4cosa.cos (60-a) .cos (60 + a) = cos3a?
Useremo
rarr2cosAcosB=cos(A+B)+cos(A-B)→2cosAcosB=cos(A+B)+cos(A−B)
LHS=4cosxcos(60^@-x)cos(60^@+x)LHS=4cosxcos(60∘−x)cos(60∘+x)
=2cosx*[2cos(60^@+x)cos(60^@-x)]=2cosx⋅[2cos(60∘+x)cos(60∘−x)]
=2cosx*[cos(60^@+x+60^@-x)+cos(60^@+x-60^@+x)]=2cosx⋅[cos(60∘+x+60∘−x)+cos(60∘+x−60∘+x)]
=2cosx[cos120^@+cos2x]=2cosx[cos120∘+cos2x]
=2cosx[cos2x-1/2]=2cosx[cos2x−12]
=cancel(2)cosx[(2cos2x-1)/cancel(2)]
=2cos2x*cosx-cosx
=cos(2x+x)+cos(2x-x)-cosx
=cos3xcancel(+cosx)cancel(-cosx)=cos3x=RHS