If α,β are two different vlaues of θ lying between 0 and 2π which satisfy the equations 6 cos \theta + 8 sin\theta= 9, find the value of sin (α+β).
We have,
6 cos θ + 8 sin θ = 9 ... (i)
⇒8sinθ=9−6cosθ⇒(8sinθ)2=(9−6cosθ)2
[∵ Squaring both sides]
⇒64sin2θ=81+36cos2θ−108cosθ⇒64sin2θ=81+36cos2θ−108cosθ⇒64(1−cos2θ)=81+36cos2θ−108cosθ⇒64−64cos2θ+64cos2θ−108cosθ+81−64=0⇒100cos2θ−108cosθ+17=0 ...(ii),
Therefore, cosα and cosβ are roots of equation (ii)
∴cosα+cosβ=17100 ...(iii)
Again, 6 cosθ+8sinθ=9
⇒(6cosθ)2=(9−8sinθ)2
[∵ Squaring both sides]
⇒36cos2θ=81+64sin2θ−144sinθ⇒36(1−sin2θ)=81+64sin2θ−144cosθ⇒36−36sin2θ=81+64sin2θ−144sinθ⇒64sin2θ+36sin2θ−144sinθ+81−36=0
⇒100sin2θ−144sinθ+45=0 ...(iv)
It is given that α,β are roots of equation (ii). So, sinα and sinβ are the roots of eqaution (iv)
∴sinα×sinβ=45100
Now, cos(α+β)=cosαcosβ−sinαsinβ
= 17100−45100
[Using equation(iii) and (v)]
= −28100=−725
Now, sin(α+β)=√1−(cosθ)2=√1−49625=√625−49625=√576625=√2425∴sin(α+β)=2425