Find the values of α and β , 0 < α , β < π 2 satisfying the following equation cos α cos β cos ( α + β ) = - 1 8 .

Given, cos α cos β cos ( α + β ) = - 1 8 ⇒ 8 cos α cos β cos ( α + β ) = - 1 ⇒ 4 cos ( α + β ) + cos ( α - β ) cos ( α + β ) = - 1 ⇒ 4 cos 2 ( α + β ) + 4 cos ( α - β ) cos ( α + β ) + 1 = 0 ⇒ 2 cos ( α + β ) + cos ( α - β ) 2 - cos 2 ( α - β ) + 1 = 0 ⇒ 2 cos ( α + β ) + cos ( α - β ) 2 = - sin 2 ( α - β ) This gives 2 cos ( α + β ) + cos ( α - β ) = 0 and sin ( α - β ) = 0 So, α = β and α + β = 2 π 3 ∵ 0 < α , β < π 2 ⇒ α = β = π 3

If cos ( A - B ) = 1 2 and sin ( A + B ) = 1 2 , find the smallest positive values of A and B .

Given, cos A - B = 1 2 ⇒ A - B = cos - 1 1 2 = π 3 . . . . ( 1 ) Also given that, sin ( A + B ) = 1 2 ⇒ A + B = sin - 1 1 2 = π 6 or 5 π 6 Taking A + B = π 6 . . . . ( 2 ) From equation ( 1 ) & ( 2 ) , we get A = π 4 and B = - π 12 but we have to calculate the smallest positive values So, now taking A + B = 5 π 6 . . . . ( 3 ) From equation ( 1 ) & ( 3 ) , we get A = 7 π 12 and B = π 4