Question #5176d

1 Answer
Aug 11, 2016

See the Proof in Explanation Section.

Explanation:

Observe that there are #n# terms of #cos#, and, in #2^n#, the no. of #2# is

also #n#. So, we utilise each #2# with every term of #cos#.

We will also use the identity # : sin2theta=2sinthetacostheta#. Thus,

The#L.H.S.=(2cosx)(2cos2x)(2cos4x)(2cos8x)...(2cos2^(n-1)x)#

#={(2sinxcosx)(2cos2x)(2cos4x)...........(2cos2^(n-1)x)}/sinx#

#={(sin2x)(2cos2x)(2cos4x).........(2cos2^(n-1)x)}/sinx#

#={(2sin2xcos2x)(2cos4x).......(2cos2^(n-1)x)}/sinx#

#={(sin4x)(2cos4x)..........(2cos2^(n-1)x)}/sinx#

#={(2sin4xcos4x)..........(2cos2^(n-1)x)}/sinx#
#vdots#
#vdots#
#vdots#
#=sin(2*2^(n-1)x)/sinx#

#=sin(2^nx)/sinx#

But, #x=pi/(2^n+1) rArr 2^nx+x=pi, or, 2^nx=pi-x#, so that,

#sin(2^nx)=sin(pi-x)=sinx#. Therefore,

The #L.H.S.=sinx/sinx=1=The R.H.S#

Hence, the Proof. Enjoy Maths.!