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Prove that: `4"cos"12^0cos48^0cos72^0=cos36^0`

Answer» `L.H.S. = 4cos12^@cos48^@cos72^@`
`=2*(2cos12^@cos48^@)cos72^@`
We know, `2cosAcosB = cos(A+B)+cos(A-B)`
So, our expression becomes,
`=2*(cos60^@+cos36^@)cos72^@`
`=2cos60^@cos72^@+2cos36^@cos72^@`
`=2(1/2)cos72^@+cos108^@+cos36^@`
`=cos72^@+cos(pi-72)^@+cos36^@`
`=cos72^@-cos72^@+cos36^@`
`=cos36^@`
`=R.H.S.`


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