Question

The value of $\cos \left( \dfrac{\pi }{{{2}^{2}}} \right).\cos \left( \dfrac{\pi }{{{2}^{3}}} \right)......\cos \left( \dfrac{\pi }{{{2}^{10}}} \right)\times \sin \left( \dfrac{\pi }{{{2}^{10}}} \right)$ is
(a) $\dfrac{1}{256}$
(b) $\dfrac{1}{2}$
(c) $\dfrac{1}{512}$
(d) $\dfrac{1}{1024}$

Answer 1

We have to find the value of $\cos \left( \dfrac{\pi }{{{2}^{2}}} \right).\cos \left( \dfrac{\pi }{{{2}^{3}}} \right)......\cos \left( \dfrac{\pi }{{{2}^{10}}} \right).\sin \left( \dfrac{\pi }{{{2}^{10}}} \right)$
We have $\cos \theta .\cos \left( 2\theta \right).....\cos \left( {{2}^{n-1}}\theta \right)=\dfrac{\sin \left( {{2}^{n}}\theta \right)}{{{2}^{n}}\sin \theta }$
So we will take $\theta =\dfrac{\pi }{{{2}^{10}}}$ and then our original equation will become as follows. $\cos \theta .\cos \left( 2\theta \right)......\cos \left( {{2}^{8}}\theta \right)\times \sin \left( \theta \right)$ , now we will apply the above mentioned rule in order to simplify $\cos$ terms. And at last we will take back $\theta$ as $\dfrac{\pi }{{{2}^{10}}}$ and simplify it to get our solution.

Complete step-by-step answer:
We have been asked in the given question that we have to find the value of the expression, that is, $\cos \left( \dfrac{\pi }{{{2}^{2}}} \right).\cos \left( \dfrac{\pi }{{{2}^{3}}} \right)......\cos \left( \dfrac{\pi }{{{2}^{10}}} \right).\sin \left( \dfrac{\pi }{{{2}^{10}}} \right)$
And we know the relation that,
$\cos \theta .\cos \left( 2\theta \right).....\cos \left( {{2}^{n-1}}\theta \right)=\dfrac{\sin \left( {{2}^{n}}\theta \right)}{{{2}^{n}}\sin \theta }$
Now Let us consider the value of the term, $\dfrac{\pi }{{{2}^{10}}}$ as $\theta$
$\Rightarrow \theta =\dfrac{\pi }{{{2}^{10}}}$
Then $\dfrac{\pi }{{{2}^{9}}}=\dfrac{\pi }{{{2}^{10}}}\times 2$
$\text{ }\dfrac{\pi }{{{2}^{9}}}=2\theta \text{ }\left[ as\text{ }\theta =\dfrac{\pi }{{{2}^{10}}} \right]$
Similarly we will get,
$\begin{aligned} & \dfrac{\pi }{{{2}^{8}}}=\dfrac{\pi }{{{2}^{10}}}\times {{2}^{2}} \\\ & \text{ }={{2}^{2}}\theta \\\ \end{aligned}$
And so on upto,
$\dfrac{\pi }{{{2}^{2}}}=\dfrac{\pi }{{{2}^{10}}}\times {{2}^{8}}\text{ }\left[ as\text{ }{{2}^{10-8}}={{2}^{2}} \right]$
So, we will get,
$\dfrac{\pi }{{{2}^{2}}}={{2}^{8}}\theta$
So our equation which was given in the question that is,
$\cos \left( \dfrac{\pi }{{{2}^{2}}} \right).\cos \left( \dfrac{\pi }{{{2}^{3}}} \right)......\cos \left( \dfrac{\pi }{{{2}^{10}}} \right).\sin \left( \dfrac{\pi }{{{2}^{10}}} \right)$
Becomes as follows.
$\cos \left( {{2}^{8}}\theta \right).\cos \left( {{2}^{7}}\theta \right).\cos \left( {{2}^{6}}\theta \right)........\cos \left( \theta \right).\sin \theta$
And on rearranging the above expression a little bit we will get,
$\cos \left( \theta \right).\cos \left( 2\theta \right).\cos \left( {{2}^{2}}\theta \right).........\cos \left( {{2}^{8}}\theta \right).\sin \theta ----\left( 1 \right)$
Now we know the relation that,
$\cos \theta .\cos \left( 2\theta \right).\cos \left( {{2}^{2}}\theta \right).....\cos \left( {{2}^{n-1}}\theta \right)=\dfrac{\sin \left( {{2}^{n}}\theta \right)}{{{2}^{n}}\sin \theta }$
So, on applying this,
$\cos \theta .\cos \left( 2\theta \right)......\cos \left( {{2}^{8}}\theta \right)$
We have $n-1=8\text{ }\Rightarrow n=8+1=9$
So we will get,
$\cos \theta .\cos \left( 2\theta \right).\cos \left( {{2}^{2}}\theta \right).....\cos \left( {{2}^{8}}\theta \right)=\dfrac{\sin \left( {{2}^{9}}\theta \right)}{{{2}^{9}}\left( \sin \theta \right)}$
Now, on putting this value in $\left( 1 \right)$ we will get,
$\cos \theta .\cos \left( 2\theta \right).\cos \left( {{2}^{2}}\theta \right).....\cos \left( {{2}^{8}}\theta \right).\sin \theta =\dfrac{\sin \left( {{2}^{9}}\theta \right)}{{{2}^{9}}\left( \sin \theta \right)}\times \sin \theta$
So on simplifying it further we will get,
$\cos \theta .\cos \left( 2\theta \right).\cos \left( {{2}^{2}}\theta \right).....\cos \left( {{2}^{8}}\theta \right).\sin \theta =\dfrac{\sin \left( {{2}^{9}}\theta \right)}{{{2}^{9}}}$
Now, on putting $\theta$ back as $\dfrac{\pi }{{{2}^{10}}}$ , we will get the equation as,

& \cos \left( \dfrac{\pi }{{{2}^{10}}} \right).\cos \left( 2\dfrac{\pi }{{{2}^{10}}} \right).\cos \left( {{2}^{2}}\dfrac{\pi }{{{2}^{10}}} \right).....\cos \left( {{2}^{8}}\dfrac{\pi }{{{2}^{10}}} \right).\sin \dfrac{\pi }{{{2}^{10}}}=\dfrac{\sin \left( {{2}^{9}}\dfrac{\pi }{{{2}^{10}}} \right)}{{{2}^{9}}} \\\ & \Rightarrow \cos \left( \dfrac{\pi }{{{2}^{10}}} \right).\cos \left( 2\dfrac{\pi }{{{2}^{10}}} \right).\cos \left( {{2}^{2}}\dfrac{\pi }{{{2}^{10}}} \right).....\cos \left( {{2}^{8}}\dfrac{\pi }{{{2}^{10}}} \right).\sin \dfrac{\pi }{{{2}^{10}}}=\dfrac{\sin \left( \dfrac{\pi }{2} \right)}{{{2}^{9}}} \\\ \end{aligned}$$ We know that, $$\sin \left( \dfrac{\pi }{2} \right)=1$$ so we will get, $$\Rightarrow \cos \left( \dfrac{\pi }{{{2}^{10}}} \right).\cos \left( 2\dfrac{\pi }{{{2}^{10}}} \right).\cos \left( {{2}^{2}}\dfrac{\pi }{{{2}^{10}}} \right).....\cos \left( {{2}^{8}}\dfrac{\pi }{{{2}^{10}}} \right).\sin \dfrac{\pi }{{{2}^{10}}}=\dfrac{1}{{{2}^{9}}}$$ $$\Rightarrow \cos \left( \dfrac{\pi }{{{2}^{10}}} \right).\cos \left( 2\dfrac{\pi }{{{2}^{10}}} \right).\cos \left( {{2}^{2}}\dfrac{\pi }{{{2}^{10}}} \right).....\cos \left( {{2}^{8}}\dfrac{\pi }{{{2}^{10}}} \right).\sin \dfrac{\pi }{{{2}^{10}}}=\dfrac{1}{512}$$ Hence we get that the correct option is option (c), that is, $$\dfrac{1}{512}$$ **So, the correct answer is “Option c”.** **Note:** We have the equation, $\cos \theta .\cos \left( 2\theta \right).\cos \left( {{2}^{2}}\theta \right).....\cos \left( {{2}^{n-1}}\theta \right)=\dfrac{\sin \left( {{2}^{n}}\theta \right)}{{{2}^{n}}\sin \theta }$ So, errors like, taking $n-1$ as $n$ usually can happen while solving this question. Apply this on, $$\cos \theta .\cos \left( 2\theta \right).\cos \left( {{2}^{2}}\theta \right).....\cos \left( {{2}^{8}}\theta \right)$$ We will have ${{2}^{n-1}}={{2}^{8}}$ $\begin{aligned} & \text{ }\Rightarrow n-1=8 \\\ & \text{ }\Rightarrow n=8+1 \\\ & \text{ }=9 \\\ \end{aligned}$ Also, mistakes like, taking $n$ directly as $8$ will lead to getting an incorrect solution.