 QUESTION

# For all $n\in N,\cos \theta \cos 2\theta \cos 4\theta .............cos{{2}^{n-1}}\theta$ equal toa) $\dfrac{\sin {{2}^{n}}\theta }{{{2}^{n}}\sin \theta }$ b) $\dfrac{\sin {{2}^{n}}\theta }{\sin \theta }$ c) $\dfrac{\cos {{2}^{n}}\theta }{{{2}^{n}}\cos 2\theta }$ d) $\dfrac{\cos {{2}^{n}}\theta }{{{2}^{n}}\sin \theta }$

Hint: Multiply and divide by $\sin \theta$ to the given expression. Use the trigonometric identity of $\sin {2}\theta$, which is given as
$\sin 2\theta =2\sin \theta \cos \theta$
Observe the given relation in the problem and put $\theta =2\theta ,{{2}^{2}}\theta ,{{2}^{3}}\theta {{..............2}^{n-1}}\theta$ in the expression mentioned above i.e.$\sin 2\theta =2\sin \theta \cos \theta$, to get values of $\sin {{2}^{2}}\theta ,\sin {{2}^{3}}\theta .........\sin {{2}^{n}}\theta$.

Let the value of the given expression in the problem be ‘M’. so, we have
$M=\cos \theta .\cos 2\theta .\cos 4\theta ...........\cos {{2}^{n-1}}\theta$ ………….. (i)
As we can observe the series with the angles of the cosine function that angles are increasing in powers of 2. It means the given series can be written as
$M=\cos {{2}^{0}}\theta \cos {{2}^{1}}\theta \cos {{2}^{2}}\theta \cos {{2}^{3}}\theta ............\cos {{2}^{n-1}}\theta$ …………….. (ii)
Now, we know that we do not have any direct identity for solving this expression, so let us multiply and divide the whole relation by $\sin \theta$ and try to convert the whole expression in a fraction. So, on multiplying and dividing the equation (ii) by $\sin \theta$, we get
$M=\dfrac{\sin \theta \cos \theta \cos 2\theta \cos {{2}^{2}}\theta ...........\cos {{2}^{n-1}}\theta }{\sin \theta }$ ……………… (iii)
Now, we know the identity of $\sin \theta$ can be given as
$\sin \theta =2\sin \theta \cos \theta$ ……………. (iv)
So, if we multiply the numerator by ‘2’, then we can replace the term $2\sin \theta \cos \theta$ by $\sin 2\theta$ using the identity given in the equation (iii). So, let us multiply and divide the equation (iii) by 2 and hence, we get
$M=\dfrac{2\sin \theta \cos \theta \cos 2\theta \cos {{2}^{2}}\theta ...........\cos {{2}^{n-1}}\theta }{2\sin \theta }$
Replace $2\sin \theta \cos \theta$ by $\sin \theta$ using equation (iv). So, we get
$M=\dfrac{\sin 2\theta \cos 2\theta \cos {{2}^{2}}\theta ..........\cos {{2}^{n-1}}\theta }{2\sin \theta }$ ………………. (v)
Now, we know that the result (iv) becomes$\sin 4\theta =2\sin 2\theta \cos 2\theta$, if we replace $\theta$ by $2\theta$ in the whole equation. So, we get
$\sin 4\theta =2\sin \theta \cos \theta$ ………………(vi)
Now, multiply and divide the equation by ‘2’ and hence convert the term $2\sin 2\theta \cos 2\theta$ by $\sin 4\theta$ using equation (vi). So, we get
\begin{align} & M=\dfrac{2\sin \theta \cos 2\theta \cos {{2}^{2}}\theta .............\cos {{2}^{n-1}}\theta }{2\times 2\sin \theta } \\ & M=\dfrac{\sin 4\theta .\cos {{2}^{2}}\theta ..........\cos {{2}^{n-1}}\theta }{2\times 2\sin \theta } \\ \end{align}
$M=\dfrac{\sin {{2}^{2}}\theta \cos {{2}^{2}}\theta ..........\cos {{2}^{n-1}}\theta }{{{2}^{2}}\sin \theta }$ ………… (vii)
Now, we can replace $\theta$ by $4\theta ,{{2}^{2}}\theta$in the equation (iv) and get that
$\sin 80=\sin {{2}^{3}}\theta =2\sin {{2}^{2}}\theta \cos {{2}^{2}}\theta$ ………………. (viii)
So, similarly, multiplying and divide the equation (vii) by ‘2’ and hence, we get
$M=\dfrac{2\sin {{2}^{2}}\theta \cos {{2}^{2}}\theta \cos {{2}^{3}}\theta .............\cos {{2}^{n-1}}\theta }{{{2}^{3}}\sin \theta }$
$M=\dfrac{\sin {{2}^{3}}\theta \cos {{2}^{3}}\theta .............\cos {{2}^{n-1}}\theta }{{{2}^{3}}\sin \theta }$………………(ix)
Now, we can proceed in the similar way to the nth term as well i.e. up to the $\cos {{2}^{n-1}}\theta$. And hence, only the power of 2in the denominator will change and we can continuously solve the numerator by using the identity (iv). So, we can get the result at last term as
$M=\dfrac{\sin {{2}^{n-1}}\theta \cos {{2}^{n-1}}\theta }{{{2}^{n-1}}\sin \theta }$ …………..(x)
Now, again multiply and divide the expression b ‘2’ and put $\theta ={{2}^{n-1}}$ in the expression (iv) and hence, get that
$\sin {{2}^{n}}\theta =2\sin {{2}^{n-1}}\theta \cos {{2}^{n-1}}\theta$
So, we get equation (x) as
$M=\dfrac{2.\sin {{2}^{n-1}}\theta \cos {{2}^{n-1}}\theta }{{{2.2}^{n-1}}\sin \theta }$
or
$M=\dfrac{\sin {{2}^{n}}\theta }{{{2}^{n}}\sin \theta }$……………. (xi)
Hence, value of the given expression in the problem is $\dfrac{sin{{2}^{n}}\theta }{{{2}^{n}}\sin \theta }$
So, option (a) is the correct answer.

Note: Multiplying and dividing by $\sin \theta$ to the given expression and converting it to the form of $\sin 2\theta =2\sin \theta \cos \theta$ is the key point of the question.
One may get the correct answer from the option by putting values of n to the expression and options. So, it can be another approach for getting write answers from the given options for these kinds of questions.
Observe the power of ‘2’ in denominator at the last step i.e.
$M=\dfrac{\sin {{2}^{n-1}}\theta \cos {{2}^{n-1}}\theta }{{{2}^{n-1}}\sin \theta }$ , very carefully.
One may go wrong if he/she puts power of 2 in the denominator as ‘n’ or ‘n – 2’ which are wrong. So, observe the power of ‘2’ in each step and don’t get confused with this step.