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