find the value of sin6°sin42°sin66°sin78°
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Answer is 1/16 i guesss.....i took 4 long pages to work it out
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Sin 6° Sin 66° Sin 42° Sin 78° = 1/16 [ 2 Cos 36° + 1 ] [ 1 - 2 Cos 72°] = 1/16 [ 2 Cos 36° - 2 cos 72° + 1 - 4 Cos 36° Cos 72° ] = 1/16 + 1/8 [ Cos 36° - Cos 72° - 2 Cos 36° Cos 72° ] = 1/16 + 1/8 [ Cos 36° - Cos 72° - ( Cos (36+72) + Cos (72-36) ) ] = 1/16 - 1/8 [ Cos 72° + Cos 108° ] = 1/16 - 1/8 [ Cos 72° - Cos (180° - 108°) ] = 1/16 - 1/8 [ Cos 72° - Cos 72° ] = 1/16
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