Find sin6°. sin42°. sin66°. sin78° =
Find sin6°. sin42°. sin66°. sin78° =
kiran , 7 Years ago
Grade 12
General Physics> Find sin6degre. sin42degree.sin66degree.s...
1 Answers
Arun
We use the trigonometric rules :
2 Sin A Sin B = Cos (A-B) - Cos (A+B)
2 Cos A Cos B = Cos (A-B) + Cos (A+B)
2 Sin A Sin B = Cos (A-B) - Cos (A+B)
2 Cos A Cos B = Cos (A-B) + Cos (A+B)
Sin 42° Sin 78° = 1/2 * [ Cos (42° - 78°) - Cos (78° + 42°) ]
= 1/2 * [ Cos 36° - Cos 120°]
= 1/2 [ Cos 36° + 1/2 ]
Sin 6° Sin 66° = 1/2 * [ Cos (6-66) - Cos (6+66) ]
= 1/2 [ Cos 60° - Cos 72° ]
= 1/2 [ 1/2 - Cos 72° ]
Hence, 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
= 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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