If α, β are two different values of x lying between 0 and 2π, which satisfy the equation 6 cos x + 8 sin x = 9, find the value of sin (α + β).
Solution
Given:
6 cosx + 8 sinx = 9
⇒ 6 cosx = 9 - 8 sinx
⇒ 36 cos2x = (9 - 8 sinx)2
⇒ 36(1 - sin2x) = 81 + 64 sin2x - 144 sinx
⇒100 sin2x - 144 sinx + 45 = 0
Now, α and β are the roots of the given equation; therefore, cos α and cos β are the roots of the above equation.
`=> sinalpha sinbeta = 45/100` `("Product of roots of a quadratic equation" ax^2+bx+c=0 "is" c/a.)`
Again, 6 cosx + 8 sinx = 9
⇒ 8 sinx = 9 - 6 cosx
⇒ 64 sin2x = (9 - 6 cosx)2
⇒ 64(1 - cos2x) = 81 + 36cos2x - 108 cosx
⇒ 100 cos2x - 108 cosx + 17 = 0
Now, α and β are the roots of the given equation; therefore, sin α and sin β are the roots of the above equation.
Therefore, cos α cos β = `17/100`
Hence, cos(α + β) = cos α cos β - sin α sin β
`=17/100-45/100`
`=-28/100`
`=-7/25`
\[\sin \left( \alpha + \beta \right) = \sqrt{1 - \cos^2 \left( \alpha + \beta \right)}\]
\[ = \sqrt{1 - \left( \frac{- 7}{25} \right)^2}\]
\[ = \sqrt{\frac{576}{625}}\]
\[ = \frac{24}{25}\]
