4cosa.cos (60-a) .cos (60 + a) = cos3a?

Useremo

rarr2cosAcosB=cos(A+B)+cos(A-B)2cosAcosB=cos(A+B)+cos(AB)

LHS=4cosxcos(60^@-x)cos(60^@+x)LHS=4cosxcos(60x)cos(60+x)

=2cosx*[2cos(60^@+x)cos(60^@-x)]=2cosx[2cos(60+x)cos(60x)]

=2cosx*[cos(60^@+x+60^@-x)+cos(60^@+x-60^@+x)]=2cosx[cos(60+x+60x)+cos(60+x60+x)]

=2cosx[cos120^@+cos2x]=2cosx[cos120+cos2x]

=2cosx[cos2x-1/2]=2cosx[cos2x12]

=cancel(2)cosx[(2cos2x-1)/cancel(2)]

=2cos2x*cosx-cosx

=cos(2x+x)+cos(2x-x)-cosx

=cos3xcancel(+cosx)cancel(-cosx)=cos3x=RHS

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