Question

$\cos A\cos 2A\cos 4A...\cos {2^{n - 1}}A$ equals
A) $\dfrac{{\sin {2^n}A}}{{{2^n}\sin A}}$
B) $\dfrac{{{2^n}\sin {2^n}A}}{{\sin A}}$
C) $\dfrac{{{2^n}\sin A}}{{\sin {2^n}A}}$
D) $\dfrac{{\sin A}}{{{2^n}\sin {2^n}A}}$

Answer 1

Hint:
We can take the given expression as x. Then we can multiply both sides of the equation with $2\sin A$ . Then we can apply the rule that $2\sin A\cos A = \sin 2A$. Then we can multiply both sides with 2 to make the RHS of the form $2\sin A\cos A$. Then we can again apply the equation $2\sin A\cos A = \sin 2A$. Thus, we can obtain a general result. Then we can write the equation in terms of x to obtain the required solution.

Complete step by step solution:
Let $x = \cos A\cos 2A\cos 4A...\cos {2^{n - 1}}A$ .. (1)
We can multiply both sides of the equation with$2\sin A$. Then the equation will become,
 $ \Rightarrow 2\sin A\,x = 2\sin A\cos A\cos 2A\cos 4A...\cos {2^{n - 1}}A$
We know that $2\sin A\cos A = \sin 2A$. On applying this condition on the RHS, we get,
 $ \Rightarrow 2\sin A\,x = \sin 2A\cos 2A\cos 4A...\cos {2^{n - 1}}A$
Now we can multiply both sides with 2. Then the equation will become,
 $ \Rightarrow {2^2}\sin A\,x = 2\sin 2A\cos 2A\cos 4A...\cos {2^{n - 1}}A$
Again, we can apply the relation $2\sin A\cos A = \sin 2A$ . So, the equation will become,
 $ \Rightarrow {2^2}\sin A\,x = \sin 4A\cos 4A...\cos {2^{n - 1}}A$
This procedure can be repeated n times until the last term becomes, $2\sin {2^{n - 1}}A\cos {2^{n - 1}}A = \sin {2^n}A$
So, the equation will become,
 $ \Rightarrow {2^n}\sin A\,x = \sin {2^n}A$
On dividing throughout with ${2^n}\sin A\,$ , we get,
 $ \Rightarrow \,x = \dfrac{{\sin {2^n}A}}{{{2^n}\sin A}}$ …. (2)
Therefore, from equation (1) and equation (2), we can obtain,
 $ \Rightarrow \cos A\cos 2A\cos 4A...\cos {2^{n - 1}}A = \dfrac{{\sin {2^n}A}}{{{2^n}\sin A}}$

So, the correct answer is option A.

Note:
Alternate solution is given by,
We know that $2\sin A\cos A = \sin 2A$ .
On rearranging, we get,
 $ \Rightarrow \cos A = \dfrac{{\sin 2A}}{{2\sin A}}$
On multiplying both sides with $\cos 2A$ , we get.
 \[ \Rightarrow \cos A\cos 2A = \dfrac{{\sin 2A\cos 2A}}{{2\sin A}}\]
Now we can multiply and divide the RHS of the equation by 2.
 \[ \Rightarrow \cos A\cos 2A = \dfrac{{2\sin 2A\cos 2A}}{{{2^2}\sin A}}\]
We know that $2\sin A\cos A = \sin 2A$ . On applying this, we get,
 \[ \Rightarrow \cos A\cos 2A = \dfrac{{\sin 4A}}{{{2^2}\sin A}}\]
On multiplying both sides with $\cos 4A$ , we get.
 \[ \Rightarrow \cos A\cos 2A\cos 4A = \dfrac{{\sin 4A\cos 4A}}{{{2^2}\sin A}}\]
Now we can multiply and divide the RHS of the equation by 2.
 \[ \Rightarrow \cos A\cos 2A\cos 4A = \dfrac{{2\sin 4A\cos 4A}}{{{2^3}\sin A}}\]
We know that $2\sin A\cos A = \sin 2A$ . On applying this, we get,
 \[ \Rightarrow \cos A\cos 2A\cos 4A = \dfrac{{\sin 8A}}{{{2^3}\sin A}}\]
We can write the equation as powers of 2.
 \[ \Rightarrow \cos A\cos {2^1}A\cos {2^2}A = \dfrac{{\sin {2^3}A}}{{{2^3}\sin A}}\]
From the powers of 2, we can write the equation generally as,
 \[ \Rightarrow \cos A\cos {2^1}A\cos {2^2}A....\cos {2^{n - 1}}A = \dfrac{{\sin {2^n}A}}{{{2^n}\sin A}}\]
Thus, we obtained the required solution.