Q13/098) If α and β are two different values of θ ( between 0 and 2π ) which satisfy the equation .................? . 6 cos θ + 8 sin θ = 9, Find the value of ..... sin ( α + β )

α and ß are the roots of the equation : 6 cosΘ + 8 sinΘ = 9
=> 6 cosα + 8 sinα = 9 ------------- (1) and
.....6 cos ß + 8 sin ß = 9 ------------- (2)
Subtracting (2) from (1), we have :
6(cosα - cos ß) + 8(sin α - sin ß) = 0
=> 4(sinα - sin ß) = 3(cos ß - cos α)
=> (sin α - sin ß)/(cos ß - cos α) = 3/4
=> [2 cos{(α + ß)/2} sin {(α - ß)/2}] / [2 sin {(α + ß)/2} sin {(α - ß)/2}] = 3/4
=> cot {(α + ß)/2} = 3/4
=> cot²{(α + ß)/2} = 9/16
=> 1 + cot²{(α + ß)/2} = 1 + 9/16 = 25/16
=> cosec²{(α + ß)/2} = 25/16
=> sin² {(α + ß)/2} = 16/25
=> sin {(α + ß)/2} = ± 4/5 --------- (3)
=> 1 - sin²{(α + ß)/2} = 1 - 16/25 = 9/25
=> cos²{(α + ß)/2} = 9/25
=> cos {(α + ß)/2} = ± 3/5 ---------- (4)
=> sin (α + ß) = 2 sin {(α + ß)/2} cos {(α + ß)/2} = ± 2(4/5)*(3/5) = ± 24/25

However as 6 cosα + 8 sinα = 9

so cosα and sinα both > 0 ( as if one is -ve then sum < 8) and so α is in 1st quadrant and similarly β

so ( α + β ) is either in 1st or 2nd quadrant so sin ( α + β )is positive and hence -24/25 need to be ruled out and hence it is 24/25.

Another solution

6 cos θ + 8 sin θ = 9
=> (3 / 5) cos θ + (4 / 5) sin θ = 9 / 10
=> sin (θ + φ) = 9/10

Clearly if
sin x = sin a
=> x = a or π - a
for given range

=> θ + φ = α + φ or π - (α + φ)
Clearly one of them is equal to ß + φ
=> π - (α + φ) = ß + φ
=> α + ß = π - 2φ
=> sin (α + ß) = sin (π - 2φ)
=> sin (α + ß) = sin 2φ
=> sin (α + ß) = 2 sin φ cos φ = 24 / 25