Prove that #sin(pi/18)sin((5pi)/6)sin((5pi)/18)sin((7pi)/18)=1/16#?

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Answer 1 · 2017-11-13T13:42:49

#LHS=sin(pi/18)sin((5pi)/6)sin((5pi)/18)sin((7pi)/18)#

#=sin(pi/2-(8pi)/18)sin(pi/2-(4pi)/18)sin(pi/2-(2pi)/18)sin(pi-pi/6)#

#=cos((8pi)/18)cos((4pi)/18)cos((2pi)/18)sin(pi/6)#

#=16/(16sin((2pi)/18))sin((2pi)/18)cos((2pi)/18)cos((4pi)/18)cos((8pi)/18)*1/2#

#=1/(16sin((2pi)/18))*sin((16pi)/18)#

#=1/(16sin((2pi)/18))*sin(pi-(2pi)/18)#

#=1/(16sin((2pi)/18))*sin((2pi)/18)#

#=1/16=RHS#