Question

The value of $\cos \dfrac{\pi }{65}\cos \dfrac{2\pi }{65}\cos \dfrac{4\pi }{65}....\cos \dfrac{32\pi }{65}$ .
$\begin{align}
  & a)\dfrac{1}{32} \\
 & b)\dfrac{1}{64} \\
 & c)-\dfrac{1}{32} \\
 & d)-\dfrac{1}{64} \\
\end{align}$

Answer 1

Hint: Now to the given expression we will multiply $2\sin \dfrac{\pi }{65}$ to the numerator and denominator. Now we will use the formula $2\sin \theta \cos \theta =\sin 2\theta $ to simplify the equation. Now repeatedly we will multiply and divide the expression with 2 and use $2\sin \theta \cos \theta =\sin 2\theta $ till finally the expression simplifies to one term in the numerator. Now we know that $\sin \left( \pi -\theta \right)=\sin \theta $ hence using this property we will get the value of the given expression.

Complete step by step answer:
Now consider the given equation $\cos \dfrac{\pi }{65}\cos \dfrac{2\pi }{65}\cos \dfrac{4\pi }{65}....\cos \dfrac{32\pi }{65}$
Now let us multiply and divide the whole equation by $2\sin \dfrac{\pi }{65}$ hence we get,
$\dfrac{1}{2\sin \dfrac{\pi }{65}}\times 2\sin \dfrac{\pi }{65}\cos \dfrac{\pi }{65}\cos \dfrac{2\pi }{65}\cos \dfrac{4\pi }{65}....\cos \dfrac{32\pi }{65}$ .
Now we know that $2\sin \theta \cos \theta =\sin 2\theta $ Hence using this we can say that $2\sin \dfrac{\pi }{65}\cos \dfrac{\pi }{65}=\sin \dfrac{2\pi }{65}$ . Now substituting this in the above equation we get,
$\dfrac{1}{2\sin \dfrac{\pi }{65}}\times \sin \dfrac{2\pi }{65}\cos \dfrac{2\pi }{65}\cos \dfrac{4\pi }{65}....\cos \dfrac{32\pi }{65}$ .
Now let us multiply and divide the whole expression by 2. Hence we get,
$\dfrac{1}{2\times 2\sin \dfrac{\pi }{65}}\times 2\sin \dfrac{2\pi }{65}\cos \dfrac{2\pi }{65}\cos \dfrac{4\pi }{65}....\cos \dfrac{32\pi }{65}$
Now again using $2\sin \theta \cos \theta =\sin 2\theta $ we can say that $2\sin \dfrac{2\pi }{65}\cos \dfrac{2\pi }{65}=\sin \dfrac{4\pi }{65}$ .
Let us substitute this equation in the above expression. Hence we get,
$\dfrac{1}{{{2}^{2}}\sin \dfrac{\pi }{65}}\times \sin \dfrac{4\pi }{65}\cos \dfrac{4\pi }{65}....\cos \dfrac{32\pi }{65}$
Now again we will multiply and divide the expression by 2. With this we get,
$\dfrac{1}{{{2}^{3}}\sin \dfrac{\pi }{65}}\times 2\sin \dfrac{4\pi }{65}\cos \dfrac{4\pi }{65}\cos \dfrac{8\pi }{65}....\cos \dfrac{32\pi }{65}$
Again using $2\sin \theta \cos \theta =\sin 2\theta $ we get,
$\dfrac{1}{{{2}^{3}}\sin \dfrac{\pi }{65}}\times \sin \dfrac{8\pi }{65}\cos \dfrac{8\pi }{65}....\cos \dfrac{32\pi }{65}$
Again let us divide and multiply by 2. Now the expression above becomes,
$\dfrac{1}{{{2}^{4}}\sin \dfrac{\pi }{65}}\times 2\sin \dfrac{8\pi }{65}\cos \dfrac{8\pi }{65}....\cos \dfrac{32\pi }{65}$
Again using the formula $2\sin \theta \cos \theta =\sin 2\theta $ we simplify the above expression.
$\dfrac{1}{{{2}^{4}}\sin \dfrac{\pi }{65}}\times \sin \dfrac{16\pi }{65}\cos \dfrac{16\pi }{65}\cos \dfrac{32\pi }{65}$
Let us again multiply and divide the equation by 4 and use the formula $2\sin \theta \cos \theta =\sin 2\theta $
\[\begin{align}
  & \dfrac{1}{{{2}^{6}}\sin \dfrac{\pi }{65}}\times 2\times 2\sin \dfrac{16\pi }{65}\cos \dfrac{16\pi }{65}\cos \dfrac{32\pi }{65} \\
 & =\dfrac{1}{{{2}^{6}}\sin \dfrac{\pi }{65}}\times 2\sin \dfrac{32}{65}\cos \dfrac{32\pi }{65} \\
 & =\dfrac{1}{{{2}^{6}}\sin \dfrac{\pi }{65}}\times \sin \dfrac{64}{65} \\
\end{align}\]
Now we know that \[\dfrac{64\pi }{65}=\dfrac{65\pi -64\pi }{65}=\pi -\dfrac{\pi }{65}\]
Hence substituting the value in the above equation we get,
$\dfrac{1}{{{2}^{6}}\sin \dfrac{\pi }{65}}\times \sin \left( \pi -\dfrac{\pi }{65} \right)$
Now again we know that $\sin \left( \pi -\theta \right)=\sin \theta $ using this property we get,
$\dfrac{1}{{{2}^{6}}\sin \dfrac{\pi }{65}}\times \sin \left( \dfrac{\pi }{65} \right)=\dfrac{1}{{{2}^{6}}}=\dfrac{1}{64}$
Hence $\cos \dfrac{\pi }{65}\cos \dfrac{2\pi }{65}\cos \dfrac{4\pi }{65}....\cos \dfrac{32\pi }{65}=\dfrac{1}{64}$
Hence the value of the given expression is $\dfrac{1}{64}$

So, the correct answer is “Option b”.

Note: Now note that with each step we multiply the denominator by 2. If we observe the pattern we can directly write the last step. Hence in such sum instead of solving each step look for the pattern and write the expression accordingly.