Can you simplify # cos(x)cos(2x)cos(4x)cos(8x)cos(16x) ... cos(2^n x) # ?,

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Øko Share
Feb 8, 2018

Answer:

#sin(2^(n+1x))/(2^(n+1)sin(x))#

Explanation:

We want to simplify

#cos(x)cos(2x)cos(4x)cos(8x)...cos(2^nx)#

The trick is to use the double-angle identity repeatedly

  • #sin(2x)=2cos(x)sin(x)#

Consider the following example

#sin(16x)=2cos(8x)sin(8x)#

#=2^2cos(8x)cos(4x)sin(4x)#

#=2^3cos(8x)cos(4x)cos(2x)sin(2x)#

#=2^4cos(8x)cos(4x)cos(2x)cos(x)sin(x)#

#=>cos(x)cos(2x)cos(4x)cos(8x)=sin(16x)/(2^4sin(x))#

If we generalize it

#sin(2^(n+1)x)=2cos(2^nx)sin(2^nx)#

#=2^2cos(2^(n)x)cos(2^(n-1)x)sin(2^(n-1)x)#

#.#
#.#
#.#

#=2^(n)cos(2^(n)x)cos(2^(n-1)x)...cos(4x)cos(2x)sin(2x)#

#=2^(n+1)cos(2^(n)x)cos(2^(n-1)x)...cos(2x)cos(x)sin(x)#

#=>cos(x)cos(2x)cos(4x)cos(8x)...cos(2^nx)=sin(2^(n+1)x)/(2^(n+1)sin(x))#

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