How do you evaluate #sin{pi/7} sin{(2pi)/7} sin{(3pi)/7}#?

Question

27600 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-16T15:42:52

Let #x=sin(pi/7)sin((2pi)/7)sin((3pi)/7)#
and

#y=cos(pi/7)cos((2pi)/7)cos((3pi)/7)#

So

#8xy=2sin(pi/7)cos(pi/7) xx2sin((2pi)/7)cos((2pi)/7)xx2sin((3pi)/7)cos((3pi)/7)#

#=sin((2pi)/7)sin((4pi)/7)sin((6pi)/7)#

#=sin((2pi)/7)sin(pi-(3pi)/7)sin(pi-pi/7)#

#=sin((2pi)/7)sin((3pi)/7)sin(pi/7)=x#

Hence #y=1/8#

Now let

#cos(pi/7)=a,cos((2pi)/7)=b,cos((3pi)/7)=c#

So

#y=cos(pi/7)cos((2pi)/7)cos((3pi)/7)=abc=1/8#

Again

#8x^2=8sin^2(pi/7)sin^2((2pi)/7)sin^2((3pi)/7)#

#=[1-cos((2pi)/7)][1-cos((4pi)/7)][1-cos((6pi)/7)]#

#=[1-cos((2pi)/7)][1+cos((3pi)/7)][1+cos(pi/7)]#

#=(1-b)(1+c)(1+a)#

#=1+a-b+c+ac-ab-bc-abc#

#=1+a-b+c+ac-ab-bc-1/8#

#=>8x^2=7/8+a-b+c+ac-ab-bc#

Now

#ac-ab-bc=1/2(2ac-2ab-2bc)#

#=1/2[2cos(pi/7)cos((3pi)/7) -2cos(pi/7)cos((2pi)/7)-2cos((2pi)/7)cos((3pi)/7)]#

#=1/2[cos((4pi)/7)+cos((2pi)/7) -cos((3pi)/7)-cos(pi/7)-cos((5pi)/7)-cos(pi/7)]#

#=1/2[-cos((3pi)/7)+cos((2pi)/7) -cos((3pi)/7)+cos((2pi)/7)-2cos(pi/7)]#

#=1/2[2cos((2pi)/7) -2cos((3pi)/7)-2cos(pi/7)]#

#=b-a-c#

#=>ac-ab-bc+a-b+c=0#

Hence we get

#=>8x^2=7/8+a-b+c+ac-ab-bc#

#=>8x^2=7/8+0#

#=>x^2=7/64#

#=>x=sqrt7/8#

#=>sin(pi/7)sin((2pi)/7)sin((3pi)/7)=sqrt7/8#