Question

Prove that:
`sin6^(@)sin42^(@)sin66^(@)sin78^(@)=1/16`

Answer 1

LHS `=sin6^(@)sin42^(@)sin66^(@)sin78^(@)`
`=1/4(2sin66^(@)sin6^(@)).(2sin78^(@)sin42^(@))`
`=1/4[cos(66^(@)-6^(@))-cos(66^(@)+6^(@))] [cos(78^(@)-42^(@))-cos(78^(@)+42^(@))]`
`=1/4[cos60^(@)-cos72^(@)][cos36^(@)-cos120^(@)]`
`=1/4[1/2-cos(90^(@)-18^(@))][cos36^(@)-cos(90^(@)+30^(@))]`
`=1/4[1/2-sin18^(@)][cos36^(@)+sin30^(@)]`
`=1/4[1/2-(sqrt(5)-1)/(4)][(sqrt(5)+1)/(4)+1/2]`
`=1/4[(2-sqrt(5)+1)/(4)][(sqrt(5)+1+2)/(4)]`
`=3-sqrt(5)(3+sqrt(5))/(64)=(9-5)/(64)`
`=4/64=1/16`= RHS Hence Proved.