Prove that `cos6^@ . cos42^@. cos66^@ cos78^@= 1/16`

Explanation

`{LHS}=cos 6^{circ} cos 42^{circ} cos 66^{circ} cos 78^{circ}`


By regrouping the LHS and multiplying and dividing by 4 we get, `=frac{1}{4}(2 sin 66^{circ} sin 6^{circ})(2 sin 78^{circ} sin 42^{circ})`


`But 2 sin A sin B=cos (A-B)-cos (A+B)`


Then the above equation becomes,



`=frac{1}{4}(cos (66^{circ}-6^{circ})-cos (66^{circ}+6^{circ})) `


`=frac{1}{4}(cos (78^{circ}-42^{circ})-cos (78^{circ}+42^{circ})) `


`=frac{1}{4}(cos (60^{circ})-cos (72^{circ}))(cos (36^{circ}).


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