^{2}}\theta ...\cos {{2}^{n-1}}\theta$ as the value of the series ${{2}^{n}}\cos \theta \cos 2\theta \cos {{2}^{2}}\theta ...\cos {{2}^{n-1}}\theta =1$ is equal to 1. Now how to prove that let us see below:
Taking the equation ${{2}^{n}}\cos \theta \cos 2\theta \cos {{2}^{2}}\theta ...\cos {{2}^{n-1}}\theta =1$, we can write the value ${{2}^{n}}$ as $2\times 2\times 2\times ..\times 2$ and placing them as:
$\Rightarrow 2\cos \theta \times 2\cos 2\theta \times 2\cos {{2}^{2}}\theta ...\times 2\cos {{2}^{n- 1}}\theta =1$

Using the trigonometric conversion we change the above equation as:
$\Rightarrow \dfrac{1}{\sin \theta }\left[ 2\cos \theta \sin \theta \times 2\cos 2\theta \times 2\cos {{2}^{2}}\theta ...\times 2\cos {{2}^{n-1}}\theta \right]$
$\Rightarrow \dfrac{1}{\sin \theta }\left[ \sin 2\theta \times 2\cos 2\theta \times 2\cos {{2}^{2}}\theta...\times 2\cos {{2}^{n-1}}\theta \right]$
$\Rightarrow \dfrac{1}{\sin \theta }\left[ \sin {{2}^{2}}\theta \times 2\cos {{2}^{2}}\theta ...\times 2\cos {{2}^{n-1}}\theta \right]$
While solving with each value the final product we get is:
$\Rightarrow \dfrac{1}{\sin \theta }\left[ 2\sin {{2}^{n-1}}\theta \times \cos {{2}^{n-1}}\theta \right]$
$\Rightarrow \dfrac{\sin {{2}^{n}}\theta }{\sin \theta }$
$\Rightarrow \dfrac{\sin \left( \pi -{{2}^{n}}\theta \right)}{\sin \theta }$
$\Rightarrow \dfrac{\sin \left( \dfrac{{{2}^{n}}\pi +\pi -{{2}^{n}}\pi }{{{2}^{n}}+1} \right)}{\sin \theta }$
Placing the value of $\theta =\dfrac{\pi }{{{2}^{n}}-1}$ as:
$\Rightarrow \dfrac{\sin \left( \theta \right)}{\sin \theta }=1$
Therefore, the value of the series ${{2}^{n}}\cos \theta \cos 2\theta \cos {{2}^{2}}\theta ...\cos {{2}^{n- 1}}\theta =1$ and the placing the value of the $\cos \theta \cos 2\theta \cos {{2}^{2}}\theta ...\cos {{2}^{n-1}}\theta$ as $\dfrac{1}{{{2}^{n}}}$.
Now as given in the question $\cos \theta \cos 2\theta \cos {{2}^{2}}\theta ...\cos {{2}^{n-1}}\theta =\dfrac{1}{{{a}^{n}}}$,
We will equate $\dfrac{1}{{{2}^{n}}}=\dfrac{1}{{{a}^{n}}}$ giving us the value of $a$ as 2.

Note: Students may go wrong if they try to solve the question by replacing the value θ in the question from the start as doing so will only make the question difficult although the answer will come but the question will get more complicated therefore first solve the question and then lastly place the value θ.