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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 =?

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Solution

## 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

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