Question

If α and β are two different values of θ lying between 0 and 2n, which satisfy the equation 6 cos θ = 8 sin θ = 9, find the value of sin (α + β).

Answer 1

$$\begin{gathered} \text{Given:} \\ 6\cos \theta +8\sin \theta =9 \\ \Rightarrow 6\cos \theta =9-8\sin \theta \\ \Rightarrow 36{\cos }^2\theta =(9-8\sin \theta {)}^2 \\ \Rightarrow 36\left(1-{\sin }^2\theta \right)=81+64{\sin }^2\theta -144\sin \theta \\ \Rightarrow 100{\sin }^2\theta -144\sin \theta +45=0 \\ \mathrm{Now},\alpha \mathrm{and}\beta \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation};\mathrm{therefore},\cos \alpha \mathrm{and}\cos \beta \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\text{the}\mathrm{above}\mathrm{equation}. \\ \Rightarrow \sin \alpha \sin \beta =\frac{45}{100}(\mathrm{Product}\mathrm{of}\mathrm{roots}\text{of}\text{a}\mathrm{quadratic}\mathrm{equation}a{x}^2+bx+c=0\mathrm{is}\frac{c}{a}.) \\ \mathrm{Again},6\cos \theta +8\sin \theta =9 \\ \Rightarrow 8\sin \theta =9-6cos\theta \\ \Rightarrow 64{\sin }^2\theta =(9-6cos\theta {)}^2 \\ \Rightarrow 64(1-{\cos }^2\theta )=81+36{\cos }^2\theta -108\cos \theta \\ \Rightarrow 100{\cos }^2\theta -108\cos \theta +17=0 \\ \mathrm{Now},\alpha \mathrm{and}\beta \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation};\mathrm{therefore},\sin \alpha \mathrm{and}\sin \beta \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\text{the}\mathrm{above}\mathrm{equation}. \\ \mathrm{Therefore},\cos \alpha \cos \beta =\frac{17}{100} \\ \mathrm{Hence},\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ =\frac{17}{100}-\frac{45}{100} \\ =-\frac{28}{100} \\ =-\frac{7}{25} \end{gathered}$$

$$\begin{gathered} \sin \left(\alpha +\beta \right)=\sqrt{1-{\cos }^2\left(\alpha +\beta \right)} \\ =\sqrt{1-{\left(\frac{-7}{25}\right)}^2} \\ =\sqrt{\frac{576}{625}} \\ =\frac{24}{25} \end{gathered}$$