Question

Can you simplify \[\cos (x)\ \cos (2x)\ \cos (4x)\ \cos (8x)\ \ldots..\ \left( {cos\ }2^{n}x \right)\] ?

Answer 1

Hint: In this question, we need to simplify the given expression. Sine, cosine and tangent are known as the basic trigonometric function. In order to simplify the given expression ,we need to use the concepts of trigonometric identities. Cosine function is nothing but a ratio of the adjacent side of a right angle to the hypotenuse of the right angle. With the help of double angle identity,first , we need to find \[\sin (16x)\] then we need to find \[\sin (8x)\] . By proceeding this, we can easily simplify the given expression.
Double angle identity :
\[\sin (2x)\ = 2\sin (x)\cos (x)\]

Complete step-by-step solution:
Given,
\[\cos (x)\ \cos (2x)\ \cos (4x)\ \cos (8x)\ \ldots..\ \left( {cos\ }2^{n}x \right)\]
Let us consider the given expression as \[f(x)\],
\[f(x)\ = \cos (x)\ \cos (2x)\ \cos (4x)\ \cos (8x)\ \ldots..\ ({cos\ }2^{n}x)\]
By using double angle identity, \[\sin (2x)\ = 2\sin (x)\cos (x)\]
When
\[\sin (16x)\ = 2\sin (8x)\ \cos (8x)\]
Now we need to find \[\sin (8x)\] ,
We get
\[= \ 2 \times 2\sin \left( 4x \right)\cos\left( 4x \right)\cos\left( 8x \right)\]
Now we can find \[\sin (4x)\],
We get,
\[= \ 2 \times 2 \times 2\sin \left( 2x \right)\cos\left( 2x \right)\cos\left( 4x \right)\cos\left( 8x \right)\]
Then finally we need to find \[\sin (2x)\] ,
\[= \ 2 \times 2 \times 2 \times 2\sin (x)\ \cos (x)\ \cos (2x)\ \cos (4x)\ \cos (8x)\]
On simplifying,
We get,
\[\sin \left( 16x \right) = 2^{4}\sin\left( x \right)\cos\left( x \right)\cos\left( 2x \right)\cos\left( 4x \right)\cos\left( 8x \right)\]
\[= \ \cos (x)\ \cos (2x)\ \cos (4x)\ \cos (8x) = \dfrac{\sin\left( 16x \right)}{2^{4}\sin\left( x \right)}\]
If we generalize the double angle identity ,
We get,
\[sin\left( 2^{n + 1}x \right) = 2sin\left( 2^{n}x \right)\cos\left( 2^{n}x \right)\]
Now we need to find \[\sin\left( 2^{n}x \right)\] ,
\[= \ 2 \times 2sin\left( 2^{n – 1}x \right)\cos\left( 2^{n – 1}x \right)\cos\left( 2^{n}x \right)\]
On proceeding this,
.
.
.
We get,
\[= \ 2^{n}\cos\left( 2^{n}x \right)cos(2^{n – 1})\cos\left( 4x \right)\cos\left( 2x \right)\sin\left( 2x \right)\]
Now we can find \[sin(2x)\] ,
\[= \ 2^{n + 1}\cos\left( 2^{n}x \right)cos(2^{n – 1})\ldots\ cos\left( 4x \right)\cos\left( 2x \right)\cos\left( x \right)\sin\left( x \right)\]
On solving
We get,
\[\Rightarrow \ 2^{n + 1}\sin \left( x \right)\cos\left( x \right)\cos\left( 2x \right)\cos\left( 4x \right)\cos\left( 8x \right)\ldots..\left( cos\ 2^{n}x \right)\]
Thus,
\[\sin \left( 2^{n + 1}x \right) = \ 2^{n + 1}\sin\left( x \right)\cos\left( x \right)\cos\left( 2x \right)\cos\left( 4x \right)\cos\left( 8x \right)\ldots..\left( cos\ 2^{n}x \right)\]
\[\Rightarrow \ cos\left( x \right)\cos\left( 2x \right)\cos\left( 4x \right)\cos\left( 8x \right)\ldots..\left( cos\ 2^{n}x \right) = \dfrac{\sin\left( 2^{n + 1}x \right)}{2^{n + 1}\sin\left( x \right)}\]
Thus we get \[\cos (x)\ \cos (2x)\ \cos (4x)\ \cos (8x)\ \ldots..\left( cos\ 2^{n}x \right)\] is equal to \[\dfrac{\sin\left( 2^{n + 1}x \right)}{2^{\left( n + 1 \right)}\ \sin(x)}\]
Final answer :
\[\cos (x)\ \cos (2x)\ \cos (4x)\ \cos (8x)\ \ldots..\left( cos\ 2^{n}x \right)\] is equal to \[\dfrac{\sin\left( 2^{n + 1}x \right)}{2^{\left( n + 1 \right)}\ \sin(x)}\]

Note: The concept used to simplify the given expression is trigonometric identities. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. Sine is nothing but it is defined as a ratio of the opposite side of a right angle to the hypotenuse of the right angle. The common technique used in this problem is the substitution rule with the use of trigonometric identities such as double angle identity.