Question

Question 1- sin 6 degree * sin 42 degree * sin 66 degree * sin 78 degree = ?

Question 2 - cos 6 degree* cos 42 degree * cos 66 degree * cos 78 degree =?

Answer 1

1 .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)


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


2. (cos 6) (cos 42)(cos 66) (cos 78)
= (1/4) * [2cos6 cos66] * [2cos42 cos78]
= (1/4) * (cos72 + cos60) * (cos120 + cos36)
= (1/4) (cos72 + 1/2) * (- 1/2 + cos36)
= (1/4) [- 1/4 + (1/2) (cos36 - cos72) + cos36cos72]
= (1/4) [- 1/4 + sin54sin18 + sin54 sin18]
= (1/4) [- 1/4 + 2sin54 sin18 cos18 / cos18]
= (1/4) [- 1/4 + 2sin54 sin36 / cos18]
= (1/4) [- 1/4 + (cos18 - cos90) / 2cos18]
= (1/4) [- 1/4 + 1/2]
= 1/16