We will use the following Identities :=
# (1) : 2sin(theta/2)cos(theta/2)=sintheta.#
#(2) : cos(theta/2)=sqrt((1+costheta)/2)=((1+costheta)/2)^(1/2).#
#(3) : lim_(thetararr0) sintheta/theta=1#.
The Exp.#=cos(x/2)cos(x/4)cos(x/8)#.
In the First Step, we multiply by #2sin(x/2)# in #Nr. & Dr.#, to get,
The Exp.#={(2cos(x/2)sin(x/2))cos(x/4)cos(x/8)}/(2sin(x/2)#
#=(sinxcos(x/4)cos(x/8))/(2sin(x/2))................[by, (1)]#
We continue in the same fashion, and, find that,
The Exp.#=((sinx)(2sin(x/4)cos(x/4))cos(x/8))/((2sin(x/2)2sin(x/4))#
#=(sinxcancelsin(x/2)cos(x/8))/(4cancelsin(x/2)sin(x/4))#
#=(sinxcos(x/8))/(4sin(x/4)#
#=(sinx(2sin(x/8)cos(x/8)))/(8sin(x/4)sin(x/8))#
#=(sinxcancelsin(x/4))/(8cancelsin(x/4)sin(x/8))#
#=sinx/(8sin(x/8))#
Thus, we have proved that,
#cos(x/2)cos(x/4)cos(x/8)=sinx/(8sin(x/8))#
Generalising,
#cos(x/2)cos(x/2^2)cos(x/2^3)...cos(x/2^n)=sinx/(2^nsin(x/2^n)#
#={(sinx/x)/(sin(x/2^n)/(x/2^n))}................(star)#
Now, as #nrarroo, 2^nrarroo, x/2^nrarr0, sin(x/2^n)/(x/2^n)rarr1#.
#:.", by "(star),#
#cos(x/2)cos(x/2^2)cos(x/2^3)...cos(x/2^n)......=sinx/x#
To get the desired result, let us take, #x=pi/2# in this last eqn.,
#cos(pi/4)cos(pi/8)cos(pi/16).......=sin(pi/2)/(pi/2)=2/pi#, whereas,
On the. L.H.S., #cos(pi/4)=sqrt2/2,#
#cos(pi/8)=sqrt((1+cos(pi/4))/2)=sqrt((2+sqrt2)/4)=(sqrt(2+sqrt2))/2#
Therefore,
#sqrt2/2*(sqrt(2+sqrt2))/2*............=2/pi#
Enjoy maths.!
