Question #f1264

Question

781 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-21T09:13:15

Please refer to the explanation below.

#(pi/9)=20°, (2pi)/9=40°, (4pi)/9=80°#

#LHS=sin(pi/9)*sin((2pi)/9)*sin((4pi)/9)#

#=sin20°sin40°sin80°#

#=(2sin20°sin40°sin80°)/2#

#=((cos(40°-20°)-cos(40°+20°))*sin80°)/2#

#=((cos20°-cos60°)*sin80°)/2#

#=((cos20°-1/2)*sin80°)/2#

#=(((2cos20°-1)/2)*sin80°)/2#

#=((2cos20°-1)*sin80)/4#

#=(2cos20°sin80°-sin80°)/4#

#=(sin(80°+20°)+sin(80°-20°)-sin80°)/4#

#=(sin100°+sin60°-sin80°)/4#

#=(sin(180°-100°)+sin60°-sin80°)/4#

#=(cancel(sin80°)+sin60°cancel(-sin80°))/4#

#=(sqrt(3)/2)/4=sqrt(3)/8=RHS#